Thermodynamic Concepts: General Principles for Energy and Heat Transfer
Introduction to Thermodynamics
Thermodynamics is the branch of science that studies the relationships between heat, work, and energy, focusing on how energy is transferred and transformed in physical and chemical processes. The principles of thermodynamics are essential for understanding how energy flows within systems and how it influences matter's properties.
In chemical thermodynamics, we primarily focus on enthalpy (heat energy), entropy (disorder), and free energy (useful energy for work), which help us predict the spontaneity, stability, and feasibility of chemical reactions.
Key Thermodynamic Concepts
1. System and Surroundings:
- In thermodynamics, a system is the part of the universe under study, while the surroundings are everything outside the system. Systems can be classified as:
- Open System: Exchanges both energy and matter with surroundings (e.g., boiling water in an open pot).
- Closed System: Exchanges only energy, not matter (e.g., a sealed container with gas).
- Isolated System: Does not exchange energy or matter with surroundings (e.g., an insulated thermos bottle).
2. Types of Energy:
- Kinetic Energy: Energy due to motion, present in particles moving within the system.
- Potential Energy: Stored energy based on the position or structure of particles.
- Internal Energy (U): The total energy (kinetic + potential) within a system, including molecular motion and bonds.
3. Heat (q):
- Heat is the transfer of thermal energy between the system and surroundings due to a temperature difference. Heat flows from a region of higher temperature to lower temperature and can be absorbed or released by the system.
- Endothermic Process: Heat is absorbed by the system (\(q > 0\)).
- Exothermic Process: Heat is released by the system (\(q < 0\)).
4. Work (W):
- Work is the energy transferred when an external force acts on a system. In thermodynamics, work often refers to pressure-volume work, where a system expands or compresses against an external pressure.
\[
W = - P \Delta V
\]
Where \(P\) is the external pressure and \(\Delta V\) is the change in volume.
- Expansion Work: The system does work on the surroundings as it expands (\(W < 0\)).
- Compression Work: Work is done on the system by the surroundings as it compresses (\(W > 0\)).
5. Internal Energy Change (\(\Delta U\)):
- The change in internal energy of a system is the sum of heat (\(q\)) added to the system and the work (\(W\)) done on or by the system.
\[
\Delta U = q + W
\]
For a closed system, this equation represents the First Law of Thermodynamics, which states that energy can neither be created nor destroyed, only transferred or transformed.
Thermodynamic Laws
1. First Law of Thermodynamics (Law of Energy Conservation):
- Energy cannot be created or destroyed; it can only change forms. This principle applies to energy transformations between the system and surroundings.
\[
\Delta U = q + W
\]
- In an isolated system, the total energy remains constant.
2. Second Law of Thermodynamics:
- The entropy (disorder) of an isolated system always tends to increase over time. In spontaneous processes, systems move toward greater entropy.
- This law explains why certain reactions or processes are naturally favorable (spontaneous) without needing external energy.
3. Third Law of Thermodynamics:
- As a system approaches absolute zero (0 K), the entropy of a perfect crystalline substance approaches zero. This provides a baseline for measuring entropy in chemical systems.
Enthalpy (H) and Heat Transfer
Enthalpy (H) is a measure of the heat content of a system at constant pressure and is defined as:
\[
H = U + PV
\]
Where \(U\) is the internal energy, \(P\) is pressure, and \(V\) is volume.
1. Change in Enthalpy (\(\Delta H\)):
- The change in enthalpy represents the heat absorbed or released by the system at constant pressure.
\[
\Delta H = q_p
\]
Where \(q_p\) is the heat exchanged at constant pressure.
2. Endothermic vs. Exothermic Reactions:
- Endothermic Reactions: Reactions where \(\Delta H > 0\) (heat absorbed), such as photosynthesis.
- Exothermic Reactions: Reactions where \(\Delta H < 0\) (heat released), such as combustion.
3. Standard Enthalpy Change (\(\Delta H^\circ\)):
- The enthalpy change measured under standard conditions (298 K, 1 atm pressure, and 1 M concentration). This standardization allows for consistent comparison of thermodynamic data across different reactions.
Entropy (S) and Spontaneity
Entropy (S) is a measure of the disorder or randomness in a system. It is a key factor in determining the spontaneity of a process. Processes that increase the system’s entropy tend to occur spontaneously.
1. Change in Entropy (\(\Delta S\)):
- A positive \(\Delta S\) indicates an increase in disorder, which is favorable for spontaneity.
2. Spontaneous Processes:
- Spontaneity depends on both enthalpy and entropy. While exothermic reactions (\(\Delta H < 0\)) are generally spontaneous, endothermic reactions can also be spontaneous if they lead to an increase in entropy.
3. Second Law of Thermodynamics and Entropy:
- The second law states that the entropy of an isolated system always increases in spontaneous processes, implying that the natural direction of change is towards greater disorder.
Gibbs Free Energy (G) and Reaction Spontaneity
Gibbs Free Energy (G) combines enthalpy and entropy to predict the spontaneity of a process at constant temperature and pressure.
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G\) is the change in Gibbs free energy,
- \(\Delta H\) is the change in enthalpy,
- \(T\) is the temperature (in Kelvin),
- \(\Delta S\) is the change in entropy.
1. Interpretation of \(\Delta G\):
- \(\Delta G < 0\): The process is spontaneous.
- \(\Delta G = 0\): The system is in equilibrium.
- \(\Delta G > 0\): The process is non-spontaneous.
2. Standard Free Energy Change (\(\Delta G^\circ\)):
- The free energy change under standard conditions. If \(\Delta G^\circ < 0\), the reaction tends to proceed spontaneously under standard conditions.
Types of Thermodynamic Processes
1. Isothermal Process:
- Occurs at constant temperature. In an isothermal process, any heat added to the system is used to perform work.
2. Adiabatic Process:
- No heat exchange between the system and surroundings. In an adiabatic process, \(\Delta Q = 0\), and changes in internal energy are due solely to work.
3. Isobaric Process:
- Occurs at constant pressure. Enthalpy changes are commonly measured under isobaric conditions.
4. Isochoric Process:
- Occurs at constant volume, meaning no work is done (\(W = 0\)), and any heat added to the system changes the internal energy directly.
Heat Transfer in Thermodynamics
1. Conduction:
- The transfer of heat through direct contact between particles in a solid, from a region of higher temperature to lower temperature.
2. Convection:
- The transfer of heat through the movement of fluids (liquids or gases). Heated fluid particles move from a region of higher temperature to lower temperature, transferring energy in the process.
3. Radiation:
- The transfer of energy through electromagnetic waves, such as infrared radiation, without needing a medium. Radiation can occur in a vacuum, as with the heat from the sun reaching Earth.
Conclusion
Thermodynamic principles govern the flow of energy and heat in chemical and physical systems, dictating the feasibility, stability, and spontaneity of reactions. Understanding concepts such as enthalpy, entropy, Gibbs free energy, and types of thermodynamic processes is essential for predicting how and why chemical reactions occur.
Entropy (S): The Measure of Disorder
Definition of Entropy
Entropy (S) is a thermodynamic property that quantifies the degree of disorder or randomness in a system. It reflects how energy is distributed among the particles of a system and how spread out or disordered the system’s particles are. Entropy plays a crucial role in determining the direction and spontaneity of physical and chemical processes.
The concept of entropy helps explain why certain processes happen naturally (spontaneously) and why some require energy input to proceed. In general, the greater the disorder in a system, the higher its entropy.
Entropy and the Second Law of Thermodynamics
The Second Law of Thermodynamics states that in any spontaneous process, the total entropy of an isolated system always increases or, at best, remains constant. This means that systems naturally evolve toward a state of greater disorder unless energy is applied to maintain or decrease order.
\[
\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \geq 0
\]
Where:
- \(\Delta S_{\text{total}}\) is the change in total entropy,
- \(\Delta S_{\text{system}}\) is the entropy change of the system,
- \(\Delta S_{\text{surroundings}}\) is the entropy change of the surroundings.
If \(\Delta S_{\text{total}} > 0\), the process is spontaneous.
Factors Affecting Entropy
1. State of Matter:
- Solids have the lowest entropy due to the highly ordered arrangement of particles.
- Liquids have higher entropy than solids because particles can move more freely.
- Gases have the highest entropy because particles move independently and are widely dispersed.
Example:
- When ice melts to form water, entropy increases because water molecules in the liquid state are more disordered than in the solid state.
2. Temperature:
- As temperature increases, entropy increases because particles move more rapidly, creating greater disorder.
Example:
- Heating a gas increases its entropy as the particles move faster and occupy a larger range of positions.
3. Number of Particles:
- Increasing the number of particles in a system increases its entropy, as more particles lead to a greater number of possible arrangements.
Example:
- When 1 mole of \(N_2O_4\) dissociates into 2 moles of \(NO_2\) gas, entropy increases because there are more gas molecules (and hence, more disorder) in the products.
4. Mixing of Substances:
- Mixing different substances, like gases or solutions, usually increases entropy because the particles become more randomly distributed.
Example:
- When salt dissolves in water, entropy increases as the ions (Na\(^+\) and Cl\(^-\)) spread out in the solvent, creating a more disordered system.
Calculating Entropy Changes (\(\Delta S\))
The change in entropy (\(\Delta S\)) of a process is given by:
\[
\Delta S = \frac{q_{\text{rev}}}{T}
\]
Where:
- \(q_{\text{rev}}\) is the heat absorbed or released in a reversible process,
- \(T\) is the absolute temperature in Kelvin.
This equation shows that entropy change depends on both the amount of heat transferred and the temperature.
Example:
- Melting ice at 0°C in a reversible process has an entropy change:
\[
\Delta S = \frac{q_{\text{rev}}}{273 \, \text{K}}
\]
Where \(q_{\text{rev}}\) is the heat required for the phase transition at 0°C.
Entropy and Spontaneity of Reactions
For a reaction to be spontaneous, the total entropy of the system and its surroundings must increase. However, the spontaneity of a reaction also depends on enthalpy (ΔH) and temperature (T), which is described by Gibbs free energy (ΔG):
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G < 0\) indicates a spontaneous reaction,
- \(\Delta G = 0\) indicates an equilibrium state,
- \(\Delta G > 0\) indicates a non-spontaneous reaction.
Entropy change (\(\Delta S\)) influences the sign of \(\Delta G\), meaning that even endothermic reactions (ΔH > 0) can be spontaneous if they lead to a sufficient increase in entropy.
Example:
- The dissolution of ammonium nitrate \((NH_4NO_3)\) in water is endothermic (ΔH > 0) but is spontaneous because it leads to a large increase in entropy (\(\Delta S > 0\)).
Examples of Entropy Changes in Physical and Chemical Processes
1. Physical Changes:
- Melting and Vaporization: Entropy increases in phase changes from solid to liquid (melting) and liquid to gas (vaporization) because the particles become more disordered.
Example:
- Water vapor has much higher entropy than liquid water because gaseous molecules are more disordered and widely spaced than liquid molecules.
2. Mixing of Gases:
- When two different gases are allowed to mix, entropy increases because the particles spread out and occupy a greater number of possible positions.
Example:
- Mixing oxygen and nitrogen gases increases entropy because each gas now occupies the entire volume, increasing disorder.
3. Dissolution:
- When a solid dissolves in a liquid, entropy often increases because the solid particles disperse in the solvent, leading to a more disordered state.
Example:
- Dissolving salt (\(NaCl\)) in water increases entropy as the \(Na^+\) and \(Cl^-\) ions spread out and interact with water molecules.
4. Chemical Reactions:
- Reactions that produce more moles of gas tend to increase entropy, as gases have higher entropy than liquids or solids.
Example:
- The decomposition of calcium carbonate (\(CaCO_3 \rightarrow CaO + CO_2\)) increases entropy because it produces an additional mole of gas (\(CO_2\)).
Microstates and Entropy (Boltzmann's Entropy Equation)
The number of microstates (W) of a system is the number of possible arrangements of particles and energy levels. The more microstates available, the greater the entropy of the system.
Boltzmann’s Equation relates entropy to the number of microstates:
\[
S = k \ln W
\]
Where:
- \(S\) is entropy,
- \(k\) is Boltzmann’s constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \(W\) is the number of microstates.
This equation shows that entropy increases with the number of ways particles can be arranged in a system.
Example:
- In a mixture of gases, there are more possible microstates than in pure gases because the particles of each gas can spread out among each other, increasing \(W\) and therefore \(S\).
Conclusion
Entropy is a fundamental thermodynamic concept that measures the degree of disorder in a system. Higher entropy generally favors spontaneity, as systems tend to evolve toward states of greater disorder. Understanding entropy helps explain the behavior of physical and chemical processes, particularly in terms of spontaneity, phase transitions, and energy distribution.
Entropy plays an indirect but essential role in determining reaction rates. While entropy itself doesn’t directly influence the rate of a reaction, it contributes to the activation energy and the probability of productive molecular collisions, both of which are critical in determining how quickly a reaction proceeds.
Here’s how entropy can affect reaction rates:
1. Entropy and Activation Energy
The activation energy (Ea) of a reaction is the minimum energy required for reactants to successfully transform into products. Entropy can impact activation energy in several ways, especially when comparing reactions with high or low entropy changes:
- Higher Entropy in Transition States: Reactions that involve an increase in entropy between the reactants and the transition state generally have a lower activation energy because the transition state has more accessible arrangements and microstates. This increased disorder in the transition state makes it easier for reactants to achieve the necessary configuration for reaction, leading to faster rates.
- Reactions Involving Decreased Entropy: Reactions where the transition state is more ordered than the reactants (e.g., association reactions or condensation reactions where two molecules combine) tend to have higher activation energies, as reaching a more organized state requires more precise collisions. This makes these reactions slower.
Example:
In the combustion of hydrocarbons, the transition state has higher entropy than the structured arrangement of reactant molecules, leading to a relatively lower activation energy and a faster reaction rate when enough heat (thermal energy) is supplied.
2. Entropy and Collision Theory
Collision theory states that for a reaction to occur, molecules must collide with proper orientation and sufficient energy. Entropy influences this by affecting how freely particles can move and collide:
- Gaseous or Dissolved States (High Entropy): Reactions involving gases or dissolved species typically proceed faster than reactions in the solid state because these phases allow for more random motion, frequent collisions, and various orientations, all contributing to higher entropy. The increased randomness makes it more likely that molecules will collide in the correct orientation, improving the chances of productive reactions.
- Solid State Reactions (Low Entropy): In solids, atoms and molecules are held in fixed positions, resulting in lower entropy and fewer possible collisions. This restricted motion reduces the reaction rate since particles can’t easily overcome the structural constraints to collide effectively.
Example:
Reactions between gases, such as hydrogen gas (H\(_2\)) and chlorine gas (Cl\(_2\)) to form hydrogen chloride (HCl), occur rapidly due to high entropy and frequent collisions in the gas phase. In contrast, reactions involving solid-state ionic lattices are slower due to limited particle mobility.
3. Entropy and the Arrhenius Equation
The Arrhenius equation provides insight into how factors like temperature affect reaction rates:
\[
k = A \, e^{\frac{-E_a}{RT}}
\]
Where:
- \(k\) is the rate constant,
- \(A\) is the frequency factor, reflecting the orientation and frequency of collisions,
- \(E_a\) is the activation energy,
- \(R\) is the gas constant,
- \(T\) is the temperature.
Entropy indirectly influences the frequency factor (A), which incorporates both the collision frequency and the effectiveness of these collisions. A high-entropy system with a greater number of available configurations (microstates) generally has a higher frequency factor, increasing the likelihood of effective collisions and thus the reaction rate.
Example:
In aqueous reactions (reactions in solution), where particles are more randomly distributed than in the solid state, the frequency factor \(A\) is typically higher due to increased particle mobility and orientation possibilities, contributing to faster reaction rates.
4. Entropy and Transition State Theory
In transition state theory, the formation of an activated complex (transition state) is essential for reactions to proceed. The entropy of the transition state affects the Gibbs free energy of activation (\(\Delta G^\ddagger\)), which in turn affects reaction rates:
\[
\Delta G^\ddagger = \Delta H^\ddagger - T \Delta S^\ddagger
\]
Where:
- \(\Delta G^\ddagger\) is the free energy of activation,
- \(\Delta H^\ddagger\) is the enthalpy of activation,
- \(\Delta S^\ddagger\) is the entropy of activation.
- Positive Entropy of Activation (\(\Delta S^\ddagger > 0\)): If the transition state has higher entropy than the reactants, the free energy of activation decreases (\(\Delta G^\ddagger\) becomes smaller), making the reaction proceed more easily, thus increasing the rate.
- Negative Entropy of Activation (\(\Delta S^\ddagger < 0\)): If the transition state is more ordered, this raises the free energy of activation, slowing the reaction.
