Chapter 4: Chemical Thermodynamics
Comprehensive Theory, Derivations, Numericals & Fully Solved Board PYQs
1. Basic Terms in Thermodynamics
Thermodynamics is the branch of science that deals with the quantitative relationship between heat and other forms of energy.
System and Surroundings
- System: A specific part of the universe under thermodynamic investigation.
- Surroundings: The rest of the universe outside the system. (Universe = System + Surroundings).
Types of Systems
| Type of System | Exchange of Matter | Exchange of Energy | Example |
|---|---|---|---|
| Open System | Yes | Yes | Hot tea in an open cup. |
| Closed System | No | Yes | Hot tea in a closed metallic flask. |
| Isolated System | No | No | Hot tea in a perfectly insulated thermos flask. |
Extensive Properties
A property whose value depends on the quantity (mass) of matter present in the system.
Examples: Mass, Volume, Internal Energy, Heat Capacity, Enthalpy, Entropy, Gibbs Energy.
Intensive Properties
A property whose value is independent of the quantity (mass) of matter present in the system.
Examples: Temperature, Pressure, Density, Refractive Index, Melting/Boiling Point.
State Functions and Path Functions
State Function: A thermodynamic property whose value depends only on the current state of the system and is independent of the path followed to reach that state (e.g., $P, V, T, U, H, S, G$).
Path Function: A property whose value depends on the path followed during a process (e.g., Work ($W$) and Heat ($q$)).
2. Thermodynamic Processes
A transition from one equilibrium state to another is called a process. Important processes include:
- Isothermal Process: Temperature of the system remains constant throughout ($\Delta T = 0, \Delta U = 0$). Heat is exchanged with surroundings.
- Isobaric Process: Pressure of the system remains constant ($\Delta P = 0$). Usually occurs in open containers.
- Isochoric Process: Volume of the system remains constant ($\Delta V = 0$). No expansion work is done ($W = 0$).
- Adiabatic Process: No heat is exchanged between system and surroundings ($q = 0$). The system is perfectly insulated. Temperature changes.
- Reversible Process: A process conducted infinitely slowly such that the driving force is infinitesimally greater than the opposing force. It can be reversed at any point. It yields maximum work.
3. Nature of Work and Maximum Work
Pressure-Volume (PV) Work
Work done by a system during expansion against a constant external pressure ($P_{ex}$) is given by:
$$ W = -P_{ex} \cdot \Delta V $$Where $\Delta V = V_2 - V_1$.
- If gas expands ($V_2 > V_1$), $\Delta V$ is positive, so $W$ is negative (Work done by the system).
- If gas is compressed ($V_2 < V_1$), $\Delta V$ is negative, so $W$ is positive (Work done on the system).
- Free Expansion: Expansion of a gas in a vacuum ($P_{ex} = 0$). Here, $W = 0$.
Expression for Maximum Work ($W_{max}$)
Maximum work is obtained in an isothermal and reversible expansion of an ideal gas. Consider $n$ moles of an ideal gas expanding isothermally and reversibly from volume $V_1$ to $V_2$ at constant temperature $T$.
Small work done $dW = -P_{ex} dV$. In a reversible process, $P_{ex} = P - dP$.
$$ dW = -(P - dP)dV = -PdV + dPdV $$
Since $dPdV$ is negligibly small, $dW = -PdV$.
Total work $W_{max} = \int_{V_1}^{V_2} -P dV$. Using Ideal Gas Equation $P = \frac{nRT}{V}$:
$$ W_{max} = -\int_{V_1}^{V_2} \frac{nRT}{V} dV = -nRT \ln\left(\frac{V_2}{V_1}\right) $$
Converting to base 10 logarithm:
$$ W_{max} = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right) $$
Since at constant $T$, $P_1V_1 = P_2V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$, we can also write:
$$ W_{max} = -2.303 nRT \log_{10} \left(\frac{P_1}{P_2}\right) $$
4. First Law of Thermodynamics
Statement: Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of the universe is constant.
