Thermodynamics -- Practice Problems

Thermodynamics — Practice Problems

10 MCQ questions covering the laws of thermodynamics, entropy, Gibbs free energy, chemical equilibrium, phase diagrams, and electrochemistry. Select an option to check your answer.

Intuition

Thermodynamics is nature’s bookkeeping for energy: The laws of thermodynamics are not abstract rules — they describe what is physically possible. Energy cannot be created (first law), and processes always increase total disorder (second law). These constraints shape everything from engine design to biological processes.

Why it matters: Thermodynamics determines whether fuels will burn, whether drugs are stable, and whether power plants can generate electricity efficiently.

The key insight: A negative $\Delta G$ means a reaction is spontaneous, but spontaneous does not mean fast — thermodynamics tells you the destination, kinetics tells you the speed.

Worked Examples

Example 1: Calculating Work for Isothermal Expansion

Problem: 2.00 mol of an ideal gas expands reversibly from 5.00 L to 15.0 L at 300 K. Calculate the work done.

Solution:

For a reversible isothermal expansion of an ideal gas:

$w = -nRT \ln\frac{V_f}{V_i}$

Substituting values:

$w = -(2.00 \text{ mol})(8.314 \text{ J mol}^{-1}\text{K}^{-1})(300 \text{ K}) \ln\frac{15.0}{5.00}$

$w = -(2.00)(8.314)(300) \ln(3.00)$

$w = -(4988.4)(1.0986)$

$w = -5481 \text{ J} = -5.48 \text{ kJ}$

Physical interpretation: The negative sign indicates work is done by the gas on the surroundings. The gas pushes against the external pressure as it expands. Since the process is isothermal and the gas is ideal, $\Delta U = 0$, so $q = -w = +5.48$ kJ (heat absorbed from surroundings).

Example 2: Entropy Change for Irreversible Process

Problem: 1.00 mol of an ideal gas at 300 K is expanded irreversibly against a constant external pressure of 1.00 bar from 10.0 L to 20.0 L. Calculate $\Delta S_{\text{system}}$, $\Delta S_{\text{surroundings}}$, and $\Delta S_{\text{total}}$.

Solution:

Step 1: Calculate $\Delta S_{\text{system}}$

Entropy is a state function. We design a reversible path between the same states:

$\Delta S_{\text{system}} = nR \ln\frac{V_f}{V_i} = (1.00)(8.314) \ln\frac{20.0}{10.0} = 8.314 \times 0.6931 = +5.76 \text{ J K}^{-1}$

Step 2: Calculate $q_{\text{irrev}}$

For expansion against constant external pressure:

$w = -P_{\text{ext}} \Delta V = -(1.00 \times 10^5 \text{ Pa})(20.0 - 10.0) \times 10^{-3} \text{ m}^3 = -1000 \text{ J}$

Since $\Delta U = 0$ (isothermal, ideal gas): $q = -w = +1000$ J

Step 3: Calculate $\Delta S_{\text{surroundings}}$

$\Delta S_{\text{surr}} = \frac{-q_{\text{irrev}}}{T_{\text{surr}}} = \frac{-1000}{300} = -3.33 \text{ J K}^{-1}$

Step 4: Calculate $\Delta S_{\text{total}}$

$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 5.76 - 3.33 = +2.43 \text{ J K}^{-1}$

Key insight: $\Delta S_{\text{total}} > 0$ confirms the process is irreversible (spontaneous). The entropy generated ($+2.43$ J K$^{-1}$) represents the “cost” of irreversibility.

Example 3: Gibbs Free Energy and Spontaneity

Problem: At 298 K, the standard Gibbs free energy of formation are: $\Delta_f G^\circ(\text{H}_2\text{O}, l) = -237.1$ kJ/mol and $\Delta_f G^\circ(\text{H}_2\text{O}_2, l) = -120.4$ kJ/mol. For the reaction $\text{H}_2\text{O}_2(l) \rightarrow \text{H}_2\text{O}(l) + \frac{1}{2}\text{O}_2(g)$, calculate $\Delta_r G^\circ$ and determine spontaneity.

Solution:

$\Delta_r G^\circ = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})$

$\Delta_r G^\circ = \left[(-237.1) + \frac{1}{2}(0)\right] - \left[(-120.4)\right]$

$\Delta_r G^\circ = -237.1 + 120.4 = -116.7 \text{ kJ/mol}$

Since $\Delta_r G^\circ < 0$, the decomposition of hydrogen peroxide is spontaneous under standard conditions.

Physical interpretation: The large negative $\Delta_r G^\circ$ explains why $\text{H}_2\text{O}_2$ is unstable and decomposes readily. The driving force is the formation of very stable water molecules.

Example 4: Van’t Hoff Equation Application

Problem: For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $K_p = 0.148$ at 298 K and $\Delta H^\circ = +57.2$ kJ/mol. Calculate $K_p$ at 350 K.