Example:
Reactions where large molecules dissociate into smaller fragments (such as decomposition reactions) often have positive entropy changes in the transition state, leading to faster rates. Conversely, association reactions where two molecules come together usually have a lower entropy in the transition state, resulting in slower rates.
5. Entropy and Catalysis
Catalysts lower the activation energy of a reaction, often by providing an alternative pathway with a different transition state that may have higher entropy:
- Increased Entropy in Catalyst-Assisted Pathways: Catalysts, especially those in solution or as surfaces, allow more accessible and flexible collision sites, increasing the entropy of the transition state. This often results in a lower activation energy and a higher reaction rate.
Example:
In enzymatic reactions, enzymes provide an organized active site that reduces the system’s overall activation energy, effectively speeding up biochemical reactions without requiring large entropy changes. However, the flexibility at the enzyme’s active site allows for a high-entropy transition state, further assisting the reaction rate.
Summary
Entropy affects reaction rates through several mechanisms:
- By influencing activation energy, where high-entropy transition states tend to lower activation barriers, increasing rates.
- Through collision frequency and orientation, with higher entropy states (like gases) supporting more frequent and effective collisions.
- Via transition state theory, where positive entropy changes in the transition state favor lower activation energy and faster rates.
- Catalysts also harness entropy by creating transition states that are more disordered, lowering activation energy and speeding up reactions.
Although entropy is not a direct factor in the kinetic rate expression, its influence on the distribution of energy and the arrangement of particles plays a significant role in reaction kinetics, making it an essential factor in understanding reaction mechanisms and speed.
Temperature has a direct and profound impact on entropy. As temperature increases, the entropy of a system also increases, because higher temperatures lead to greater molecular motion, increased randomness, and more accessible energy states. This relationship is fundamental to thermodynamics and affects physical processes, phase changes, and chemical reactions.
Here’s how temperature impacts entropy:
1. Increased Molecular Motion
As temperature rises, molecules gain kinetic energy and move faster. This increase in molecular motion creates more disorder or randomness in the system, which translates to higher entropy.
- At low temperatures, molecules have limited motion, especially in solids, where particles are held in fixed positions and can only vibrate.
- At higher temperatures, molecular vibrations, rotations, and translations (especially in gases and liquids) increase, allowing particles to access a wider range of positions and velocities.
Example:
In a gas, heating the system increases the speed of gas molecules, making them more widely dispersed and occupying more possible states, thus increasing the entropy of the gas.
2. Greater Access to Microstates
Temperature impacts entropy because higher temperatures allow molecules to access a greater number of microstates—specific ways in which energy can be distributed within the system. According to Boltzmann’s entropy equation:
\[
S = k \ln W
\]
Where:
- \(S\) is the entropy,
- \(k\) is Boltzmann's constant,
- \(W\) is the number of microstates.
With increased thermal energy, particles can adopt more energy distributions, thus increasing \(W\) and thereby increasing entropy \(S\).
Example:
Consider a solid heating up to become a liquid. As it transitions, molecules can move more freely and adopt a much wider range of positions and arrangements, resulting in a significant increase in \(W\) and thus in entropy.
3. Entropy Change in Phase Transitions
Phase changes illustrate the impact of temperature on entropy very clearly:
- Melting (Solid to Liquid): As temperature reaches the melting point, a solid becomes a liquid, increasing entropy significantly. In the liquid phase, molecules have more freedom to move, creating a more disordered state.
- Vaporization (Liquid to Gas): At the boiling point, a liquid changes to a gas, leading to an even larger increase in entropy. In the gaseous state, molecules move freely and occupy a much larger volume, resulting in maximum disorder.
During these transitions, the entropy change (\(\Delta S\)) can be calculated as:
\[
\Delta S = \frac{q_{\text{rev}}}{T}
\]
Where:
- \(q_{\text{rev}}\) is the heat absorbed during the phase change,
- \(T\) is the temperature at which the change occurs.
Example:
When water (liquid) vaporizes at 100°C (373 K), it absorbs heat and undergoes a large increase in entropy because the water molecules move from a relatively ordered liquid state to a highly disordered gas state.
4. Entropy Change with Temperature in Reversible Processes
In thermodynamic processes, such as heating a gas reversibly at constant volume, the entropy change (\(\Delta S\)) is given by:
\[
\Delta S = \int \frac{\delta q_{\text{rev}}}{T}
\]
This relationship shows that as temperature \(T\) increases, the addition of a small amount of heat \(\delta q_{\text{rev}}\) has a smaller impact on entropy than it would at lower temperatures. Therefore:
- At high temperatures, adding heat causes a smaller change in entropy because the system already has significant thermal energy and disorder.
- At low temperatures, adding heat causes a larger change in entropy because the particles are relatively more ordered, so even a small increase in energy significantly increases disorder.
5. Temperature and Entropy in the Third Law of Thermodynamics
The Third Law of Thermodynamics states that as a system approaches absolute zero (0 K), the entropy of a perfect crystalline structure approaches zero. At 0 K, molecular motion theoretically stops, resulting in a single possible microstate (\(W = 1\)), which gives:
\[
S = k \ln 1 = 0
\]
As temperature increases from absolute zero, even slightly, particles begin to gain energy and move, increasing the number of accessible microstates and thus increasing entropy. This principle highlights how closely temperature and entropy are linked.
Example:
In a crystal at very low temperatures, only vibrational motions are possible, resulting in low entropy. As the crystal is heated, vibrational motions increase, and eventually, rotational and translational motions are added, increasing entropy as temperature rises.
Summary: Key Points of Temperature’s Impact on Entropy
- Higher temperatures increase entropy due to more molecular motion and energy dispersal.
- Phase changes (e.g., melting, vaporization) exhibit significant entropy increases due to higher molecular freedom.
- Microstates and molecular disorder increase as temperature rises, leading to greater entropy.
- Low temperatures cause greater entropy changes per unit of heat added than high temperatures do, as per the formula \(\Delta S = \frac{q_{\text{rev}}}{T}\).
- Third Law of Thermodynamics implies that at absolute zero, entropy is minimal, and as temperature increases from 0 K, entropy increases as molecular motion begins.
In essence, temperature and entropy are directly proportional: as temperature rises, so does entropy, because increased thermal energy enables molecules to occupy more states, move more freely, and increase overall system disorder.
|
Q 1. The law of thermodynamics which leads to the conclusion that it is impossible to convert 'all' the heat extracted from a hot source into work is :; |
|
(A) zeroth law; |
|
(B) first law; |
|
(C) second law; |
|
(D) third law; |
|
Q 2. A heat engine has an efficiency \(\eta\). Temperatures of source and sink are each decreased by \(100 \mathrm{~K}\). Then, the efficiency of the engine.; |
|
(A) Increases; |
|
(B) Decreases; |
|
(C) Remains constant; |
|
(D) Becomes 1; |
|
Q 3. An ideal gas heat engine operates in Carnot cycle between \(227^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\). It absorbs \(6 \times 10^{4}\) cal of heat at high temperature. Amount of heat converted to work is:; |
|
(A) \(1.2 \times 10^{4} \mathrm{cal}\); |
|
(B) \(4.8 \times 10^{4} \mathrm{cal}\); |
|
(C) \(6 \times 10^{4} \mathrm{cal}\); |
|
(D) \(2.4 \times 10^{4} \mathrm{cal}\); |
|
Q 4. Which statement is incorrect?; |
|
(A) All reversible cycles have same efficiency; |
|
(B) Reversible cycle has more efficiency than an irreversible one; |
|
(C) Carnot cycle is a reversible one; |
|
(D) Carnot cycle has the maximum efficiency in all cycles; |
|
Q 5. A cyclic process ABCA is shown in the \(V-T\) diagram. Process on the \(P-V\) diagram is; |
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(D); |
|
(C); |
|
(B); |
|
(A); |
|
Q 6. In an isochoric process, if \(T_{1}=27^{\circ} \mathrm{C}\) and \(T_{2}=127^{\circ} \mathrm{C}\), then \(\frac{p_{1}}{p_{2}}\) will be equal to; |
|
(D) \(\frac{1}{3}\); |
|
(C) \(\frac{3}{4}\); |
|
(B) \(\frac{2}{3}\); |
|
(A) \(\frac{9}{59}\); |
|
Q 7. The first law of thermodynamic is expressed as:; |
|
(D) None of the above; |
|
(C) \(\mathrm{W}=\mathrm{q}+\Delta \mathrm{E}\); |
|
(B) \(q=\Delta E-W\); |
|
(A) \(\Delta \mathrm{E}=\mathrm{q}-\mathrm{W}\); |
|
Q 8. Which of the following parameters can define the thermodynamic state of a system?; |
|
(D) Velocity; |
|
(C) Gravity; |
|
(B) Temperature; |
|
(A) Work; |
|
Q 9. Which of the following statement is correct?; |
|
(D) diamond is more stable than graphite; |
|
(C) graphite is more stable than diamond; |
|
(B) formation of graphite is endothermic; |
|
(A) formation of diamond is exothermic; |
|
Q 10. A given mass of gas expands from state A to state B by three paths 1,2 and 3 as shown in the figure. If \(\mathrm{w}_{1}, \mathrm{w}_{2}\) and \(\mathrm{w}_{3}\) respectively are be the works done by the gas along three paths, then; |
|
(D) \(\mathrm{w}_{1}<\mathrm{w}_{2} ; \mathrm{w}_{1}>\mathrm{w}_{3}\); |
|
(C) \(\mathrm{w}_{1}=\mathrm{w}_{2}=\mathrm{w}_{3}\); |
|
(B) \(\mathrm{w}_{1}<\mathrm{w}_{2}<\mathrm{w}_{3}\); |
|
(A) \(\mathrm{w}_{1}>\mathrm{w}_{2}>\mathrm{w}_{3}\); |
Work Done (W): Energy Transfer Involving Mechanical Actions
Definition of Work Done
In thermodynamics, work (W) is a form of energy transfer that occurs when a force acts on an object or system over a distance. It represents mechanical action through which energy is transferred between the system and its surroundings. Work is commonly performed in thermodynamic systems by compressing, expanding, or moving parts within the system, often involving pressure-volume (PV) work in gases.
Mathematically, work is defined as:
\[
W = F \cdot d
\]
Where:
- \(W\) is the work done,
- \(F\) is the force applied,
- \(d\) is the distance over which the force is applied.
Types of Work in Thermodynamic Systems
1. Pressure-Volume (PV) Work:
- PV work occurs when the volume of a gas changes under external pressure. For instance, when a gas expands against a piston, it performs work on the surroundings, and when it compresses, work is done on the gas.
- PV work is calculated as:
\[
W = - P_{\text{ext}} \Delta V
\]
Where:
- \(P_{\text{ext}}\) is the external pressure,
- \(\Delta V\) is the change in volume of the gas.
- The negative sign indicates that work done by the system (expansion) is negative, while work done on the system (compression) is positive.
2. Electrical Work:
- Electrical work is performed in systems involving electrochemical reactions or electric circuits, where energy transfer occurs through the movement of electrons across a potential difference.
- In chemical cells, for example, ions moving across electrodes generate an electric current, doing work on the external circuit.
3. Surface Work:
- In systems where the surface area changes (e.g., liquid films), surface tension performs work, as energy is required to expand or contract the surface area.
Work in Different Thermodynamic Processes
The work done by or on a gas varies depending on the type of thermodynamic process:
1. Isothermal Process (Constant Temperature):
- In an isothermal process, the temperature remains constant. For an ideal gas expanding or compressing isothermally, the work done is given by:
\[
W = - nRT \ln \frac{V_f}{V_i}
\]
Where:
- \(n\) is the number of moles of gas,
- \(R\) is the gas constant,
- \(T\) is the absolute temperature,
- \(V_i\) and \(V_f\) are the initial and final volumes, respectively.
- In this process, as the gas expands, it absorbs heat from the surroundings to maintain a constant temperature, allowing it to perform work.
2. Adiabatic Process (No Heat Exchange):
- In an adiabatic process, no heat is exchanged between the system and surroundings (\(q = 0\)). Any energy change within the system is due to work alone:
\[
\Delta U = W
\]
- Here, the work done results in a change in internal energy, often leading to temperature changes within the system. For an ideal gas undergoing adiabatic expansion or compression, the relationship between \(P\) and \(V\) is:
\[
P V^{\gamma} = \text{constant}
\]
Where \(\gamma\) (gamma) is the heat capacity ratio \(\left(\frac{C_p}{C_v}\right)\).
3. Isochoric Process (Constant Volume):
- In an isochoric process, the volume remains constant (\(\Delta V = 0\)), so no PV work is done:
\[
W = 0
\]
- Any energy change in an isochoric process is due solely to heat transfer, as no work is done on or by the system.
4. Isobaric Process (Constant Pressure):
- In an isobaric process, the pressure remains constant. For a gas expanding or compressing at constant pressure, the work done is:
\[
W = - P \Delta V
\]
- In an isobaric process, as the gas changes volume, it does work on the surroundings (expansion) or has work done on it (compression), with the energy transfer also affecting the system’s internal energy.
Sign Convention for Work
- Work Done by the System: When a system (e.g., a gas) expands, it does work on the surroundings, so \(W\) is considered negative in thermodynamic sign convention.
- Work Done on the System: When work is done on the system (e.g., compressing a gas), \(W\) is considered positive because energy is transferred into the system.
This convention aligns with the First Law of Thermodynamics:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat added to the system,
- \(W\) is the work done on or by the system.
Importance of Work in Thermodynamics
1. Energy Transfer and System Changes:
- Work is one of the primary ways that energy is transferred into or out of a system, alongside heat. This energy transfer can alter the internal energy, temperature, and volume of a system.
2. Practical Applications in Engines and Compressors:
- Work is fundamental in mechanical devices, such as engines, compressors, and turbines, where gases expand or compress to generate usable energy.
- For example, in internal combustion engines, fuel combustion generates gas expansion, which does work on pistons to produce motion.
3. PV Diagrams and Thermodynamic Cycles:
- PV diagrams (plots of pressure versus volume) are used to visualize work done in various thermodynamic cycles, such as the Carnot cycle or Otto cycle. The area under a PV curve represents the work done by or on the system during each cycle.
Examples of Work in Thermodynamic Processes
1. Expanding Gas in a Cylinder:
- When gas in a piston expands at constant pressure, it performs work on the piston, pushing it outward. This work done by the gas is:
\[
W = - P \Delta V
\]
- If the initial volume is 2.0 L and expands to 4.0 L under a pressure of 1 atm, the work done by the gas is:
\[
W = - (1 \, \text{atm})(4.0 \, \text{L} - 2.0 \, \text{L}) = -2.0 \, \text{L atm} \approx -202.6 \, \text{J}
\]
2. Compression of an Ideal Gas:
- In an adiabatic compression, such as a gas compressed in a sealed container, work is done on the gas, increasing its internal energy and temperature. This work results in no heat exchange, so:
\[
\Delta U = W
\]
- For example, compressing air in a bicycle pump increases the temperature due to work done on the gas.
3. Lifting a Weight:
- In mechanics, when a force lifts a weight, work is done against gravity. If a 5 kg weight is lifted 2 meters, the work done is:
\[
W = F \cdot d = (mg) \cdot d = (5 \, \text{kg})(9.8 \, \text{m/s}^2)(2 \, \text{m}) = 98 \, \text{J}
\]
- This example illustrates work done in a gravitational field, where energy is transferred to increase the potential energy of the weight.