Mathematical Expression
If $q$ is the heat supplied to a system and $W$ is the work done on the system, the change in internal energy ($\Delta U$) is:
$$ \Delta U = q + W $$First Law for Different Processes
| Process | Condition | Modified Equation ($\Delta U = q + W$) | Conclusion |
|---|---|---|---|
| Isothermal | $\Delta U = 0$ | $0 = q + W \implies q = -W$ | Heat absorbed equals work done by the system. |
| Adiabatic | $q = 0$ | $\Delta U = W$ | Increase in internal energy equals work done on the system. |
| Isochoric | $\Delta V = 0 \implies W = 0$ | $\Delta U = q_v$ | Heat absorbed at constant volume increases internal energy. |
| Isobaric | $P = \text{constant}$ | $\Delta U = q_p - P\Delta V$ | $q_p = \Delta U + P\Delta V$ (which defines Enthalpy change). |
5. Enthalpy ($H$)
Enthalpy is a state function defined as the sum of the internal energy of the system and its pressure-volume energy.
$$ H = U + PV $$The change in enthalpy at constant pressure is given by: $$ \Delta H = \Delta U + P\Delta V $$
Since $\Delta U = q_p - P\Delta V$ (from the First Law), we get: $\Delta H = q_p$.
Conclusion: The heat absorbed or evolved at constant pressure is exactly equal to the change in enthalpy of the system.
Relationship between $\Delta H$ and $\Delta U$ for Chemical Reactions
For reactions involving gases, let $n_1$ be moles of gaseous reactants and $n_2$ be moles of gaseous products.
From ideal gas law: $PV_1 = n_1RT$ and $PV_2 = n_2RT$.
$P\Delta V = P(V_2 - V_1) = n_2RT - n_1RT = (n_2 - n_1)RT = \Delta n_g RT$
Substituting in $\Delta H = \Delta U + P\Delta V$:
$$ \Delta H = \Delta U + \Delta n_g RT $$
Where $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.
6. Thermochemistry and Hess's Law
Thermochemistry deals with the enthalpy changes accompanying chemical reactions.
- Standard Enthalpy of Reaction ($\Delta_r H^\circ$): The enthalpy change accompanying a reaction when all reactants and products are in their standard states (1 bar pressure, 298 K).
- Standard Enthalpy of Formation ($\Delta_f H^\circ$): The enthalpy change when one mole of a pure compound is formed from its constituent elements in their standard states. (Note: $\Delta_f H^\circ$ of pure elements like $O_2(g), C(graphite), H_2(g)$ is zero).
- Standard Enthalpy of Combustion ($\Delta_c H^\circ$): The enthalpy change when one mole of a substance is completely burnt in excess of oxygen in standard state. (Always negative, exothermic).
Calculation of $\Delta_r H^\circ$ from Formation Enthalpies:
$$ \Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) $$Hess's Law of Constant Heat Summation
Statement: The overall enthalpy change for a chemical reaction is the same, regardless of whether the reaction takes place in one single step or in a series of steps.
Application: It is used to calculate the enthalpies of reactions which cannot be determined experimentally (e.g., enthalpy of formation of methane, extremely slow reactions).
7. Entropy ($S$) and Spontaneity (Second Law)
Spontaneous Process: A process that occurs on its own without the intervention of any external agency. They are irreversible and tend towards an equilibrium state.
Entropy ($S$): A thermodynamic state function that measures the degree of randomness or disorder in a system. The greater the disorder, the higher the entropy.
Order: $S_{solid} < S_{liquid} < S_{gas}$.
The change in entropy ($\Delta S$) is given by:
$$ \Delta S = \frac{q_{rev}}{T} $$Where $q_{rev}$ is the heat absorbed reversibly at constant temperature $T$. Unit: $\text{J K}^{-1} \text{mol}^{-1}$.
Second Law of Thermodynamics
Statement: The total entropy of the universe (system + surroundings) increases continuously in a spontaneous process.
$$ \Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} $$- If $\Delta S_{total} > 0$: Process is spontaneous.