Solution:

Using the integrated van’t Hoff equation:

$\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

$\ln\frac{K_2}{0.148} = -\frac{57200}{8.314}\left(\frac{1}{350} - \frac{1}{298}\right)$

$\ln\frac{K_2}{0.148} = -(6880.5)(0.002857 - 0.003356)$

$\ln\frac{K_2}{0.148} = -(6880.5)(-0.000499) = +3.433$

$\frac{K_2}{0.148} = e^{3.433} = 30.95$

$K_2 = 0.148 \times 30.95 = 4.58$

Key insight: For this endothermic reaction ($\Delta H^\circ > 0$), increasing temperature increases $K_p$ (from 0.148 to 4.58). This is consistent with Le Chatelier’s principle: adding heat favors the endothermic direction.

Laws of Thermodynamics and Entropy

Q1. For an ideal gas undergoing reversible isothermal expansion from volume $V_i$ to $V_f$, the work done by the gas is:

A. $w = -nRT\\ln(V_f/V_i)$ B. $w = -P\\Delta V$ C. $w = -nRT\\Delta T$ D. $w = -nC_V\\Delta T$’,

Answer: A

Answer: A — For a reversible expansion of an ideal gas: $w = -\\int_\{V_i\}^\{V_f\} P\\,dV = -\\int_\{V_i\}^\{V_f\} \\frac\{nRT\}\{V\}dV = -nRT\\ln\\frac\{V_f\}\{V_i\}$. The negative sign indicates work done by the system. For $V_f > V_i$, $w < 0$ (work is done by the gas). Since the process is isothermal and the gas is ideal, $\\Delta U = 0$, so $q = -w = nRT\\ln(V_f/V_i)$.

Q2. The entropy change for a reversible process is defined as $\\Delta S = q_\{\\text\{rev\}\}/T$. For an irreversible process, how is $\\Delta S$ calculated?

A. Design a reversible path between the same initial and final states and calculate $\\Delta S = \\int dq_\{\\text\{rev\}\}/T$ along that path B. $\\Delta S = q_\{\\text\{irrev\}\}/T$ C. $\\Delta S = 0$ for all irreversible processes D. $\\Delta S$ cannot be calculated for irreversible processes’,

Answer: A

Answer: A — Entropy is a state function, so $\\Delta S$ depends only on initial and final states, not on the path. For an irreversible process, we design a reversible path connecting the same states and compute $\\Delta S = \\int dq_\{\\text\{rev\}\}/T$. We cannot use $q_\{\\text\{irrev\}\}/T$ because the inequality $dS > dq_\{\\text\{irrev\}\}/T$ holds.

Q3. For the reaction $\\text\{H\}_2(g) + \\text\{I\}_2(g) \\rightleftharpoons 2\\text\{HI\}(g)$ at equilibrium with $K_p = 50.0$, if the initial partial pressures of $\\text\{H\}_2$ and $\\text\{I\}_2$ are both 0.10$ bar and no HI is present, what is the equilibrium partial pressure of HI?

A. 0.155 bar B. 0.100 bar C. 0.200 bar D. 0.050 bar

Answer: A

Answer: A — $K_p = \\frac\{(p_\{\\text\{HI\}\})^2\}\{p_\{\\text\{H\}*2\} \\cdot p*\{\\text\{I\}*2\}\} = 50.0$. Let $x$ = amount of HI formed. Then $p*\{\\text\{H\}*2\} = p*\{\\text\{I\}*2\} = 0.10 - x/2$ and $p*\{\\text\{HI\}\} = x$. So $\\frac\{x^2\}\{(0.10-x/2)^2\} = 50.0$, giving $\\frac\{x\}\{0.10-x/2\} = \\sqrt\{50\} \\approx 7.07$, so $x = 0.707(0.10-x/2) = 0.0707 - 0.354x$, thus $1.354x = 0.0707$, $x \\approx 0.0522$ bar. But let me re-check: $x/(0.10 - 0.5x) = 7.07$, $x = 0.707 - 3.535x$, $4.535x = 0.707$, $x = 0.156$ bar.

Q4. The relationship between the equilibrium constant $K$ and the standard Gibbs free energy change is:

A. \\Delta_r G^\\circ = -RT\\ln K B. \\Delta_r G^\\circ = RT\\ln K C. $\\Delta_r G = -RT\\ln K$ D. \\Delta_r G^\\circ = -nRT\\ln K’,

Answer: A

Answer: A — The fundamental relationship is \\Delta_r G^\\circ = -RT\\ln K where $K$ is the dimensionless thermodynamic equilibrium constant. A large $K$ ($K \\gg 1$) means \\Delta_r G^\\circ \\ll 0 (products strongly favored). At equilibrium, $\\Delta_r G = 0$. For non-standard conditions: \\Delta_r G = \\Delta_r G^\\circ + RT\\ln Q where $Q$ is the reaction quotient.