Conclusion
Work is a fundamental concept in thermodynamics representing energy transfer through mechanical action. In thermodynamic systems, work commonly involves pressure-volume changes as gases expand or compress. The amount of work done depends on the type of process, such as isothermal, adiabatic, or isobaric, and the work sign convention indicates whether the system gains or loses energy. Understanding work is essential for analyzing energy flow in engines, compressors, and various physical systems, making it a key concept in physics and engineering.
|
Q 1. A refrigerator with the coefficient of performance \(\frac{1}{3}\) releases \(200 \mathrm{~J}\) of heat to a hot reservoir. Then the work done on the working substance is; |
|
(A) \(\frac{100}{3} \mathrm{~J}\); |
|
(B) \(100 \mathrm{~J}\); |
|
(C) \(\frac{200}{3} \mathrm{~J}\); |
|
(D) \(150 \mathrm{~J}\); |
|
Q 2. The equation of a state of a gas is given by \(P(V-b)=n R T\). If 1 mole of a gas is isothermally expanded from volume \(V\) and \(2 V\), the work done during the process is; |
|
(A) \(R T \ln \left|\frac{2 V-b}{V-b}\right|\); |
|
(B) \(R T \ln \left|\frac{V-b}{V}\right|\); |
|
(C) \(R T \ln \left|\frac{V-b}{2 V-b}\right|\); |
|
(D) \(R T \ln \left|\frac{V}{V-b}\right|\); |
|
Q 3. Determine the work done by an ideal gas undergoing a cyclic process from \(1 \rightarrow 4 \rightarrow 3 \rightarrow 2 \rightarrow 1\). Given \(P_{1}=10^{5} \mathrm{~Pa}, P_{0}=3 \times 10^{5} \mathrm{~Pa}, P_{3}=4 \times 10^{5} \mathrm{~Pa}\) and \(V_{2}-V_{1}=10 \mathrm{~L}\); |
|
(A) \(740 \mathrm{~J}\); |
|
(B) \(750 \mathrm{~J}\); |
|
(C) \(730 \mathrm{~J}\); |
|
(D) \(745 \mathrm{~J}\); |
|
Q 4. A sample of ideal gas \((\gamma=1.4)\) is heated at constant pressure. If \(100 \mathrm{~J}\) of heat is supplied to the gas the work done by the gas is; |
|
(A) \(28.57 \mathrm{~J}\); |
|
(B) \(56.54 \mathrm{~J}\); |
|
(C) \(38.92 \mathrm{~J}\); |
|
(D) \(65.38 \mathrm{~J}\); |
|
Q 5. The magnitude of work done by a gas that undergoes a reversible expansion along the path \(\mathrm{ABC}\) shown in the figure is; |
|
(A)48 J; |
|
(B)52 J; |
|
(C)69 J; |
|
(D)44 J; |
|
Q 6. An ideal gas is taken around the cycle \(\mathrm{ABCA}\) as shown in \(\mathrm{P}-\mathrm{V}\) diagram. The net work done during the cycle is equal to -; |
|
(A) \(12 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(B) \(6 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(C) \(5 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(D) \(\mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
Q 7. An ideal gas is taken around the cycle \(\mathrm{ABCA}\) as shown in \(\mathrm{P}-\mathrm{V}\) diagram. The net work done during the cycle is equal to -; |
|
(A) \(12 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(B) \(6 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(C) \(5 \mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
(D) \(\mathrm{P}_{1} \mathrm{~V}_{1}\); |
|
Q 8. The magnitude of work done by a gas that undergoes a reversible expansion along the path \(\mathrm{ABC}\) shown in the figure is; |
|
(A)48 J; |
|
(B)52 J; |
|
(C)69 J; |
|
(D)44 J; |
|
Q 9. A sample of ideal gas \((\gamma=1.4)\) is heated at constant pressure. If \(100 \mathrm{~J}\) of heat is supplied to the gas the work done by the gas is; |
|
(A) \(28.57 \mathrm{~J}\); |
|
(B) \(56.54 \mathrm{~J}\); |
|
(C) \(38.92 \mathrm{~J}\); |
|
(D) \(65.38 \mathrm{~J}\); |
|
Q 10. Determine the work done by an ideal gas undergoing a cyclic process from \(1 \rightarrow 4 \rightarrow 3 \rightarrow 2 \rightarrow 1\). Given \(P_{1}=10^{5} \mathrm{~Pa}, P_{0}=3 \times 10^{5} \mathrm{~Pa}, P_{3}=4 \times 10^{5} \mathrm{~Pa}\) and \(V_{2}-V_{1}=10 \mathrm{~L}\); |
|
(A) \(740 \mathrm{~J}\); |
|
(B) \(750 \mathrm{~J}\); |
|
(C) \(730 \mathrm{~J}\); |
|
(D) \(745 \mathrm{~J}\); |
Entropy (S): Measure of Disorder or Randomness in Chemical Thermodynamics
Definition of Entropy in Chemical Thermodynamics
In chemical thermodynamics, entropy (S) is a thermodynamic property that quantifies the degree of disorder or randomness in a system. Entropy is a measure of how energy is dispersed within a system at a given temperature. In chemical reactions, entropy is a key factor in predicting the spontaneity of processes and helps determine whether a reaction is energetically favorable.
Entropy increases as a system becomes more disordered. For example, gases, with their highly dispersed particles, have higher entropy than liquids and solids. Entropy changes in a reaction influence the system’s free energy and its tendency to move toward a more stable state.
Key Aspects of Entropy in Chemical Processes
1. Entropy and Molecular Disorder:
- In chemical thermodynamics, entropy is associated with the distribution of particles and energy in a system. Systems tend toward higher entropy because more configurations are possible when energy is more spread out.
- Entropy is higher in systems where molecules are more disordered, such as gases, which have dispersed molecules moving randomly, compared to structured solids.
2. Entropy and Energy Dispersal:
- Entropy represents how energy is distributed among molecules. Greater dispersal (spread of energy across molecules) corresponds to greater entropy.
- For example, in an exothermic reaction where energy is released to the surroundings, the surroundings' entropy increases due to the added thermal energy.
3. Units of Entropy:
- Entropy is expressed in Joules per Kelvin (J/K), as it is a measure of energy distribution per unit temperature.
Entropy Change in Chemical Reactions
1. Standard Entropy Change (\(\Delta S\)):
- The entropy change of a chemical reaction can be calculated as the difference between the entropy of the products and the reactants:
\[
\Delta S = S_{\text{products}} - S_{\text{reactants}}
\]
- If \(\Delta S\) is positive, the reaction leads to increased disorder. If negative, the reaction leads to decreased disorder.
2. Entropy in Phase Transitions:
- Entropy changes significantly during phase changes:
- Solid to Liquid (Melting): Entropy increases as the solid becomes a liquid, allowing molecules more freedom to move.
- Liquid to Gas (Vaporization): Entropy greatly increases as a liquid becomes a gas, with molecules dispersing widely.
- Gas to Liquid/Solid (Condensation/Freezing): Entropy decreases as molecules become more ordered in the condensed phase.
Example:
- Water (H\(_2\)O) has a positive entropy change when it vaporizes: \(\Delta S > 0\) as it changes from liquid to gas, allowing water molecules to spread out and move freely.
3. Entropy in Mixing and Dissolution:
- Entropy increases when substances are mixed or dissolved, as molecules or ions become more randomly distributed.
Example:
- Dissolving salt (NaCl) in water increases entropy because the ions (Na\(^+\) and Cl\(^-\)) disperse among water molecules, leading to greater disorder.
4. Entropy and Chemical Reactions:
- In a chemical reaction, if the products are in a more disordered state than the reactants, entropy increases, favoring spontaneity.
- Decomposition reactions often result in an increase in entropy as one compound breaks down into simpler products, increasing disorder.
Example:
- The decomposition of calcium carbonate:
\[
CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g)
\]
- Here, entropy increases because the solid \(CaCO_3\) breaks down to produce \(CO_2\) gas, a more disordered state than the solid reactants.
Calculating Entropy Change (\(\Delta S\))
The entropy change for a reversible process at constant temperature can be calculated as:
\[
\Delta S = \frac{q_{\text{rev}}}{T}
\]
Where:
- \(\Delta S\) is the entropy change,
- \(q_{\text{rev}}\) is the heat absorbed or released reversibly,
- \(T\) is the absolute temperature.
This formula is particularly useful in calculating entropy changes during phase transitions, such as melting or vaporization, where heat is added reversibly at a constant temperature.
Entropy and the Second Law of Thermodynamics
The Second Law of Thermodynamics states that the total entropy of an isolated system always increases in a spontaneous process. This principle is crucial for predicting the direction of chemical reactions:
- For any spontaneous reaction, the total entropy (system + surroundings) increases.
- The universe (the system plus surroundings) always evolves toward greater disorder, favoring reactions that increase the total entropy.
Example:
In an exothermic reaction, such as combustion, energy is released to the surroundings, increasing the surroundings’ entropy and contributing to the overall entropy increase.
Entropy and Free Energy (\(\Delta G\))
In chemical thermodynamics, entropy helps determine the spontaneity of a reaction when combined with enthalpy (\(H\)) in the Gibbs free energy equation:
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G\) is the Gibbs free energy change,
- \(\Delta H\) is the enthalpy change,
- \(T\) is the temperature,
- \(\Delta S\) is the entropy change.
1. Positive \(\Delta S\) (Increase in Entropy):
- A positive \(\Delta S\) (increase in entropy) contributes to a negative \(\Delta G\), favoring spontaneity at higher temperatures.
2. Negative \(\Delta S\) (Decrease in Entropy):
- A negative \(\Delta S\) (decrease in entropy) opposes spontaneity, and reactions are less likely to proceed spontaneously unless \(\Delta H\) is sufficiently negative (exothermic).
Example:
- Dissolution of ammonium nitrate (NH\(_4\)NO\(_3\)) is an endothermic reaction with positive \(\Delta S\), and it dissolves spontaneously because the increase in entropy (disorder of ions in solution) favors spontaneity despite the positive \(\Delta H\).
Entropy and Equilibrium
At equilibrium, the entropy of a system is maximized for the given conditions, meaning the system has reached a state of maximum disorder without further net change.
In chemical reactions, equilibrium is reached when the forward and reverse reaction rates are equal, and the system’s entropy no longer changes significantly. At this point, the Gibbs free energy change \(\Delta G\) is zero:
\[
\Delta G = 0 \Rightarrow \Delta H = T \Delta S
\]
Example:
- In a reversible reaction, such as the dissociation of acetic acid (\(CH_3COOH\)) in water, the system reaches equilibrium when the dissociation and recombination rates balance, resulting in a stable entropy for the system at that temperature.
Conclusion
In chemical thermodynamics, entropy (S) is a fundamental property that reflects the degree of disorder in a system. It is essential for predicting the direction and spontaneity of chemical reactions, as reactions tend to proceed toward states with higher entropy. Entropy changes affect phase transitions, dissolution, chemical equilibria, and reaction spontaneity through the Gibbs free energy relationship.
|
Q 1. When you make ice cubes, the entropy of water; |
|
(A) Does not change; |
|
(B) Increases; |
|
(C) Decreases; |
|
(D) May either increase or decrease depending on the process used; |
|
Q 2. When you make ice cubes, the entropy of water; |
|
(A) Does not change; |
|
(B) Increases; |
|
(C) Decreases; |
|
(D) May either increase or decrease depending on the process used; |
|
Q 3. Considering entropy \((\mathrm{S})\) as a thermodynamic parameter, the criterion for the spontaneity of any process is; |
|
(a) \(\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surroundirgs }}>0\); |
|
(b) \(\Delta \mathrm{S}_{\text {system }}-\Delta \mathrm{S}_{\text {surroundirgs }}>0\); |
|
(c) \(\Delta \mathrm{S}_{\text {system }}>0\) only; |
|
(d) \(\Delta \mathrm{S}_{\text {surroundirgs }}>0\) only; |
|
Q 4. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?; |
|
(a) Exothermic and increasing disorder; |
|
(b) Exothermic and decreasing disorder; |
|
(c) Endothermic and increasing disorder; |
|
(d) Endothermic and decreasing disorder; |
|
Q 5. In which of the following reactions, standard entropy change \(\left(\Delta \mathrm{S}^{\circ}\right)\) is positive and standard Gibb's energy change \(\left(\Delta \mathrm{G}^{\circ}\right)\) decreases sharply with increasing temperature ?; |
|
(a) \(\mathrm{C}\) graphite \(+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})\); |
|
(b) \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\); |
|
(c) \(\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{MgO}(\mathrm{s})\); |
|
(d) \(\frac{1}{2}\) C graphite \(+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{CO}_{2}(\mathrm{~g})\); |
|
Q 6. In which of the following entropy decreases?; |
|
(a) Crystallization of sucrose solution; |
|
(b) Rusting of iron; |
|
(c) Melting of ice; |
|
(d) Vaporization of camphor; |
|
Q 7. Identify the correct statement regarding entropy; |
|
(a) At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero.; |
|
(b) At absolute zero temperature, the entropy of a perfectly crystalline substance is positive.; |
|
(c) Absolute entropy of a substance cannot be determined.; |
|
(d) At \(0^{\circ} \mathrm{C}\), the entropy of a perfectly crystalline substance is taken to be zero; |
|
Q 8. Unit of entropy is; |
|
(a) \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\); |
|
(b) \(\mathrm{J} \mathrm{mol}^{-1}\); |
|
(c) \(\mathrm{J}^{-1} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\); |
|
(d) \(\mathrm{JK} \mathrm{mol}^{-1}\); |
|
Q 9. Assertion : Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature Reason : Entropy of the system increases with increase in temperature.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 10. Assertion : An exothermic process which is nonspontaneous at high temperature may become spontaneous at a low temperature Reason : There occurs a decrease in entropy factor as the temperature is decreased.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
Specific Heat Capacity (C)
Definition of Specific Heat Capacity
Specific heat capacity (C) is a property of a substance that measures the amount of heat energy required to raise the temperature of one gram of the substance by one degree Celsius (°C) or one Kelvin (K). It reflects a material's ability to absorb and store thermal energy. The specific heat capacity depends on the nature of the substance and is an important concept in thermodynamics, particularly in understanding heat transfer and temperature changes in chemical reactions and physical processes.
The specific heat capacity is commonly expressed in units of J/g·°C or J/g·K.
Mathematical Expression for Specific Heat Capacity
The heat (\(q\)) required to change the temperature of a given mass of a substance is given by the formula:
\[
q = m \cdot C \cdot \Delta T
\]
Where:
- \(q\) is the heat absorbed or released (in joules),
- \(m\) is the mass of the substance (in grams),
- \(C\) is the specific heat capacity (in J/g·°C or J/g·K),
- \(\Delta T\) is the change in temperature (in °C or K).
This formula allows us to calculate how much heat energy is needed to change the temperature of a given mass of a substance by a certain amount. Conversely, it can be used to find the specific heat capacity of an unknown substance by measuring \(q\), \(m\), and \(\Delta T\).
Importance of Specific Heat Capacity in Chemical Thermodynamics
1. Heat Absorption and Temperature Change:
- Specific heat capacity is a measure of how much heat a substance can absorb before experiencing a significant temperature change. Substances with high specific heat capacities absorb more heat without large changes in temperature, making them effective at storing thermal energy.
Example:
- Water has a high specific heat capacity (4.18 J/g·°C), which is why it heats up and cools down slowly. This property of water is important in moderating temperatures in natural environments and in chemical reactions where water is used as a solvent.
2. Heat Transfer in Chemical Reactions:
- In exothermic and endothermic reactions, the specific heat capacity of the substances involved affects the temperature changes observed. A substance with a low specific heat capacity will undergo larger temperature changes for the same amount of heat transferred, while one with a high specific heat capacity will have smaller temperature changes.
Example:
- In a calorimetry experiment, a known mass of water with a high specific heat capacity is used to absorb the heat released by a reaction, allowing for accurate measurement of heat changes.
3. Energy Requirements for Heating and Cooling:
- Calculating the specific heat capacity helps determine the energy requirements for heating or cooling substances in chemical processes, industrial applications, and thermal management systems.
Specific Heat Capacity in Different States of Matter
1. Solids:
- In solids, particles are closely packed, and specific heat capacities are generally lower because energy goes into vibrating particles within a fixed structure.
Example:
- Metals typically have low specific heat capacities, which is why they heat up and cool down quickly compared to materials like water or air.
2. Liquids:
- Liquids generally have higher specific heat capacities than solids, as particles are free to move, and more energy is required to raise their temperature.
Example:
- Water has one of the highest specific heat capacities among common substances, making it an effective coolant and temperature stabilizer in biological and chemical systems.
3. Gases:
- Gases typically have variable specific heat capacities depending on whether they are under constant pressure (\(C_p\)) or constant volume (\(C_v\)) conditions. This is due to the work done by or on the gas as it expands or compresses.
Example:
- Carbon dioxide has different specific heat capacities at constant pressure and constant volume because gas expansion does work, affecting heat absorption.
Specific Heat Capacity in Calorimetry
In calorimetry, specific heat capacity plays a critical role in calculating the heat absorbed or released by a substance in a chemical or physical change.
The heat transferred in a calorimetry experiment can be calculated using:
\[
q = m \cdot C \cdot \Delta T
\]
If a reaction occurs within a known mass of water or another solution, the specific heat capacity of that solvent (often water) allows us to measure the heat exchange, since:
- Endothermic reactions absorb heat, causing the temperature of the surrounding water to decrease.
- Exothermic reactions release heat, causing the temperature of the surrounding water to increase.
Example:
In a reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) in an aqueous solution, the temperature change in water is measured. Using water’s specific heat capacity (4.18 J/g·°C), the heat released by the reaction can be calculated, providing valuable thermodynamic data.
Factors Affecting Specific Heat Capacity
1. Molecular Structure:
- The specific heat capacity of a substance is affected by its molecular structure and the types of bonds present. Substances with complex molecular structures, or substances that exhibit hydrogen bonding (like water), generally have higher specific heat capacities because more energy is required to raise their temperature.
2. Phase of Matter:
- The specific heat capacity varies with the phase of matter due to the different degrees of freedom for molecular movement. For example, gases under constant pressure have higher specific heat capacities than the same gases under constant volume conditions because they can expand and do work on their surroundings.