- If $\Delta S_{total} < 0$: Process is non-spontaneous.
- If $\Delta S_{total} = 0$: Process is at equilibrium.
8. Gibbs Free Energy ($G$)
To predict spontaneity without analyzing the surroundings, J.W. Gibbs introduced Free Energy. It is defined as the maximum amount of energy available from a system that can be converted into useful work.
$$ G = H - TS $$The change in Gibbs energy at constant temperature and pressure is the Gibbs-Helmholtz equation:
$$ \Delta G = \Delta H - T\Delta S $$Criterion for Spontaneity based on $\Delta G$
- If $\Delta G < 0$ (Negative): Process is spontaneous.
- If $\Delta G > 0$ (Positive): Process is non-spontaneous.
- If $\Delta G = 0$: Process is at equilibrium.
Relationship between Standard Gibbs Energy and Equilibrium Constant ($K$)
$$ \Delta G^\circ = -RT \ln K $$
$$ \Delta G^\circ = -2.303 RT \log_{10} K $$
9. Solved Textbook Numericals
Problem 1: Maximum Work ($W_{max}$)
2 moles of an ideal gas are expanded isothermally and reversibly from 10 L to 20 L at 300 K. Calculate the maximum work done. ($R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$)
Solution:
Given: $n = 2 \text{ mol}, T = 300 \text{ K}, V_1 = 10 \text{ L}, V_2 = 20 \text{ L}$
Formula: $W_{max} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$
$W_{max} = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}\left(\frac{20}{10}\right)$
$W_{max} = -2.303 \times 4988.4 \times \log_{10}(2)$
$W_{max} = -11488.28 \times 0.3010$
$W_{max} = -3458 \text{ J} = -3.458 \text{ kJ}$
Answer: The maximum work done is $-3.458 \text{ kJ}$. (The negative sign indicates work is done by the system).
Problem 2: Enthalpy vs Internal Energy
Calculate $\Delta U$ at 298 K for the reaction:
$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)$
Given $\Delta H = -1410 \text{ kJ}$. ($R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$)
Solution:
Relation: $\Delta H = \Delta U + \Delta n_g RT \implies \Delta U = \Delta H - \Delta n_g RT$
Calculate $\Delta n_g$: (moles of gaseous products) - (moles of gaseous reactants)
Gas products = $2$ (from $2CO_2$). Note: $H_2O$ is liquid, so ignore.
Gas reactants = $1 + 3 = 4$ (from $C_2H_4$ and $3O_2$)
$\Delta n_g = 2 - 4 = -2 \text{ mol}$
Convert $R$ to kJ: $R = 8.314 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1}$
$\Delta U = -1410 - (-2 \times 8.314 \times 10^{-3} \times 298)$
$\Delta U = -1410 - (-4.955)$
$\Delta U = -1410 + 4.955 = -1405.045 \text{ kJ}$
Answer: The change in internal energy ($\Delta U$) is $-1405.045 \text{ kJ}$.
Problem 3: Gibbs Free Energy
For a certain reaction, $\Delta H = -224 \text{ kJ}$ and $\Delta S = -153 \text{ J K}^{-1}$. Calculate $\Delta G$ at 298 K and predict whether the reaction is spontaneous.
Solution:
Given: $\Delta H = -224 \text{ kJ} = -224,000 \text{ J}$, $\Delta S = -153 \text{ J K}^{-1}$, $T = 298 \text{ K}$
Formula: $\Delta G = \Delta H - T\Delta S$
$\Delta G = -224,000 - (298 \times -153)$
$\Delta G = -224,000 - (-45,594)$
$\Delta G = -224,000 + 45,594 = -178,406 \text{ J} = -178.4 \text{ kJ}$
Answer: $\Delta G = -178.4 \text{ kJ}$. Since $\Delta G$ is negative, the reaction is spontaneous.
10. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the mathematical expression for the first law of thermodynamics for an isothermal process. (March 2013, Oct 2015)
Answer: For an isothermal process, $\Delta U = 0$. Since $\Delta U = q + W$, the expression becomes $0 = q + W$ or $q = -W$.