3. Temperature Range:
- Specific heat capacity can vary slightly with temperature, especially at very high temperatures, where molecular vibrations and rotations increase significantly, requiring more energy.
Examples of Specific Heat Capacities of Common Substances
- Water (liquid): 4.18 J/g·°C – High specific heat capacity, making it a common substance for thermal regulation in chemical processes.
- Ethanol (liquid): 2.44 J/g·°C – Lower than water, heats up faster with the same amount of energy.
- Iron (solid): 0.45 J/g·°C – Metals generally have low specific heat capacities, allowing them to conduct and lose heat rapidly.
- Aluminum (solid): 0.90 J/g·°C – Still low, but higher than iron, so aluminum heats up slower than iron for the same energy input.
Applications of Specific Heat Capacity
1. Chemical Reactions and Industrial Processes:
- Understanding specific heat capacity helps control and optimize energy use in processes requiring heating or cooling, such as chemical reactors, where temperature control is essential for reaction rates and yield.
2. Biological and Environmental Systems:
- Water’s high specific heat capacity is crucial in biology, as it stabilizes temperatures in living organisms and in large bodies of water, influencing climate and ecosystems.
3. Energy Storage and Management:
- Materials with high specific heat capacities are used in thermal energy storage systems (e.g., molten salts) for applications like solar energy storage, where heat needs to be absorbed and released efficiently over time.
Conclusion
Specific heat capacity (C) is a fundamental property that quantifies a substance’s ability to absorb and store thermal energy. It plays a crucial role in chemical thermodynamics by influencing temperature changes during chemical reactions and physical processes. Specific heat capacity is essential for calculating energy requirements in heating or cooling, especially in calorimetry experiments where heat exchange is monitored.
|
Q 1. A certain ideal gas undergoes a polytropic process \(P V^{n}=\) constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be \(\gamma\), then the range of values of \(n\) will be; |
|
(A) n lies between 0, \(\gamma\); |
|
(B) n lies between 1, \(\gamma\); |
|
(C) \(n=\gamma\); |
|
(D) \(n>\gamma\); |
|
Q 2. A certain ideal gas undergoes a polytropic process \(P V^{n}=\) constant such that the molar specific heat during the process is negative. If the ratio of the specific heats of the gas be \(\gamma\), then the range of values of \(n\) will be; |
|
(A) n lies between 0, \(\gamma\); |
|
(B) n lies between 1, \(\gamma\); |
|
(C) \(n=\gamma\); |
|
(D) \(n>\gamma\); |
|
Q 3. Among the following, the intensive properties are (i) molar conductivity (ii) electromotive force (iii) resistance (iv) heat capacity; |
|
(a) (ii) and (iii); |
|
(b) (i), (ii) and (iii); |
|
(c) (i) and (iv); |
|
(d) (i) only; |
|
Q 4. Which of the following is an example of extensive property?; |
|
(a) Temperature; |
|
(b) Density; |
|
(c) Mass; |
|
(d) Pressure; |
|
Q 5. Which of the following factors do not affect heat capacity?; |
|
(a) Size of system; |
|
(b) Composition of system; |
|
(c) Nature of system; |
|
(d) Temperature of the system; |
|
Q 6. The heat required to raise the temperature of body by \(1 \mathrm{C}^{\circ}\) is called; |
|
(a) specific heat; |
|
(b) thermal capacity; |
|
(c) water equivalent; |
|
(d) None of these.; |
|
Q 7. Equal volumes of two monoatomic gases, \(\mathrm{A}\) and \(\mathrm{B}\), at same temperature and pressure are mixed The ratio of specific heats \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)\) of the mixture will be :; |
|
(a) 0.83; |
|
(b) 1.50; |
|
(c) 3.3; |
|
(d) 1.67; |
|
Q 8. Equal volumes of two monoatomic gases, \(A\) and \(B\), at same temperature and pressure are mixed. The ratio of specific heats \(\left(C_{p} / C_{v}\right)\) of the mixture will be : [2012 M]; |
|
(a) 0.83; |
|
(b) 1.50; |
|
(c) 3.3; |
|
(d) 1.67; |
|
Q 9. Heat required to raise the temperature of 1 mole of substance by \(1^{\circ}\) is called; |
|
(a) specific heat.; |
|
(b) molar heat capacity.; |
|
(c) water equivalent.; |
|
(d) specific gravity.; |
Adiabatic Process in Chemical Thermodynamics
Definition of an Adiabatic Process
In chemical thermodynamics, an adiabatic process is one in which no heat is exchanged between the system and its surroundings. During an adiabatic process, any change in a system’s internal energy is a result of work done on or by the system, as there is no heat (\(q\)) entering or leaving the system. This concept is especially important in studying reactions and phase changes in isolated systems or adiabatic calorimeters, where heat transfer is intentionally minimized or eliminated.
For an adiabatic process:
\[
q = 0
\]
Thus, according to the First Law of Thermodynamics, the change in internal energy (\(\Delta U\)) is given by:
\[
\Delta U = W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done on or by the system.
Characteristics of Adiabatic Processes in Chemical Thermodynamics
1. No Heat Transfer:
- By definition, adiabatic processes involve no heat exchange with surroundings. Any energy change within the system arises solely from work interactions. For example, in an adiabatic reaction vessel, temperature changes are due to work done, not heat absorbed or released.
2. Temperature Change:
- In an adiabatic process, the temperature of the system changes due to work being done on or by the system. This temperature change directly impacts reaction rates and equilibrium in chemical reactions.
- Adiabatic Compression: Temperature increases as the system does work on itself.
- Adiabatic Expansion: Temperature decreases as the system does work on the surroundings.
3. Reversible and Irreversible Adiabatic Processes:
- Reversible Adiabatic Process: Occurs very slowly, allowing the system to remain in equilibrium throughout.
- Irreversible Adiabatic Process: Happens rapidly (e.g., rapid gas expansion), where the system does not maintain equilibrium.
Mathematical Expressions for Adiabatic Processes in Ideal Gases
For an ideal gas undergoing an adiabatic process, the relationship between pressure (P), volume (V), and the heat capacity ratio (\(\gamma\)) is given by:
\[
P V^{\gamma} = \text{constant}
\]
Where:
- \(\gamma\) (gamma) is the heat capacity ratio (\(\frac{C_p}{C_v}\)),
- \(C_p\): Heat capacity at constant pressure,
- \(C_v\): Heat capacity at constant volume.
Alternatively, the relationship can be expressed as:
\[
T V^{\gamma - 1} = \text{constant}
\]
Where:
- \(T\) is the temperature of the gas,
- \(V\) is the volume.
These relationships show how pressure, volume, and temperature change in an adiabatic process, which is particularly useful in calculating the thermodynamic behavior of gases in chemical processes.
Applications of Adiabatic Processes in Chemical Thermodynamics
1. Adiabatic Calorimetry:
- Adiabatic calorimetry is used to measure the heat of reactions or specific heat capacities of substances under adiabatic conditions. Here, the reaction vessel is insulated to prevent any heat exchange with the surroundings. This setup allows accurate measurement of temperature changes solely due to the reaction, enabling calculations of reaction enthalpies and understanding reaction kinetics.
2. Adiabatic Flame Temperature:
- The adiabatic flame temperature is the maximum temperature that a flame can reach without losing heat to the surroundings. It provides insights into the combustion efficiency and potential energy release of fuels. For instance, hydrocarbon fuels can achieve high adiabatic flame temperatures under ideal combustion conditions, providing a measure of the energy released in reactions.
3. Rapid Gas Expansion and Compression:
- Adiabatic processes are essential in understanding rapid gas expansion or compression, where no heat exchange occurs, but the gas temperature changes due to work done. This principle is applied in refrigeration systems and certain types of reactors where rapid temperature changes are controlled to influence reaction rates.
4. Atmospheric and Environmental Chemistry:
- In the atmosphere, air parcels often undergo adiabatic processes as they rise or fall in the absence of heat exchange with surroundings. This adiabatic cooling and heating influence cloud formation, weather patterns, and pollutant dispersion.
Example of Adiabatic Process in Chemical Reactions
Consider an exothermic reaction taking place in an adiabatic calorimeter. Since the system is adiabatic, no heat is lost or gained from the surroundings, so the energy released by the reaction manifests as an increase in temperature within the system.
For instance, in an adiabatic neutralization reaction between a strong acid and a strong base in an insulated vessel:
1. The heat generated by the reaction increases the internal energy of the system.
2. Since no heat can leave the vessel, the temperature of the system rises, which can be used to calculate the enthalpy change of the reaction.
Adiabatic Processes and the Second Law of Thermodynamics
According to the Second Law of Thermodynamics, the entropy of an isolated system in a natural process will increase. However, in a reversible adiabatic process, the entropy of the system remains constant because no heat is exchanged and the process is idealized as fully reversible.
In contrast, for irreversible adiabatic processes, such as rapid gas expansions, entropy increases due to the generation of disorder, even though no heat is exchanged with the surroundings.
Comparison of Adiabatic and Isothermal Processes in Chemistry
- Adiabatic Process: No heat exchange occurs, so temperature changes result from work. It is particularly useful in fast or insulated reactions.
- Isothermal Process: Temperature remains constant, requiring heat exchange to offset work done. This type of process is more commonly used when temperature-sensitive reactions occur in open systems or with controlled heat sources.
Example:
In a chemical reactor, an adiabatic setup can be used for reactions where a rapid temperature rise is needed to maintain reaction rates, while an isothermal setup is better suited for reactions where temperature control is necessary for product stability.
Conclusion
In chemical thermodynamics, an adiabatic process describes a system where no heat exchange occurs, and changes in temperature or internal energy result solely from work done on or by the system. This principle is particularly valuable in calorimetry, combustion studies, and understanding temperature-driven chemical reactions. Adiabatic processes help chemists and engineers accurately measure energy changes in isolated systems, allowing precise control over reaction conditions, making this concept essential for chemical thermodynamic studies and applications.
|
Q 1. The ratio of the slopes of \(p-V\) graphs of adiabatic and isothermal is; |
|
(A) \(\frac{\gamma-1}{\gamma}\); |
|
(B) \(\gamma-1\); |
|
(C) \(\gamma\); |
|
(D) \(\gamma+1\); |
|
Q 2. Which of the following is an equivalent cyclic process corresponding to the thermodynamic cyclic given in the figure?\nWhere, \(1 \rightarrow 2\) is adiabatic. (Graphs are schematic and are not to scale); |
|
(A); |
|
(B); |
|
(C); |
|
(D); |
|
Q 3. A diatomic ideal gas is compressed adiabatically to \(\frac{1}{32}\) of its initial volume. If the initial temperature of the gas is \(T_{\mathrm{i}}\) (in kelvin) and the final temperature is \(T_{\mathrm{f}}=a T_{\mathrm{i}}\), then value of \(a\) is; |
|
(A) 4; |
|
(B) 6; |
|
(C) 5; |
|
(D) 9; |
|
Q 4. 5.6 \(\mathrm{L}\) of helium gas at \(\mathrm{STP}\) is adiabatically compressed to \(0.7 \mathrm{~L}\). Taking the initial temperature to be \(T_{1}\) the work done in the process is; |
|
(A) \(-\frac{9}{8} R T_{1}\); |
|
(B) \(\frac{3}{2} R T_{1}\); |
|
(C) \(\frac{15}{8} R T_{1}\); |
|
(D) \(\frac{9}{2} R T_{1}\); |
|
Q 5. \(1 \mathrm{Mole}\) of \(\mathrm{CO}_{2}\) gas at \(300 \mathrm{~K}\) expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done (in \(\mathrm{kJ} / \mathrm{mol})\) is: (Given \(\gamma=1.33\) and \(\mathrm{C}_{\mathrm{v}}=25.10 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\) for \(\left.\mathrm{CO}_{2}\right)\) report your answer by rounding it up to nearest whole number; |
|
(A)4; |
|
(B)6; |
|
(C)8; |
|
(D)5; |
|
Q 6. \(1 \mathrm{Mole}\) of \(\mathrm{CO}_{2}\) gas at \(300 \mathrm{~K}\) expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done (in \(\mathrm{kJ} / \mathrm{mol})\) is: (Given \(\gamma=1.33\) and \(\mathrm{C}_{\mathrm{v}}=25.10 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\) for \(\left.\mathrm{CO}_{2}\right)\) report your answer by rounding it up to nearest whole number; |
|
(A)4; |
|
(B)6; |
|
(C)8; |
|
(D)5; |
|
Q 7. 5.6 \(\mathrm{L}\) of helium gas at \(\mathrm{STP}\) is adiabatically compressed to \(0.7 \mathrm{~L}\). Taking the initial temperature to be \(T_{1}\) the work done in the process is; |
|
(A) \(-\frac{9}{8} R T_{1}\); |
|
(B) \(\frac{3}{2} R T_{1}\); |
|
(C) \(\frac{15}{8} R T_{1}\); |
|
(D) \(\frac{9}{2} R T_{1}\); |
|
Q 8. A diatomic ideal gas is compressed adiabatically to \(\frac{1}{32}\) of its initial volume. If the initial temperature of the gas is \(T_{\mathrm{i}}\) (in kelvin) and the final temperature is \(T_{\mathrm{f}}=a T_{\mathrm{i}}\), then value of \(a\) is; |
|
(A) 4; |
|
(B) 6; |
|
(C) 5; |
|
(D) 9; |
|
Q 9. Which of the following is an equivalent cyclic process corresponding to the thermodynamic cyclic given in the figure?\nWhere, \(1 \rightarrow 2\) is adiabatic. (Graphs are schematic and are not to scale); |
|
(A); |
|
(B); |
|
(C); |
|
(D); |
|
Q 10. The ratio of the slopes of \(p-V\) graphs of adiabatic and isothermal is; |
|
(A) \(\frac{\gamma-1}{\gamma}\); |
|
(B) \(\gamma-1\); |
|
(C) \(\gamma\); |
|
(D) \(\gamma+1\); |
Internal Energy (U): Total Energy of the System
Definition of Internal Energy
In chemical thermodynamics, internal energy (U) is the total energy contained within a thermodynamic system, encompassing all forms of energy stored in the system's molecules and atoms. Internal energy includes the kinetic energy from molecular motion (translations, rotations, and vibrations) and potential energy from the forces between particles. This quantity is crucial in chemical reactions and phase changes, as it represents the system’s capacity to do work or transfer heat.
Internal energy is a state function, meaning its value depends only on the current state of the system (e.g., temperature, pressure, volume), not on how the system reached that state. In practice, internal energy changes (\(\Delta U\)) are usually measured, as the absolute internal energy is difficult to determine.
Components of Internal Energy
1. Kinetic Energy of Particles:
- The kinetic energy within a system includes the translational, rotational, and vibrational energy of molecules. The degree of molecular motion depends on the temperature of the system, with higher temperatures leading to greater kinetic energy.
- For gases, kinetic energy is often the largest contributor to internal energy because gas particles have significant freedom of movement.
2. Potential Energy of Intermolecular Forces:
- Potential energy arises from interactions between atoms and molecules, such as van der Waals forces, hydrogen bonding, ionic attractions, and covalent bonds.
- In condensed phases (liquids and solids), potential energy contributes significantly to internal energy, as particles are closer together and interact more strongly than in gases.
Internal Energy in Chemical Thermodynamics
In chemical systems, internal energy changes (\(\Delta U\)) are central to understanding reaction energetics and phase changes. For example, in an exothermic reaction, internal energy decreases as energy is released to the surroundings, while in an endothermic reaction, internal energy increases as energy is absorbed.
The First Law of Thermodynamics provides a foundation for analyzing changes in internal energy:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat added to the system,
- \(W\) is the work done on or by the system.
Internal Energy in Different Thermodynamic Processes
1. Isothermal Process (Constant Temperature):
- In an isothermal process, the temperature remains constant, so the internal energy change \(\Delta U\) of an ideal gas is zero because internal energy in an ideal gas depends solely on temperature.
2. Adiabatic Process (No Heat Exchange):
- In an adiabatic process, no heat is transferred (\(q = 0\)). Therefore, any change in internal energy is due entirely to work done on or by the system:
\[
\Delta U = W
\]
3. Isochoric Process (Constant Volume):
- In an isochoric (constant volume) process, no work is done (\(W = 0\)) because volume does not change. Therefore, the change in internal energy is equal to the heat added or removed from the system:
\[
\Delta U = q_V
\]
This relationship is useful for calorimetry, where reactions occur in a constant-volume setting.
4. Isobaric Process (Constant Pressure):
- In an isobaric (constant pressure) process, internal energy change is related to both heat absorbed and work done due to volume change. However, under constant pressure, enthalpy (\(H\)) is more commonly used than internal energy.
Measuring Internal Energy Changes in Chemical Reactions
1. Calorimetry:
- Calorimetry is an experimental technique used to measure the heat absorbed or released in a chemical reaction, often under constant volume (bomb calorimeter) to directly determine changes in internal energy.
- In a bomb calorimeter, which operates at constant volume, the heat measured (\(q_V\)) represents the change in internal energy for combustion reactions.
2. Internal Energy and Bond Energies:
- In chemical reactions, changes in internal energy often involve breaking and forming bonds. Breaking bonds requires energy (endothermic), increasing internal energy, while forming bonds releases energy (exothermic), decreasing internal energy.
Example:
- For a combustion reaction, such as methane burning in oxygen:
\[
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{energy}
\]
- The reaction releases heat, leading to a decrease in the internal energy of the system, as energy is transferred to the surroundings.
Internal Energy and the Ideal Gas Law
For an ideal gas, internal energy depends solely on temperature. This can be expressed mathematically as:
\[
U = n C_v T
\]
Where:
- \(n\) is the number of moles,
- \(C_v\) is the molar heat capacity at constant volume,
- \(T\) is the temperature in Kelvin.
This relationship shows that, for an ideal gas, any increase in temperature (due to heat added or work done on the gas) directly increases internal energy. For real gases, however, intermolecular interactions also contribute to internal energy.
Applications of Internal Energy in Chemical Thermodynamics
1. Reaction Energetics:
- Internal energy changes help predict reaction energetics, indicating whether energy is absorbed or released during chemical transformations. By analyzing \(\Delta U\), chemists can better understand reaction enthalpy and spontaneity.
2. Phase Transitions:
- During phase transitions, internal energy plays a key role. For example, during melting or vaporization, internal energy increases due to the input of heat to overcome intermolecular attractions. Conversely, condensation and freezing release internal energy.
3. Thermodynamic Cycles:
- Internal energy calculations are crucial in cycles like the Carnot cycle and Otto cycle, where energy transformations between heat and work form the basis for engines and refrigeration.
4. Biochemical Systems:
- In biochemistry, changes in internal energy influence cellular processes, such as ATP hydrolysis, where energy stored in molecular bonds is released for biological functions.
Comparison of Internal Energy with Enthalpy
While internal energy (U) and enthalpy (H) are related, they serve different purposes in thermodynamics:
- Internal Energy (U): Represents the total energy in a system, including kinetic and potential energy, useful for constant volume processes.
- Enthalpy (H): Represents the heat content of a system at constant pressure and includes work done against atmospheric pressure. Enthalpy is especially relevant for reactions occurring at constant pressure, such as in open containers.
Example:
In a constant-volume calorimeter, internal energy change (\(\Delta U\)) is measured directly, while in an open, constant-pressure calorimeter, enthalpy change (\(\Delta H\)) is more relevant.
Conclusion
Internal energy (U) is a fundamental thermodynamic property that represents the total energy within a chemical system. In chemical thermodynamics, it is critical for understanding reaction energetics, phase transitions, and heat-work interactions in isolated and closed systems. Through the First Law of Thermodynamics, internal energy changes help quantify the heat absorbed or released in reactions, enabling precise control of chemical processes and applications across various scientific fields. Understanding internal energy allows chemists to predict the behavior of substances under different conditions, making it a foundational concept in thermodynamics.
|
Q 1. If amount of heat given to a system be \(50 \mathrm{~J}\) and work done on the system be \(15 \mathrm{~J}\), then change in internal energy of the system is; |
|
(A) \(35 \mathrm{~J}\); |
|
(B) \(50 \mathrm{~J}\); |
|
(C) \(65 \mathrm{~J}\); |
|
(D) \(15 \mathrm{~J}\); |
|
Q 2. At constant volume, \(4 \mathrm{~mol}\) of an ideal gas when heated from \(300 \mathrm{~K}\) to \(500 \mathrm{~K}\) changes its internal energy by \(5000 \mathrm{~J}\). The molar heat capacity at constant volume is; |
|
(A)\(6.25 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\) ; |
|
(B)\(4.25 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
(C)\(6.50 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
(D)\(6.75 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
Q 3. At constant volume, \(4 \mathrm{~mol}\) of an ideal gas when heated from \(300 \mathrm{~K}\) to \(500 \mathrm{~K}\) changes its internal energy by \(5000 \mathrm{~J}\). The molar heat capacity at constant volume is; |
|
(A)\(6.25 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\) ; |
|
(B)\(4.25 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
(C)\(6.50 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
(D)\(6.75 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}\); |
|
Q 4. If amount of heat given to a system be \(50 \mathrm{~J}\) and work done on the system be \(15 \mathrm{~J}\), then change in internal energy of the system is; |
|
(A) \(35 \mathrm{~J}\); |
|
(B) \(50 \mathrm{~J}\); |
|
(C) \(65 \mathrm{~J}\); |
|
(D) \(15 \mathrm{~J}\); |
|
Q 5. Which of the following factors affect the internal energy of the system?; |
|
(a) Heat passes into or out of the system.; |
|
(b) Work is done on or by the system.; |
|
(c) Matter enters or leaves the system.; |
|
(d) All of the above; |
|
Q 6. Assertion : At constant temperature and pressure whatever heat absorbed by the system is used in doing work Reason : Internal energy change is zero.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 7. Assertion : Absolute value of internal energy of a substance cannot be determined Reason : It is impossible to determine exact values of constitutent energies of the substances.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 8. Assertion : There is exchange in internal energy in a cyclic process Reason : Cyclic proces is the one in which the sytem returns to its initial state after a number of reactions.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 9. Assertion : Internal energy is an extensive property Reason : Internal energy depends upon the amount of the system.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 10. Read the following statements carefully and choose the correct option (i) Internal energy, \(\mathrm{U}\), of the system is a state function. (ii) \(-\mathrm{w}\) shows, that work is done on the system. (iii) \(+\mathrm{w}\) shows, that work is done by the system; |
|
(a) (i) and (ii) are correct; |
|
(b) (ii) and (iii) are correct; |
|
(c) (i) and (iii) are correct; |
|
(d) Only (i) is correct; |
Gibbs Free Energy (ΔG): Determines Spontaneity of Processes
Definition of Gibbs Free Energy
In chemical thermodynamics, Gibbs free energy (G) is a thermodynamic potential that combines enthalpy (\(H\)), entropy (\(S\)), and temperature (\(T\)) to determine whether a process will occur spontaneously at constant pressure and temperature. The change in Gibbs free energy (\(\Delta G\)) for a reaction or process predicts its spontaneity: if \(\Delta G\) is negative, the process is spontaneous; if positive, the process is non-spontaneous.
The Gibbs free energy change is calculated as:
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G\): Gibbs free energy change,
- \(\Delta H\): Enthalpy change,
- \(T\): Temperature in Kelvin,
- \(\Delta S\): Entropy change.
This relationship shows that both the enthalpy (heat content) and entropy (disorder) changes, along with temperature, influence the spontaneity of a process.
Interpretation of ΔG and Spontaneity
1. Negative \(\Delta G\) (Spontaneous Process):
- If \(\Delta G < 0\), the process is spontaneous and will proceed in the forward direction under the given conditions. Spontaneous reactions release free energy, favoring product formation.
Example:
- In cellular respiration, glucose breaks down to produce energy, with a negative \(\Delta G\), driving the reaction forward and supplying energy for biological processes.
2. Positive \(\Delta G\) (Non-Spontaneous Process):
- If \(\Delta G > 0\), the process is non-spontaneous in the forward direction. External energy input is required to drive the reaction.
Example:
- Photosynthesis has a positive \(\Delta G\), meaning it is non-spontaneous and requires sunlight as an energy source.
3. Zero \(\Delta G\) (Equilibrium):
- If \(\Delta G = 0\), the system is at equilibrium, with no net change in the concentration of reactants and products. The forward and reverse reactions occur at equal rates.
Factors Affecting Gibbs Free Energy
1. Enthalpy (ΔH):
- \(\Delta H\) represents the heat absorbed or released in a reaction. Exothermic reactions (\(\Delta H < 0\)) tend to decrease \(\Delta G\), favoring spontaneity. Endothermic reactions (\(\Delta H > 0\)) increase \(\Delta G\) unless offset by a significant increase in entropy.
2. Entropy (ΔS):
- \(\Delta S\) reflects the degree of disorder or randomness in a system. Reactions that increase disorder (\(\Delta S > 0\)) tend to make \(\Delta G\) more negative, especially at higher temperatures, making the process more likely to be spontaneous.
3. Temperature (T):
- Temperature is a crucial factor in determining \(\Delta G\). For reactions where \(\Delta S\) is positive, increasing \(T\) makes \(\Delta G\) more negative, enhancing spontaneity. For reactions where \(\Delta S\) is negative, increasing \(T\) can make \(\Delta G\) positive, reducing spontaneity.
Calculating Gibbs Free Energy in Chemical Reactions
For a chemical reaction:
\[
\text{aA} + \text{bB} \rightarrow \text{cC} + \text{dD}
\]
The Gibbs free energy change can be calculated using the standard Gibbs free energies of formation (\(G_f^\circ\)) of the reactants and products:
\[
\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})
\]
Where:
- \(\Delta G_f^\circ\) values are standard Gibbs free energies of formation for each species, usually available in thermodynamic tables.
Example:
For the combustion of methane:
\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g)
\]
The \(\Delta G\) can be calculated using the standard Gibbs free energies of formation for \(CH_4\), \(O_2\), \(CO_2\), and \(H_2O\) from thermodynamic data.
Spontaneity and the Temperature Dependence of ΔG
The sign of \(\Delta G\) depends on \(\Delta H\), \(\Delta S\), and \(T\), giving rise to four possible cases:
1. \(\Delta H < 0\) and \(\Delta S > 0\):
- \(\Delta G\) is always negative, and the reaction is spontaneous at all temperatures.
2. \(\Delta H < 0\) and \(\Delta S < 0\):
- \(\Delta G\) is negative at low temperatures and positive at high temperatures. The reaction is spontaneous only at low temperatures.
3. \(\Delta H > 0\) and \(\Delta S > 0\):
- \(\Delta G\) is negative at high temperatures and positive at low temperatures. The reaction is spontaneous only at high temperatures.
4. \(\Delta H > 0\) and \(\Delta S < 0\):
- \(\Delta G\) is always positive, and the reaction is non-spontaneous at all temperatures.
These cases help predict under which conditions a reaction will proceed spontaneously.
Applications of Gibbs Free Energy in Chemical Thermodynamics
1. Predicting Reaction Spontaneity:
- By calculating \(\Delta G\), chemists can determine whether a reaction will proceed spontaneously under specific conditions, aiding in the design of chemical processes and understanding reaction feasibility.
2. Equilibrium and Reaction Direction:
- At equilibrium, \(\Delta G = 0\). The position of equilibrium depends on \(\Delta G\) under standard conditions and provides insights into the relative stability of reactants and products.
3. Electrochemical Cells:
- In electrochemistry, Gibbs free energy change relates to the maximum work (electrical energy) that can be obtained from an electrochemical cell:
\[
\Delta G = -nFE
\]
Where \(n\) is the number of moles of electrons transferred, \(F\) is Faraday’s constant, and \(E\) is the cell potential.
4. Biochemical Reactions:
- In biochemistry, Gibbs free energy is used to analyze energy transfer in metabolic pathways. Reactions with negative \(\Delta G\) drive essential biological processes, such as ATP hydrolysis, that power cellular functions.
Examples of Gibbs Free Energy in Chemical Reactions
1. Combustion of Fuels:
- The combustion of fuels (e.g., methane, gasoline) releases energy, resulting in a negative \(\Delta G\) under standard conditions, making these reactions highly spontaneous and suitable for energy generation.
2. Dissolution Processes:
- The spontaneity of dissolution depends on the balance between enthalpy and entropy changes. For example, the dissolution of ammonium nitrate is endothermic (\(\Delta H > 0\)) but occurs spontaneously because of the positive entropy change (\(\Delta S > 0\)), yielding a negative \(\Delta G\) at room temperature.
3. Photosynthesis:
- Photosynthesis has a positive \(\Delta G\) under standard conditions and is non-spontaneous, requiring sunlight as an energy source to drive the reaction.
Conclusion
Gibbs free energy (\(\Delta G\)) is a fundamental concept in chemical thermodynamics that determines the spontaneity of a process. A negative \(\Delta G\) indicates a spontaneous reaction, while a positive \(\Delta G\) indicates non-spontaneity. Gibbs free energy integrates enthalpy, entropy, and temperature, allowing chemists to predict reaction behavior, analyze equilibrium, and understand energy transformations.
|
Q 1. Which one of the following process is non-spontaneous?; |
|
(A) Dissolution of \(\mathrm{CuSO}_{4}\) in water.; |
|
(B) Reaction between \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) to form water.; |
|
(C) Water flowing down hill.; |
|
(D) Flow of electric current from low potential to high potential; |
|
Q 2. Which one of the following process is non-spontaneous?; |
|
(A) Dissolution of \(\mathrm{CuSO}_{4}\) in water.; |
|
(B) Reaction between \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) to form water.; |
|
(C) Water flowing down hill.; |
|
(D) Flow of electric current from low potential to high potential; |
|
Q 3. A reaction cannot take place spontaneously at any temperature when; |
|
(a) both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are positive; |
|
(b) both \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are negative; |
|
(c) \(\Delta \mathrm{H}\) is negative and \(\Delta \mathrm{S}\) is positive; |
|
(d) \(\Delta \mathrm{H}\) is positive and \(\Delta \mathrm{S}\) is negative; |
|
Q 4. A reaction is spontaneous at low temperature but nonspontaneous at high temperature Which of the following is true for the reaction?; |
|
(a) \(\Delta \mathrm{H}>0, \Delta \mathrm{S}>0\); |
|
(b) \(\Delta \mathrm{H}<0, \Delta \mathrm{S}>0\); |
|
(c) \(\Delta \mathrm{H}>0, \Delta \mathrm{S}=0\); |
|
(d) \(\Delta \mathrm{H}<0, \Delta \mathrm{S}<0\); |
|
Q 5. A chemical reaction is spontaneous at \(298 \mathrm{~K}\) but nonspontaneous at \(350 \mathrm{~K}\) Which one of the following is true for the reaction? & \(\Delta \mathrm{G}\) & \(\Delta \mathrm{H}\) & \(\Delta \mathrm{S}\) \\; |
|
(a) - & - & + \\; |
|
(b) + & + & + \\; |
|
(c) - & + & - \\; |
|
(d) - & - & -; |
|
Q 6. For a particular reversible reaction at temperature \(T, \Delta H\) and \(\Delta S\) were found to be both + ve If \(T_{e}\) is the temperature at equilibrium, the reaction would be spontaneous when; |
|
(a) \(T_{e}>T\); |
|
(b) \(T>T_{e}\); |
|
(c) \(T_{e}\) is 5 times \(T\); |
|
(d) \(T=T_{e}\); |
|
Q 7. Pick out the wrong statement; |
|
(a) The standard free energy of formation of all elements is zero; |
|
(b) A process accompanied by decrease in entropy is spontaneous under certain conditions; |
|
(c) The entropy of a perfectly crystalline substance at absolute zero is zero; |
|
(d) A process that leads to increase in free energy will be spontaneous; |
|
Q 8. Identify the correct statement for change of Gibbs energy for a system \(\Delta \mathrm{G}_{\text {system }}\) at constant temperature and pressure; |
|
(a) If \(\Delta \mathrm{G}_{\text {system }}=0\), the system has attained equilibrium; |
|
(b) If \(\Delta \mathrm{G}_{\text {system }}=0\), the system is still moving in a particular direction; |
|
(c) If \(\Delta \mathrm{G}_{\text {system }}<0\), the process is not spontaneous; |
|
(d) If \(\Delta \mathrm{G}_{\text {system }}>0\), the process is not spontaneous; |
|
Q 9. Identify the correct statement regarding a spontaneous process:; |
|
(a) Lowering of energy in the process is the only criterion for spontaneity.; |
|
(b) For a spontaneous process in an isolated system, the change in entropy is positive.; |
|
(c) Endothermic processes are never spontaneous.; |
|
(d) Exothermic processes are always spontaneous.; |
|
Q 10. A chemical reaction will be spontaneous if it is accompanied by a decrease of; |
|
(a) entropy of the system.; |
|
(b) enthalpy of the system.; |
|
(c) internal energy of the system.; |
|
(d) free energy of the system.; |
Second Law of Thermodynamics: Spontaneous Processes Increase Entropy
Definition of the Second Law of Thermodynamics
In chemical thermodynamics, the Second Law of Thermodynamics states that in any spontaneous process, the total entropy of a system and its surroundings always increases. Entropy, represented as \(S\), is a measure of disorder or randomness, and the Second Law establishes that natural processes tend to move toward a state of maximum entropy.
Mathematically, the Second Law can be expressed as:
\[
\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0
\]
where:
- \(\Delta S_{\text{total}}\) is the total entropy change,
- \(\Delta S_{\text{system}}\) is the entropy change in the system,
- \(\Delta S_{\text{surroundings}}\) is the entropy change in the surroundings.