Q2. State Hess’s law of constant heat summation. (March 2016, March 2020)
Answer: Hess's law states that the overall enthalpy change for a chemical reaction is constant, regardless of whether the reaction takes place in one single step or in a series of steps.
Q3. What is an intensive property? (Oct 2014, March 2019)
Answer: A thermodynamic property whose value is independent of the quantity or size of matter present in the system is called an intensive property (e.g., temperature, pressure).
2 Mark Questions (SA-I)
Q4. Derive the expression for pressure-volume ($PV$) work. (March 2015, Oct 2018)
Consider a gas enclosed in a cylinder fitted with a frictionless, weightless movable piston of area $A$. Let the gas expand against a constant external pressure $P_{ex}$.
The opposing force on the piston is $F = P_{ex} \times A$.
Let the piston move upward by a small distance $d$. The work done by the system to surroundings is $W = -F \times d$. (Negative sign because work is done by the system against surroundings).
Substitute $F$: $W = -(P_{ex} \times A) \times d$.
Since Area $\times$ distance ($A \times d$) is the change in volume ($\Delta V$),
Therefore, $W = -P_{ex} \cdot \Delta V$.
Q5. Distinguish between Reversible and Irreversible processes. (March 2017, March 2022)
- Reversible Process: Driving force is infinitesimally greater than opposing force. It takes infinite time for completion. It can be reversed at any point. Work done is maximum.
- Irreversible Process: Driving force is significantly greater than opposing force. Takes finite time for completion. Cannot be reversed easily. Work done is not maximum. (All natural processes are irreversible).
3 Mark Questions (SA-II)
Q6. Derive the expression for maximum work ($W_{max}$) done by the system in an isothermal reversible process. (Oct 2013, March 2018, March 2021)
1. Consider $n$ moles of an ideal gas enclosed in a cylinder with a weightless piston. Let it expand isothermally and reversibly from volume $V_1$ to $V_2$ at temp $T$.
2. In a reversible process, external pressure $P_{ex}$ is infinitesimally smaller than gas pressure $P$. So, $P_{ex} = P - dP$.
3. Small work done $dW = -P_{ex} dV = -(P - dP)dV = -PdV + dPdV$.
Since $dP$ and $dV$ are both very small, their product $dPdV$ is negligible. Thus, $dW = -PdV$.
4. Total work $W_{max}$ is obtained by integrating from $V_1$ to $V_2$:
$W_{max} = \int_{V_1}^{V_2} -P dV$
5. From ideal gas equation, $P = \frac{nRT}{V}$. Substitute this in integral:
$W_{max} = -\int_{V_1}^{V_2} \frac{nRT}{V} dV = -nRT \int_{V_1}^{V_2} \frac{dV}{V} = -nRT \ln \left[\frac{V_2}{V_1}\right]$
6. Converting natural log to base 10:
$$ W_{max} = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right) $$
Q7. Explain the relationship between Gibbs free energy change ($\Delta G$) and spontaneity of a chemical reaction. Define standard enthalpy of formation. (March 2014, Oct 2019)
$\Delta G$ and Spontaneity: Gibbs free energy ($G = H - TS$) is the criteria to determine spontaneity without considering the surroundings. At constant temperature and pressure, $\Delta G = \Delta H - T\Delta S$.
- If $\Delta G$ is negative ($\Delta G < 0$), the process is spontaneous in the forward direction.
- If $\Delta G$ is positive ($\Delta G > 0$), the process is non-spontaneous in the forward direction (but spontaneous in reverse).
- If $\Delta G$ is zero ($\Delta G = 0$), the system is at chemical equilibrium.
Standard Enthalpy of Formation ($\Delta_f H^\circ$): It is defined as the enthalpy change accompanying the formation of exactly one mole of a pure compound from its constituent elements, with all substances being in their standard states (1 bar pressure and specified temperature, usually 298K).
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