A positive \(\Delta S_{\text{total}}\) indicates a spontaneous process, while a negative value means the process is non-spontaneous. This law is foundational in understanding the direction and feasibility of chemical reactions and phase changes.
Key Concepts of the Second Law in Chemical Thermodynamics
1. Spontaneity of Reactions:
- The Second Law implies that for a process to be spontaneous, it must result in an increase in the total entropy of the system and surroundings. Even if a reaction appears unfavorable based solely on enthalpy (\(\Delta H\)), a sufficiently large increase in entropy (\(\Delta S\)) can make it spontaneous.
- A spontaneous process does not require external energy input and naturally occurs in the direction that increases entropy.
2. Entropy as a State Function:
- Entropy is a state function, meaning it depends only on the initial and final states of the system, not on the path taken. As reactions progress toward equilibrium, they often result in a greater degree of randomness or disorder in the final state.
3. Irreversible vs. Reversible Processes:
- Irreversible processes (e.g., spontaneous reactions) always increase the total entropy of the universe.
- Reversible processes are idealized processes where entropy remains constant in the universe (\(\Delta S_{\text{total}} = 0\)). Such processes are theoretical and do not occur naturally.
Mathematical Formulation of the Second Law
In chemical systems, the change in entropy for a process can be calculated using the formula:
\[
\Delta S = \frac{q_{\text{rev}}}{T}
\]
Where:
- \(\Delta S\) is the entropy change,
- \(q_{\text{rev}}\) is the heat absorbed or released in a reversible process,
- \(T\) is the absolute temperature (in Kelvin).
This equation is useful for calculating entropy changes during phase transitions, such as melting or vaporization, where heat is transferred reversibly.
Applications of the Second Law in Chemical Thermodynamics
1. Reaction Spontaneity and Gibbs Free Energy:
- The Second Law links entropy to Gibbs free energy (\( \Delta G \)), a concept that combines enthalpy and entropy to determine spontaneity:
\[
\Delta G = \Delta H - T \Delta S
\]
Where:
- \(\Delta G < 0\) indicates a spontaneous process, consistent with an increase in total entropy.
- This relationship shows that the Second Law drives the conditions under which chemical reactions or processes are spontaneous.
2. Phase Transitions:
- The Second Law explains why phase transitions occur spontaneously in specific directions:
- Melting: Solid to liquid, where entropy increases as particles move more freely.
- Vaporization: Liquid to gas, with a significant entropy increase as gas particles are highly disordered.
- These processes proceed spontaneously at the appropriate temperature and pressure conditions.
3. Dissolution and Mixing:
- When substances dissolve or mix, entropy increases because molecules become more dispersed, leading to greater disorder. For example, dissolving NaCl in water increases entropy as the lattice breaks apart, dispersing Na\(^+\) and Cl\(^-\) ions in the solvent.
- Entropy considerations are also crucial in understanding solubility and why certain solutes dissolve in solvents spontaneously.
4. Chemical Equilibrium:
- At equilibrium, the entropy of a system is maximized for the given conditions, and the reaction has reached a state where the forward and reverse reactions occur at equal rates. At this point, there is no further increase in the total entropy of the universe, and \(\Delta G = 0\).
- The Second Law thus helps predict the position of equilibrium in a chemical reaction, providing insights into the ratio of products and reactants at equilibrium.
Examples of Entropy Increase in Chemical Processes
1. Combustion Reactions:
- Combustion reactions, such as the burning of methane in oxygen to produce carbon dioxide and water, result in a large increase in entropy because gaseous products (e.g., \(CO_2\) and \(H_2O\)) are produced, creating a more disordered state.
\[
CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)
\]
- The production of gases from relatively ordered reactants (methane and oxygen) increases the total entropy, making combustion reactions highly spontaneous.
2. Endothermic Dissolution:
- Dissolution of certain salts, like ammonium nitrate (NH\(_4\)NO\(_3\)), is endothermic (\(\Delta H > 0\)) but spontaneous due to a large positive entropy change as the salt dissociates into ions in solution.
- The increase in entropy offsets the endothermic nature, allowing the process to occur spontaneously at room temperature.
3. Phase Change in Ice Melting:
- Ice melting at temperatures above 0°C is a spontaneous process because the entropy of liquid water is higher than that of solid ice, even though heat must be absorbed (endothermic process). The entropy increase (\(\Delta S > 0\)) drives the spontaneity of melting.
Entropy and the Universe: Total Entropy Change
The Second Law applies not only to chemical reactions but also to the universe as a whole:
\[
\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0
\]
For a process to be spontaneous, the total entropy of the system and surroundings must increase, driving the natural direction of change toward greater disorder. This is often referred to as the arrow of time in thermodynamics, as it dictates that systems move toward higher entropy and stability.
Example:
Consider an exothermic reaction, where the system releases heat to the surroundings, increasing the entropy of the surroundings and resulting in a positive \(\Delta S_{\text{universe}}\). This makes the reaction spontaneous, in alignment with the Second Law.
The Second Law and Heat Engines
The Second Law also has significant implications for heat engines and energy efficiency. In any real engine, some energy is inevitably lost as waste heat, and the engine cannot convert all the absorbed heat into work due to entropy increases. This principle establishes the Carnot efficiency limit for ideal engines, reflecting the maximum possible efficiency constrained by entropy.
Conclusion
The Second Law of Thermodynamics is a fundamental principle in chemical thermodynamics, asserting that spontaneous processes increase the entropy of the universe. In chemical systems, this law dictates reaction spontaneity, phase transitions, dissolution, and equilibrium behavior. By driving systems toward maximum entropy, the Second Law defines the natural direction of chemical processes, making it essential for understanding the behavior of matter and energy in thermodynamic systems.
|
Q 1. The third law of thermodynamics states that:; |
|
(A) the entropy of a perfectly crystalline pure substance at zero \(\mathrm{K}\) is zero.; |
|
(B) absolute entropy of hydrogen ion is zero at zero \(\mathrm{K}\).; |
|
(C) Net change in entropy in conversion \(\mathrm{H}_{2(\mathrm{~g})}(130 \mathrm{~K}) \rightarrow \mathrm{H}_{2(\mathrm{~g})}(200 \mathrm{~K})\) is zero.; |
|
(D) entropy generally decrease in combustion reactions.; |
|
Q 2. The third law of thermodynamics states that:; |
|
(A) the entropy of a perfectly crystalline pure substance at zero \(\mathrm{K}\) is zero.; |
|
(B) absolute entropy of hydrogen ion is zero at zero \(\mathrm{K}\).; |
|
(C) Net change in entropy in conversion \(\mathrm{H}_{2(\mathrm{~g})}(130 \mathrm{~K}) \rightarrow \mathrm{H}_{2(\mathrm{~g})}(200 \mathrm{~K})\) is zero.; |
|
(D) entropy generally decrease in combustion reactions.; |
|
Q 3. In an exothermic reaction (reversible) which of the following has positive value?; |
|
(a) Enthalpy; |
|
(b) Entropy; |
|
(c) Gibb's free energy; |
|
(d) None of these; |
|
Q 4. An ideal gas undergoes isothermal expansion at constant pressure During the process; |
|
(a) enthalpy increases but entropy decreases.; |
|
(b) enthalpy remains constant but entropy increases.; |
|
(c) enthalpy decreases but entropy increases.; |
|
(d) Both enthalpy and entropy remain constant.; |
|
Q 5. For which of the following processes, \(\Delta S\) is negative?; |
|
(a) \(\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})\); |
|
(b) \(\mathrm{N}_{2}(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}, 5 \mathrm{~atm})\); |
|
(c) \(\mathrm{C}\) (diamond) \(\rightarrow \mathrm{C}\) (graphite); |
|
(d) \(\mathrm{N}_{2}(\mathrm{~g}, 273 \mathrm{~K}) \rightarrow \mathrm{N}_{2}(\mathrm{~g}, 300 \mathrm{~K})\); |
Enthalpy (\(\Delta H\)): Heat Changes at Constant Pressure
Definition of Enthalpy
In chemical thermodynamics, enthalpy (H) is a state function that represents the total heat content of a system at constant pressure. The enthalpy change (\(\Delta H\)) during a process indicates the amount of heat absorbed or released when a reaction occurs at constant pressure, making it fundamental for studying reaction energetics. Enthalpy combines the system's internal energy (U) and the product of pressure (P) and volume (V):
\[
H = U + PV
\]
Where:
- \(H\) is the enthalpy,
- \(U\) is the internal energy,
- \(P\) is the pressure,
- \(V\) is the volume.
Enthalpy Change (\(\Delta H\)) in Chemical Reactions
The enthalpy change (\(\Delta H\)) for a chemical reaction measures the heat absorbed or released at constant pressure. For a reaction written as:
\[
\text{Reactants} \rightarrow \text{Products}
\]
The enthalpy change can be defined as:
\[
\Delta H = H_{\text{products}} - H_{\text{reactants}}
\]
1. \(\Delta H < 0\) (Exothermic Reaction):
- An exothermic reaction releases heat, resulting in a negative enthalpy change. Energy is transferred from the system to the surroundings, making the products have lower enthalpy than the reactants.
Example:
- The combustion of methane:
\[
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{energy}
\]
- This reaction releases heat to the surroundings, resulting in a negative \(\Delta H\).
2. \(\Delta H > 0\) (Endothermic Reaction):
- An endothermic reaction absorbs heat, resulting in a positive enthalpy change. Energy is absorbed by the system from the surroundings, making the products have higher enthalpy than the reactants.
Example:
- The decomposition of calcium carbonate:
\[
CaCO_3(s) \rightarrow CaO(s) + CO_2(g)
\]
- This reaction requires heat input, leading to a positive \(\Delta H\).
Standard Enthalpy Changes
To facilitate comparisons, enthalpy changes are often measured under standard conditions (298 K, 1 atm, 1 M concentration) and are denoted with a superscript °:
1. Standard Enthalpy of Reaction (\(\Delta H^\circ_{\text{rxn}}\)):
- This is the enthalpy change for a reaction under standard conditions.
2. Standard Enthalpy of Formation (\(\Delta H^\circ_f\)):
- The enthalpy change when 1 mole of a compound is formed from its elements in their standard states. The \(\Delta H^\circ_f\) of elements in their most stable form (e.g., \(O_2\), \(N_2\), \(H_2\)) is zero.
Example:
- For water:
\[
H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H^\circ_f = -285.8 \, \text{kJ/mol}
\]
3. Standard Enthalpy of Combustion (\(\Delta H^\circ_c\)):
- The enthalpy change when 1 mole of a substance burns completely in oxygen under standard conditions.
Example:
- The combustion of glucose:
\[
C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \quad \Delta H^\circ_c = -2803 \, \text{kJ/mol}
\]
Enthalpy and Calorimetry
Calorimetry is an experimental technique used to measure enthalpy changes by determining the heat absorbed or released during a reaction. The two main types of calorimetry are:
1. Constant-Pressure Calorimetry:
- Conducted in an open system where the pressure remains constant (e.g., a coffee-cup calorimeter), this setup measures the heat of reactions at constant pressure, allowing direct calculation of \(\Delta H\).
- The heat (\(q\)) absorbed or released at constant pressure is equal to \(\Delta H\):
\[
q_p = \Delta H
\]
2. Constant-Volume Calorimetry:
- Conducted in a closed, rigid system (e.g., bomb calorimeter) where volume remains constant, this setup measures the internal energy change (\(\Delta U\)), which can be related to enthalpy if the work term is considered.
Enthalpy in Phase Changes
Enthalpy changes are also crucial in understanding phase changes, where substances transition between solid, liquid, and gaseous states. Important enthalpy changes in phase transitions include:
1. Enthalpy of Fusion (\(\Delta H_{\text{fus}}\)):
- The heat required to convert 1 mole of a solid to a liquid at its melting point. This process is endothermic (positive \(\Delta H\)) as it requires energy to overcome intermolecular forces.
2. Enthalpy of Vaporization (\(\Delta H_{\text{vap}}\)):
- The heat required to convert 1 mole of a liquid to a gas at its boiling point. Vaporization is also endothermic, with a positive \(\Delta H\) since energy is needed to separate molecules.
3. Enthalpy of Sublimation (\(\Delta H_{\text{sub}}\)):
- The heat required to convert 1 mole of a solid directly to a gas. Sublimation involves overcoming all intermolecular forces holding the solid together, so it is also endothermic.
Example:
For water:
\[
H_2O(l) \rightarrow H_2O(g) \quad \Delta H_{\text{vap}} = +40.7 \, \text{kJ/mol}
\]
Hess’s Law and Enthalpy Calculations
Hess’s Law states that the total enthalpy change of a reaction is the same, regardless of the pathway taken. This law allows us to calculate \(\Delta H\) for complex reactions by breaking them down into steps with known enthalpies.
Mathematically:
\[
\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \ldots
\]
Example:
For the reaction:
\[
C(s) + O_2(g) \rightarrow CO_2(g)
\]
Using known reactions with their enthalpy changes, we can determine \(\Delta H\) indirectly.
Applications of Enthalpy in Chemical Thermodynamics
1. Predicting Reaction Feasibility:
- Enthalpy helps predict if reactions are energetically favorable. Exothermic reactions (\(\Delta H < 0\)) are often spontaneous, while endothermic reactions (\(\Delta H > 0\)) may require an energy source to proceed.
2. Biochemical Reactions:
- In biochemistry, enthalpy is key to understanding metabolic reactions, such as ATP hydrolysis, which releases energy necessary for cellular functions.
3. Industrial Processes:
- Enthalpy calculations are essential in designing industrial chemical processes, where efficient energy management can save costs. For instance, exothermic reactions can be harnessed for heat recovery in manufacturing.
4. Phase Transition Applications:
- Understanding enthalpy changes in phase transitions is vital in fields like cryogenics, refrigeration, and distillation, where controlled heating or cooling is necessary.
Enthalpy and the First Law of Thermodynamics
The First Law of Thermodynamics links enthalpy to internal energy (\(U\)) and work done by or on the system:
\[
\Delta H = \Delta U + P \Delta V
\]
Where:
- \(\Delta H\) represents heat change at constant pressure,
- \(\Delta U\) represents changes in internal energy,
- \(P \Delta V\) accounts for the work done by the system when it expands or compresses.
This equation underscores that enthalpy not only includes internal energy but also the work required to displace the environment under constant pressure, making it a practical thermodynamic function for reactions occurring in open systems.
Conclusion
Enthalpy (\(\Delta H\)) represents the heat change at constant pressure and is a vital concept in chemical thermodynamics. It helps determine the heat absorbed or released in chemical reactions and phase changes. Exothermic reactions release heat, while endothermic reactions absorb it, making enthalpy essential in predicting reaction feasibility, designing calorimetry experiments, and managing energy in industrial and biochemical processes. By quantifying heat changes, enthalpy provides insights into the energy dynamics within chemical systems, enabling accurate analysis and control of thermodynamic processes.
|
Q 1. The bond dissociation enthalpy of gaseous \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\) and \(\mathrm{HCl}\) are 435.243 and \(431 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. In the same unit, enthalpy of formation of \(\mathrm{HCl}\) gas is ; |
|
(B) -184; |
|
(C) -92; |
|
(D)+184; |
|
+92; |
|
Q 2. In a constant volume calorimeter, \(3.5 \mathrm{~g}\) of a gas with molecular weight 28 was burnt in excess oxygen at \(298.0 \mathrm{~K}\). The temperature of the calorimeter was found to increase from \(298.0 \mathrm{~K}\) to \(298.45 \mathrm{~K}\) due to the combustion process. Given that the heat capacity of the calorimeter is \(2.5 \mathrm{~kJ} \mathrm{~K}^{-1}\), the numerical value for the enthalpy of combustion of the gas in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is; |
|
(A)\(12\mathrm{kJmol}^{-1}\); |
|
(B)\(15\mathrm{kJmol}^{-1}\); |
|
(C)\(20 \mathrm{kJmol}^{-1}\); |
|
(D)\(9 \mathrm{kJmol}^{-1}\); |
|
Q 3. Some properties of system are given below out of which \(x\) are number of extensive property and \(y\) are the number of intensive property. find \((\mathrm{x}-\mathrm{y})\). Boiling point, surface tension, heat capacity, gibbs free energy, vapour pressure, refractive index, density, temperature, entropy, enthalpy, mass, length, volume, area, internal energy.; |
|
(A)1; |
|
(B)2; |
|
(C)3; |
|
(D)4; |
|
Q 4. Some properties of system are given below out of which \(x\) are number of extensive property and \(y\) are the number of intensive property. find \((\mathrm{x}-\mathrm{y})\). Boiling point, surface tension, heat capacity, gibbs free energy, vapour pressure, refractive index, density, temperature, entropy, enthalpy, mass, length, volume, area, internal energy.; |
|
(A)1; |
|
(B)2; |
|
(C)3; |
|
(D)4; |
|
Q 5. In a constant volume calorimeter, \(3.5 \mathrm{~g}\) of a gas with molecular weight 28 was burnt in excess oxygen at \(298.0 \mathrm{~K}\). The temperature of the calorimeter was found to increase from \(298.0 \mathrm{~K}\) to \(298.45 \mathrm{~K}\) due to the combustion process. Given that the heat capacity of the calorimeter is \(2.5 \mathrm{~kJ} \mathrm{~K}^{-1}\), the numerical value for the enthalpy of combustion of the gas in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is; |
|
(A)\(12\mathrm{kJmol}^{-1}\); |
|
(B)\(15\mathrm{kJmol}^{-1}\); |
|
(C)\(20 \mathrm{kJmol}^{-1}\); |
|
(D)\(9 \mathrm{kJmol}^{-1}\); |
|
Q 6. The bond dissociation enthalpy of gaseous \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\) and \(\mathrm{HCl}\) are 435.243 and \(431 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. In the same unit, enthalpy of formation of \(\mathrm{HCl}\) gas is ; |
|
(B) -184; |
|
(C) -92; |
|
(D)+184; |
|
+92; |
|
Q 7. Among the following the state function(s) is (are) (i) Internal energy (ii) Irreversible expansion work (iii) Reversible expansion work (iv) Molar enthalpy; |
|
(a) (ii) and (iii); |
|
(b) (i), (ii) and (iii); |
|
(c) (i) and (iv); |
|
(d) (i) only; |
|
Q 8. Enthalpy change \((\Delta \mathrm{H})\) of a system depends upon its; |
|
(a) Initial state; |
|
(b) Final state; |
|
(c) Both on initial and final state; |
|
(d) None of these; |
|
Q 9. The enthalpy change of a reaction does not depend on; |
|
(a) The state of reactants and products; |
|
(b) Nature of reactants and products; |
|
(c) Different intermediate reactions; |
|
(d) Initial and final enthalpy change of a reaction.; |
|
Q 10. The relationship between enthalpy change and internal energy change is; |
|
(a) \(\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}\); |
|
(b) \(\Delta \mathrm{H}=(\Delta \mathrm{E}+\mathrm{V} \Delta \mathrm{P})\); |
|
(c) \(\Delta \mathrm{H}=\Delta \mathrm{E}-\mathrm{P} \Delta \mathrm{V}\); |
|
(d) \(\Delta \mathrm{H}=\mathrm{P} \Delta \mathrm{V}-\Delta \mathrm{E}\); |
Closed and Isolated Systems in Chemical Thermodynamics
In chemical thermodynamics, closed and isolated systems refer to two types of thermodynamic systems that differ in their capacity to exchange energy and matter with their surroundings. Understanding these systems is essential for analyzing how reactions and processes proceed in various environments, as the level of interaction with surroundings significantly affects thermodynamic behavior and measurements.
Closed System
A closed system allows exchange of energy (heat or work) with its surroundings but does not permit the exchange of matter. This characteristic makes closed systems ideal for studying reactions and processes where energy transfer occurs but where no external substances can enter or leave.
- Example: A sealed container that allows for heat transfer through its walls but is completely closed to any inflow or outflow of gases or liquids is a closed system.
In closed systems:
1. Energy Exchange: The system can absorb or release heat (q) or perform work (W), which impacts the system’s internal energy.
- If heat is absorbed from the surroundings, the internal energy increases.
- If heat is released, the internal energy decreases.
2. No Matter Exchange: Since matter cannot cross the boundary, the total mass and the composition of the system remain constant.
3. Applications:
- Calorimetry experiments: Closed systems are often used in calorimetry (e.g., bomb calorimeters), where the heat absorbed or released during a reaction is measured without any mass exchange.
- Chemical reactions in a sealed vessel: Closed systems are common in reactions where researchers wish to control the composition and measure only energy changes.
Example of Thermodynamic Analysis in a Closed System:
For a closed system, the First Law of Thermodynamics is applied as follows:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat exchanged with surroundings,
- \(W\) is the work done on or by the system.
This equation helps calculate energy changes in reactions conducted in a closed environment, providing insights into reaction enthalpy and energy requirements.
Isolated System
An isolated system is completely insulated from its surroundings, allowing no exchange of matter or energy (neither heat nor work) with the surroundings. The system's total internal energy and composition remain constant, making it the ideal environment for studying the natural progression of reactions without external influence.
- Example: A perfectly insulated thermos bottle with a tightly sealed cap is an isolated system, as it prevents heat, work, and matter exchange with its surroundings.
In isolated systems:
1. No Energy Exchange: There is no heat transfer (\(q = 0\)) or work done (\(W = 0\)), so the internal energy of the system remains constant.
2. No Matter Exchange: The system is completely sealed from the surroundings, so its mass and chemical composition are fixed.
3. Applications:
- Adiabatic calorimetry: Isolated systems are used in adiabatic calorimetry, where the system’s heat change is studied in the absence of any heat exchange with the surroundings.
- Studying spontaneous processes: Since isolated systems allow processes to proceed without external influence, they are ideal for observing whether a reaction or phase change is spontaneous under specific initial conditions.
Example of Thermodynamic Analysis in an Isolated System:
For an isolated system, the First Law of Thermodynamics simplifies to:
\[
\Delta U = 0
\]
Since there is no heat or work exchange, the internal energy remains constant. Any change within the system, such as a chemical reaction or phase change, must occur using the energy already present within the system.
Comparison of Closed and Isolated Systems
| Property | Closed System | Isolated System |
||--||
| Energy Exchange | Yes (heat or work can cross the boundary) | No (no heat or work crosses the boundary)|
| Matter Exchange | No (mass remains constant within the system) | No (mass and energy remain constant) |
| Internal Energy | Changes with heat/work interactions | Constant, as no energy is exchanged |
| Applications | Calorimetry (closed vessels), reactions with controlled heat exchange | Adiabatic processes, adiabatic calorimetry|
Thermodynamic Implications of Closed and Isolated Systems
1. Entropy and the Second Law of Thermodynamics:
- In an isolated system, the total entropy can only increase or remain constant, consistent with the Second Law of Thermodynamics. This allows us to study spontaneous processes without external energy influence.
- In a closed system, entropy changes depend on both heat exchange with the surroundings and internal reactions.
2. Practical Examples in Chemistry:
- Closed System: A reaction in a sealed flask with controlled heating is a closed system, useful for studying enthalpy changes without mass changes.
- Isolated System: A bomb calorimeter is close to an isolated system, where all heat generated by the reaction stays within the insulated chamber, providing accurate enthalpy data.
3. Gibbs Free Energy:
- In isolated systems, since no energy or matter exchange occurs, processes must occur spontaneously if they are to proceed, following a decrease in Gibbs free energy (\(\Delta G < 0\)) for any internal changes.
Applications in Chemical Thermodynamics
1. Studying Spontaneous Reactions:
- Isolated systems allow us to observe whether reactions proceed naturally based on initial conditions. This setup is essential for studying spontaneity without interference.
2. Measuring Heat of Combustion in Closed Systems:
- Closed systems like bomb calorimeters provide a nearly isolated environment where the heat released in combustion reactions can be accurately measured, essential for calculating the enthalpy change of fuels and other substances.
3. Phase Change Experiments:
- Closed systems are used to study phase transitions, such as melting or vaporization, where heat can be controlled without losing matter.
Conclusion
In chemical thermodynamics, closed and isolated systems provide distinct environments for analyzing thermodynamic behavior:
- A closed system allows energy exchange but no matter exchange, ideal for reactions where heat transfer is required without changes in mass.
- An isolated system restricts both energy and matter exchange, making it perfect for studying spontaneous processes and natural progression without external influence.
|
Q 1. Which of the following is closed system ?; |
|
(a) Jet engine; |
|
(b) Tea placed in a steel kettle; |
|
(c) Pressure cooker; |
|
(d) Rocket engine during propulsion; |
|
Q 2. An isolated system is that system in which; |
|
(a) There is no exchange of energy with the surroundings; |
|
(b) There is exchange of mass and energy with the surroundings; |
|
(c) There is no exchange of mass or energy with the surroundings; |
|
(d) There is exchange of mass with the surroundings; |
|
Q 3. In a closed insulated container, a liquid is stirred with a paddle to increase the temperature, which of the following is true? [2002]; |
|
(a) \(\Delta E=W \neq 0, q=0\); |
|
(b) \(\Delta E=W=q \neq 0\); |
|
(c) \(\Delta E=0, W=q \neq 0\); |
|
(d) \(W=0, \Delta E=q \neq 0\); |
|
Q 4. Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures \(T_{1}\) and \(T_{2}\left(T_{1} |
|
(a); |
|
(b); |
|
(c); |
|
(d); |
|
Q 5. For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?; |
|
(a); |
|
(b); |
|
(c); |
|
(d); |
First Law of Thermodynamics: Energy Conservation
Definition of the First Law of Thermodynamics
In chemical thermodynamics, the First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. This is also known as the law of energy conservation. For a closed system, this principle implies that any change in the system’s internal energy (U) is due to the energy exchanged with the surroundings in the form of heat (q) or work (W).
Mathematically, the First Law of Thermodynamics is expressed as:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U\) is the change in internal energy of the system,
- \(q\) is the heat absorbed or released by the system,
- \(W\) is the work done on or by the system.
This equation serves as a foundation in chemical thermodynamics for understanding energy flow in reactions, phase changes, and other processes.
Understanding the Terms in the First Law
1. Internal Energy (\(U\)):
- Internal energy represents the total energy within a system, including the kinetic and potential energy of molecules. Changes in internal energy (\(\Delta U\)) depend on the heat added to or removed from the system and the work done by or on the system.
2. Heat (\(q\)):
- Heat is the energy transferred due to a temperature difference between the system and surroundings. In chemical processes:
- If \(q > 0\), the system absorbs heat (endothermic process).
- If \(q < 0\), the system releases heat (exothermic process).
3. Work (\(W\)):
- Work is the energy transferred when a force acts over a distance. In thermodynamics, work is often associated with pressure-volume (PV) work in gases, calculated as:
\[
W = - P \Delta V
\]
- A negative sign indicates work done by the system (expansion), while positive indicates work done on the system (compression).
Application of the First Law in Chemical Thermodynamics
1. Constant Volume Processes (Isochoric):
- In a constant-volume process, \(\Delta V = 0\), so no work is done (\(W = 0\)), and the First Law simplifies to:
\[
\Delta U = q_V
\]
- Here, any heat added to the system directly changes its internal energy. This setup is common in bomb calorimetry for measuring the heat of combustion.
2. Constant Pressure Processes (Isobaric):
- In processes at constant pressure, the heat added to the system equals the enthalpy change (\(\Delta H\)):
\[
q_P = \Delta H
\]
- Most reactions occurring in open containers (like beakers or test tubes) are constant pressure processes, where the enthalpy change approximates the heat change.
3. Adiabatic Processes:
- In an adiabatic process, there is no heat exchange (\(q = 0\)), so the First Law becomes:
\[
\Delta U = W
\]
- This relationship is crucial in adiabatic calorimetry and certain thermodynamic cycles, such as in engines, where work done results solely from internal energy changes.
Implications of the First Law for Chemical Reactions
1. Energy Conservation in Reactions:
- During a chemical reaction, the total energy remains constant, although energy can shift between the system and surroundings as heat and work.
- Exothermic Reactions: Release heat to surroundings, resulting in \(\Delta U < 0\).
- Endothermic Reactions: Absorb heat from surroundings, resulting in \(\Delta U > 0\).
2. Calorimetry and Heat Measurements:
- The First Law is fundamental in calorimetry, where the heat of reactions is measured by observing temperature changes in a controlled environment. The heat change, calculated as \(q = m \cdot C \cdot \Delta T\), represents the system's energy exchange.
3. Enthalpy and Internal Energy Relationship:
- At constant pressure, the enthalpy (\(\Delta H\)) can be related to internal energy by:
\[
\Delta H = \Delta U + P \Delta V
\]
- This equation links heat change at constant pressure (enthalpy) with the work term, explaining how energy transformation occurs during chemical reactions.
Applications of the First Law in Thermodynamic Processes
1. Thermodynamic Cycles:
- In thermodynamic cycles like the Carnot cycle and Otto cycle (e.g., in engines), the First Law explains energy transformations as systems perform work or absorb heat. Cyclic processes help convert thermal energy into mechanical work, demonstrating energy conservation in closed cycles.
2. Phase Transitions:
- Phase changes, such as melting, vaporization, and sublimation, involve energy exchanges without changing chemical composition. The First Law applies to these transitions by relating heat absorbed or released to changes in internal energy.
Example:
- For water evaporating at constant pressure, the heat absorbed (equal to \(\Delta H_{\text{vap}}\)) increases the internal energy as molecules enter the gas phase.
3. Biochemical Systems:
- In biochemistry, the First Law governs cellular processes where ATP hydrolysis releases energy to fuel metabolic reactions. Cells convert chemical energy from ATP into work and heat, exemplifying energy conservation.
Sign Convention in the First Law of Thermodynamics
- Heat (\(q\)): Positive when heat is added to the system, negative when released by the system.
- Work (\(W\)): Positive when work is done on the system (compression), negative when work is done by the system (expansion).
Example:
For gas compression at constant temperature, work is done on the gas (\(W > 0\)), increasing internal energy as heat is released to keep temperature constant.
The First Law in Isolated Systems
For an isolated system, where no energy (heat or work) or matter is exchanged with surroundings, \(\Delta U = 0\). The internal energy remains constant, and any process within the system must proceed by redistributing existing energy.
Example:
In an adiabatic container, any chemical reaction occurs without exchanging heat or work with the surroundings, demonstrating pure energy conservation within the isolated system.
Conclusion
The First Law of Thermodynamics establishes energy conservation as a core principle in chemical thermodynamics. It dictates that energy changes in a system occur through heat and work exchanges with surroundings, while the total energy remains constant. This law is foundational for understanding reaction energetics, calorimetry, and thermodynamic cycles. By quantifying energy transformations, the First Law provides insights into chemical processes, phase changes, and energy flow within biological and industrial systems, making it essential in both theoretical and applied thermodynamics.
|
Q 1. According to the first law of thermodynamics, \(\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}\) In special cases the statement can be expressed in different ways Which of the following is not a correct expression?; |
|
(a) At constant temperature \(\mathrm{q}=-\mathrm{W}\); |
|
(b) When no work is done \(\Delta \mathrm{U}=\mathrm{q}\); |
|
(c) In gaseous system \(\Delta \mathrm{U}=\mathrm{q}+\mathrm{P} \Delta \mathrm{V}\); |
|
(d) When work is done by the system : \(\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}\); |
|
Q 2. Assertion : First law of thermodynamics is applicable to an electric fan or a heater Reason : In an electric fan, the electrical energy is converted into mechanical work that moves the blades. In a heater, electrical energy is converted into heat energy.; |
|
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
|
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
|
(c) Assertion is correct, reason is incorrect; |
|
(d) Assertion is incorrect, reason is correct.; |
|
Q 3. According to the first law of thermodynamics which of the following quantities represents change in a state function?; |
|
(a) \(\mathrm{q}_{\mathrm{rev}}\); |
|
(b) \(\mathrm{q}_{\mathrm{rev}}-\mathrm{W}_{\text {rev }}\); |
|
(c) \(\mathrm{q}_{\mathrm{rev}} / \mathrm{W}_{\text {rev }}\); |
|
(d) \(\mathrm{q}_{\text {rev }}+\mathrm{W}_{\text {rev }}\); |
|
Q 4. Which one of the following equations does not correctly represent the first law of thermodynamics for the given process involving an ideal gas? (Assume non-expansion work is zero); |
|
(a) Cyclic process: \(q=-w\); |
|
(b) Adiabatic process: \(\Delta U=-w\); |
|
(c) Isochoric process: \(\Delta U=q\); |
|
(d) Isothermal process: \(q=-w\); |
Equilibrium Constant (\(K_p\)): Relates to Concentrations at Equilibrium
Definition of the Equilibrium Constant
In chemical thermodynamics, the equilibrium constant (\(K\)) quantifies the ratio of product and reactant concentrations at equilibrium for a given reaction. When dealing with gases, this constant is often represented as \(K_p\), where \(p\) denotes partial pressure. \(K_p\) provides insights into the extent of a reaction, allowing chemists to predict the relative amounts of reactants and products when a reaction reaches equilibrium.
For a general reaction:
\[
aA + bB \rightleftharpoons cC + dD
\]
where \(A\), \(B\), \(C\), and \(D\) are gases with respective stoichiometric coefficients \(a\), \(b\), \(c\), and \(d\), the equilibrium constant \(K_p\) is given by:
\[
K_p = \frac{(P_C)^c \, (P_D)^d}{(P_A)^a \, (P_B)^b}
\]
Where:
- \(P_A\), \(P_B\), \(P_C\), and \(P_D\) are the partial pressures of the gases at equilibrium,
- \(K_p\) is dimensionless when the reaction involves equal numbers of gas moles on both sides, otherwise, it may have units depending on the reaction.
Significance of \(K_p\) in Chemical Thermodynamics
1. Predicting Reaction Direction:
- The magnitude of \(K_p\) provides insights into which direction the reaction favors at equilibrium:
- If \(K_p \gg 1\): The equilibrium favors the products (reaction proceeds primarily to the right).
- If \(K_p \ll 1\): The equilibrium favors the reactants (reaction proceeds primarily to the left).
- \(K_p\) allows chemists to understand if a reaction will yield mostly products or remain dominated by reactants at equilibrium.
2. Quantifying Extent of Reaction:
- \(K_p\) represents the point at which the forward and reverse reaction rates are equal. At this point, the concentrations (or partial pressures) of reactants and products remain constant.
- A high \(K_p\) value indicates a reaction that proceeds nearly to completion, while a low \(K_p\) implies limited product formation at equilibrium.
3. Determining Reaction Feasibility:
- Knowing \(K_p\) helps determine if a reaction can proceed to a significant extent under certain conditions, especially important in industrial and biochemical applications where maximizing product yield is desired.
Relationship Between \(K_p\) and \(K_c\)
For reactions involving gases, the equilibrium constant can also be expressed in terms of concentration as \(K_c\). The relationship between \(K_p\) and \(K_c\) is given by:
\[
K_p = K_c \, (RT)^{\Delta n}
\]
Where:
- \(R\) is the gas constant (0.0821 L·atm·K\(^{-1}\)·mol\(^{-1}\)),
- \(T\) is the temperature in Kelvin,
- \(\Delta n\) is the difference in moles of gases between products and reactants (\(\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}\)).
This relationship shows that \(K_p\) and \(K_c\) are equal when \(\Delta n = 0\) (i.e., the moles of gaseous reactants and products are equal).
Factors Affecting the Equilibrium Constant
1. Temperature:
- \(K_p\) is temperature-dependent. Changes in temperature shift the position of equilibrium and thus affect the equilibrium constant.
- Exothermic reactions (\(\Delta H < 0\)): An increase in temperature decreases \(K_p\).
- Endothermic reactions (\(\Delta H > 0\)): An increase in temperature increases \(K_p\).
- This dependency is explained by the van’t Hoff equation:
\[
\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}
\]
2. Pressure and Concentration Changes:
- While \(K_p\) itself does not change with pressure or concentration, changing the pressure or volume of a gaseous reaction system can shift the position of equilibrium (according to Le Chatelier’s Principle), altering the partial pressures of reactants and products until equilibrium is re-established.
3. Catalysts:
- A catalyst does not affect \(K_p\); it only speeds up the rate at which equilibrium is achieved by providing an alternative pathway with lower activation energy for both forward and reverse reactions.
Using \(K_p\) to Calculate Equilibrium Composition
Given \(K_p\), we can calculate the partial pressures of reactants and products at equilibrium. For a reaction where initial pressures are known, the ICE (Initial, Change, Equilibrium) table method is often used:
1. Set up Initial Conditions with initial pressures of reactants and products.
2. Define Changes in terms of variables (like \(x\)) based on stoichiometry.
3. Solve for Equilibrium Pressures by substituting into the \(K_p\) expression and solving for \(x\).
Example:
For the reaction:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
with a given \(K_p\), we can determine equilibrium concentrations of \(N_2\), \(H_2\), and \(NH_3\) at a specific temperature by solving the \(K_p\) expression.
Applications of \(K_p\) in Chemical Thermodynamics
1. Industrial Synthesis (e.g., Haber Process):
- The production of ammonia (NH\(_3\)) in the Haber process uses knowledge of \(K_p\) to optimize reaction conditions for maximum yield, balancing temperature, pressure, and catalysts to achieve the desired equilibrium composition.
2. Predicting Gas Mixtures in Closed Systems:
- In systems like sealed reaction vessels, \(K_p\) helps predict the partial pressures of gases at equilibrium, important in gas-phase reactions and storage of reactive gases.
3. Biochemical Reactions in Closed Systems:
- For reactions involving gases, such as cellular respiration (involving O\(_2\) and CO\(_2\)), \(K_p\) calculations help in modeling the exchange and equilibrium of gases under physiological conditions.
4. Environmental Applications:
- In atmospheric chemistry, \(K_p\) helps predict the concentrations of pollutants and gases in equilibrium, aiding in understanding and managing chemical processes that impact air quality.
Sample Calculation Using \(K_p\)
Consider a reaction:
\[
H_2(g) + I_2(g) \rightleftharpoons 2HI(g)
\]
Given \(K_p = 50.0\) at a certain temperature and initial pressures \(P_{H_2} = P_{I_2} = 0.5\) atm, we can set up an ICE table and solve for the equilibrium partial pressures to determine the composition at equilibrium.
Conclusion
The equilibrium constant \(K_p\) is fundamental in chemical thermodynamics, providing a measure of the balance between reactants and products at equilibrium for gas-phase reactions. By determining the extent to which a reaction proceeds, \(K_p\) helps chemists predict reaction behavior, optimize industrial processes, and understand equilibrium dynamics in environmental and biological systems. Temperature significantly influences \(K_p\), while pressure and volume changes affect the position of equilibrium but not the value of \(K_p\). The equilibrium constant thus plays a critical role in understanding and controlling chemical reactions in various applications.
|
Q 1. What is the equilibrium constant if ATP hydrolysis by water produce standard free energy of \(-50 \mathrm{~kJ} /\) mole under normal body conditions?; |
|
(a) \(2.66 \times 10^{8}\); |
|
(b) \(5.81 \times 10^{8}\); |
|
(c) \(1.18 \times 10^{7}\); |
|
(d) \(1.98 \times 10^{8}\); |
|
Q 2. Standard entropies of \(\mathrm{X}_{2}, Y_{2}\) and \(\mathrm{XY}_{3}\) are 60,40 and \(50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively For the reaction \(\frac{1}{2} X_{2}+\frac{3}{2} Y_{2} \rightleftharpoons X Y_{3}, \Delta H=-30 \mathrm{~kJ}\) to be at equilibrium, the temperature should be: [2010]; |
|
(a) \(750 \mathrm{~K}\); |
|
(b) \(1000 \mathrm{~K}\); |
|
(c) \(1250 \mathrm{~K}\); |
|
(d) \(500 \mathrm{~K}\); |
|
Q 3. Consider the reaction equilibrium, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \leftrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \Delta H^{\circ}=-198 \mathrm{~kJ}\). On the basis of Le Chatelier's principle, the condition favorable for forward reaction is; |
|
(a) lowering of temperature as well as pressure; |
|
(b) increasing temperature as well as pressure.; |
|
(c) lowering the temperature and increasing the pressure.; |
|
(d) any value of temperature and pressure.; |
|
Q 4. The exothermic formation of \(\mathrm{ClF}_{3}\) is represented by the equation \[ \mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g}) ; \Delta_{\mathrm{r}} H=-329 \mathrm{~kJ} \] Which of the following will increase the quantity of \(\mathrm{ClF}_{3}\) in an equilibrium mixture of \(\mathrm{Cl}_{2}, \mathrm{~F}_{2}\) and \(\mathrm{ClF}_{3}\) ?; |
|
(a) Increasing the temperature.; |
|
(b) Removing \(\mathrm{Cl}_{2}\).; |
|
(c) Increasing the volume of the container.; |
|
(d) Adding \(\mathrm{F}_{2}\); |
|
Q 5. For a reversible reaction: \(\mathrm{X}(\mathrm{g})+3 \mathrm{Y}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g})\) \(\Delta H=-40 \mathrm{~kJ}\) the standard entropies of \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) are 60,40 and \(50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively The temperature at which the above reaction attains equilibrium?; |
|
(a) \(400 \mathrm{~K}\); |
|
(b) \(500 \mathrm{~K}\); |
|
(c) \(273 \mathrm{~K}\); |
|
(d) \(373 \mathrm{~K}\); |
|
Q 6. The correct relationship between free energy change in a reaction and the corresponding equilibrium constant \(K_{C}\) is; |
|
(a) \(\Delta G^{\circ}=-R T \ln K_{C}\); |
|
(b) \(\Delta G^{\circ}-=R T \ln K_{C}\); |
|
(c) \(\Delta G^{\circ}=R T \ln K_{C}\); |
|
(d) \(-\Delta G^{\circ}=R T \ln K_{C}\); |
|
Q 7. For the reaction, \(\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g}), \Delta H^{\circ}\) and \(\Delta S^{\circ}\) are, respectively, \(-29 8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(-0 100 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) at \(298 \mathrm{~K}\) The equilibrium constant for the reaction at \(298 \mathrm{~K}\) is; |
|
(a) \(1.0 \times 10^{-10}\); |
|
(b) 10; |
|
(c) 1; |
|
(d) \(1.0 \times 10^{10}\); |
|
Q 8. Which of the following lines correctly show the temperature dependence of equilibrium constant, \(K\), for an exothermic reaction?; |
|
(a) B and C; |
|
(b) C and D; |
|
(c) A and D; |
|
(d) A and B; |
Bond Dissociation Enthalpy: Energy to Break Bonds
Definition of Bond Dissociation Enthalpy
In chemical thermodynamics, bond dissociation enthalpy (also known as bond enthalpy or bond energy) is the amount of energy required to break a specific chemical bond in one mole of gaseous molecules, resulting in the formation of separated atoms in the gaseous state. It represents the strength of a chemical bond, and higher bond dissociation enthalpies indicate stronger bonds that require more energy to break.
For a bond A–B in a gaseous molecule:
\[
A-B(g) \rightarrow A(g) + B(g)
\]
The bond dissociation enthalpy, \(\Delta H\), is the enthalpy change required to break one mole of the bond in the gaseous state.
Importance of Bond Dissociation Enthalpy in Chemical Thermodynamics
1. Understanding Bond Strength:
- Bond dissociation enthalpy gives insight into bond strength. Stronger bonds require more energy to break, while weaker bonds have lower dissociation enthalpies.
2. Predicting Reaction Energetics:
- The bond dissociation enthalpies of reactants and products help calculate the enthalpy change (\(\Delta H\)) for reactions. By comparing the energy required to break bonds in reactants to the energy released when forming bonds in products, we can predict whether a reaction is endothermic or exothermic.
3. Chemical Stability:
- Molecules with high bond dissociation enthalpies are generally more chemically stable, as they require more energy input to break their bonds.
Types of Bond Dissociation Enthalpies
1. Homolytic Bond Dissociation Enthalpy:
- In homolytic bond dissociation, each atom in the bond retains one electron, resulting in the formation of two radicals. Homolytic bond dissociation enthalpies are typically used to describe nonpolar bonds.
- Example:
\[
Cl_2(g) \rightarrow 2Cl(g) \quad \Delta H = 242 \, \text{kJ/mol}
\]
2. Heterolytic Bond Dissociation Enthalpy:
- In heterolytic bond dissociation, one atom retains both electrons, forming ions instead of radicals. Heterolytic bond dissociation enthalpies are often higher than homolytic ones because forming ions usually requires more energy.
- Example:
\[
H-Cl(g) \rightarrow H^+(g) + Cl^-(g)
\]
Average Bond Enthalpy
For polyatomic molecules with the same type of bond occurring multiple times (like C–H bonds in methane), the bond enthalpy is averaged over all identical bonds, as breaking each bond in sequence requires slightly different amounts of energy.
For example, in methane (CH\(_4\)):
\[
CH_4(g) \rightarrow C(g) + 4H(g)
\]
The average C–H bond enthalpy is calculated by dividing the total enthalpy required by the number of bonds:
\[
\text{Average C-H bond enthalpy} = \frac{\text{Total energy to break 4 C-H bonds}}{4}
\]
Calculating Reaction Enthalpy Using Bond Dissociation Enthalpies
The enthalpy change of a reaction (\(\Delta H_{\text{rxn}}\)) can be estimated using bond dissociation enthalpies:
\[
\Delta H_{\text{rxn}} = \sum \Delta H_{\text{bonds broken}} - \sum \Delta H_{\text{bonds formed}}
\]
Where:
- \(\sum \Delta H_{\text{bonds broken}}\) is the total energy required to break all bonds in the reactants,
- \(\sum \Delta H_{\text{bonds formed}}\) is the total energy released when bonds in the products are formed.
Example:
For the combustion of methane:
\[
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O
\]
Using known bond enthalpies, we can calculate the overall reaction enthalpy to determine if the reaction is exothermic or endothermic.
Factors Affecting Bond Dissociation Enthalpy
1. Bond Order:
- Higher bond order (e.g., double or triple bonds) generally increases bond dissociation enthalpy. For example, a C≡C bond has a higher bond enthalpy than a C=C bond.
2. Atomic Size:
- Larger atoms have weaker bonds due to greater distance between nuclei, resulting in lower bond dissociation enthalpies.
3. Electronegativity:
- Differences in electronegativity affect bond polarity, often increasing bond strength for highly polar bonds, such as the O–H bond in water, which has a high bond enthalpy due to the strength of the polar bond.
4. Molecular Environment:
- The presence of other atoms or functional groups can influence bond strength through electronic effects like inductive or resonance effects. For example, C–H bonds in methane have slightly different bond enthalpies than C–H bonds in more complex organic molecules.
Applications of Bond Dissociation Enthalpy in Chemical Thermodynamics
1. Predicting Reaction Mechanisms:
- Bond enthalpies help predict which bonds in a molecule will break or form, aiding in the understanding of reaction mechanisms, especially for free-radical reactions.
2. Energy Requirements in Industrial Processes:
- Knowledge of bond enthalpies is crucial in designing industrial reactions, particularly those involving high-energy bonds, like the N≡N bond in nitrogen gas, which requires significant energy to break.
3. Understanding Biochemical Reactions:
- In biochemistry, bond dissociation enthalpies are important for understanding the stability and energy requirements of molecular bonds in biomolecules, such as the O–H bond in ATP, which is broken during energy release in cells.
4. Thermodynamic Calculations:
- Bond dissociation enthalpies are essential for estimating the enthalpy changes in reactions, especially for reactions involving complex organic compounds where direct calorimetry is challenging.
Examples of Bond Dissociation Enthalpies
1. Hydrogen Molecule:
- The bond dissociation enthalpy of \(H_2\) is high due to the strong H–H bond:
\[
H_2(g) \rightarrow 2H(g) \quad \Delta H = 436 \, \text{kJ/mol}
\]
2. Oxygen Molecule:
- Breaking the O=O double bond requires less energy than a triple bond but more than a single bond:
\[
O_2(g) \rightarrow 2O(g) \quad \Delta H = 498 \, \text{kJ/mol}
\]
3. Carbon-Hydrogen Bond:
- The C–H bond in methane has a bond dissociation enthalpy around 414 kJ/mol, reflecting its relative strength in hydrocarbons.
Conclusion
Bond dissociation enthalpy is a measure of the energy required to break chemical bonds in a molecule, providing insight into bond strength and stability. In chemical thermodynamics, bond enthalpies are essential for estimating reaction enthalpies, predicting reaction mechanisms, and understanding molecular stability. By comparing bond enthalpies of reactants and products, chemists can determine the energy changes in a reaction, helping to design efficient processes and analyze reaction energetics in industrial and biochemical applications.
|
Q 1. The bond dissociation energies of \(X_{2}, Y_{2}\) and \(X Y\) are in the ratio of \(1: 0 5: 1 \Delta H\) for the formation of \(X Y\) is \(-200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The bond dissociation energy of \(X_{2}\) will be [2018]; |
|
(a) \(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\); |
|
(b) \(100 \mathrm{~kJ} \mathrm{~mol}^{-1}\); |
|
(c) \(400 \mathrm{~kJ} \mathrm{~mol}^{-1}\); |
|
(d) \(800 \mathrm{~kJ} \mathrm{~mol}^{-1}\); |
Thermodynamic Concepts: General principles for energy and heat transfer. |
Work Done (W): Energy transfer involving mechanical actions. |
Entropy (S): Measure of disorder or randomness. |
Specific Heat Capacity (C): Heat required to raise the temperature of 1 mole. |
Adiabatic Process: No heat exchange (q = 0). |
Internal Energy (U): Total energy of the system. |
Gibbs Free Energy (ΔG): Determines spontaneity of processes. |
Second Law of Thermodynamics: Spontaneous processes increase entropy. |
Enthalpy (ΔH): Heat changes at constant pressure. |
Closed/Isolated System: No exchange of matter (closed) or energy (isolated). |
First Law of Thermodynamics: Energy conservation. |
Equilibrium Constant (Kp): Relates to the concentrations at equilibrium. |
Bond Dissociation Enthalpy: Energy to break bonds. |