Phase Equilibria and Electrochemistry

Q5. According to the Clausius-Clapeyron equation, the relationship between vapour pressure and temperature for a pure substance is:

A. $\\ln\\frac\{P_2\}\{P_1\} = -\\frac\{\\Delta H_\{\\text\{vap\}\}\}\{R\}\\left(\\frac\{1\}\{T_2\} - \\frac\{1\}\{T_1\}\\right)$ B. $\\ln P = \\Delta H_\{\\text\{vap\}\} \\cdot T + C$ C. $P_2/P_1 = T_2/T_1$ D. $\\ln P = -\\Delta S_\{\\text\{vap\}\}/R$’,

Answer: A

Answer: A — The Clausius-Clapeyron equation (assuming $\\Delta H_\{\\text\{vap\}\}$ is constant) relates vapour pressures at two temperatures: $\\ln(P_2/P_1) = -\\frac\{\\Delta H_\{\\text\{vap\}\}\}\{R\}(1/T_2 - 1/T_1)$. This can be used to estimate $\\Delta H_\{\\text\{vap\}\}$ from vapour pressure data at different temperatures, or to extrapolate vapour pressure to a new temperature.

Q6. In a binary liquid mixture exhibiting positive deviation from Raoult’s law, which of the following is true?

A. The A-B interactions are weaker than the average A-A and B-B interactions; the mixture has a minimum in the boiling point diagram B. The A-B interactions are stronger than average; the mixture has a maximum boiling point’, “Raoult’s law is obeyed exactly at all compositions”, ‘The mixture cannot form an azeotrope’,

Answer: A

Answer: A — Positive deviation: $P_\{\\text\{total\}\} > x_A P_A^* + x_B P_B^*$ (Raoult’s law prediction). This occurs when A-B interactions are weaker than A-A and B-B, making it easier for molecules to escape the liquid. The boiling point diagram shows a minimum (minimum-boiling azeotrope). Example: ethanol-water. Negative deviation (stronger A-B) gives maximum-boiling azeotropes.

Q7. For an electrochemical cell $\\text\{Zn\}\\,|\\,\\text\{Zn\}^\{2+\}\\,\\|\\,\\text\{Cu\}^\{2+\}\\,|\\,\\text\{Cu\}$, the standard cell potential E^\\circ_\{\\text\{cell\}\} is related to \\Delta_r G^\\circ by:

A. \\Delta_r G^\\circ = -nFE^\\circ_\{\\text\{cell\}\} where $n$ is the number of moles of electrons transferred B. \\Delta_r G^\\circ = -nRT E^\\circ_\{\\text\{cell\}\} C. \\Delta_r G^\\circ = nFE^\\circ_\{\\text\{cell\}\} D. \\Delta_r G^\\circ = -RT\\ln E^\\circ_\{\\text\{cell\}\}’,

Answer: A

Answer: A — The relationship between Gibbs free energy and cell potential is \\Delta_r G^\\circ = -nFE^\\circ_\{\\text\{cell\}\} where $n$ is moles of electrons transferred and $F = 96485$ C/mol is Faraday’s constant. A positive E^\\circ_\{\\text\{cell\}\} means \\Delta_r G^\\circ < 0 (spontaneous). The Nernst equation extends this to non-standard conditions: E = E^\\circ - (RT/nF)\\ln Q.

Q8. The Gibbs phase rule states that $F = C - P + 2$. For a one-component system (C=1) at its triple point, how many degrees of freedom are there?

A. $F = 0$ (invariant point); temperature and pressure are fixed B. $F = 1$ (temperature or pressure can vary) C. $F = 2$ (both can vary independently) D. $F = 3$’,

Answer: A

Answer: A — At the triple point, three phases coexist ($P = 3$) for a single component ($C = 1$): $F = 1 - 3 + 2 = 0$. This means the triple point is invariant: both temperature and pressure are uniquely determined (for water: 273.16 K, 611 Pa). Along a phase boundary (two phases), $F = 1$ (univariant). In a single-phase region, $F = 2$.

Q9. For an ideal solution, the chemical potential of component $i$ is \\mu_i = \\mu_i^\\circ + RT\\ln x_i. What is the Gibbs free energy of mixing for an ideal binary solution of $n_A$ moles of A and $n_B$ moles of B?

A. $\\Delta_\{\\text\{mix\}\} G = nRT(x_A \\ln x_A + x_B \\ln x_B)$, which is always negative (spontaneous) B. $\\Delta_\{\\text\{mix\}\} G = 0$ for ideal solutions C. $\\Delta_\{\\text\{mix\}\} G = nRT(x_A + x_B)$ D. $\\Delta_\{\\text\{mix\}\} G$ is positive (non-spontaneous)’,

Answer: A

Answer: A — For an ideal binary solution: $\\Delta_\{\\text\{mix\}\} G = nRT(x_A \\ln x_A + x_B \\ln x_B)$. Since $x_i < 1$, $\\ln x_i < 0$, so $\\Delta_\{\\text\{mix\}\} G < 0$ (spontaneous mixing). The entropy of mixing is $\\Delta_\{\\text\{mix\}\} S = -nR(x_A \\ln x_A + x_B \\ln x_B) > 0$. The enthalpy of mixing is zero ($\\Delta_\{\\text\{mix\}\} H = 0$), which defines an ideal solution.

Q10. The van’t Hoff equation describes how the equilibrium constant changes with temperature: \\frac\{d\\ln K\}\{dT\} = \\frac\{\\Delta H^\\circ\}\{RT^2\}. For an exothermic reaction (\\Delta H^\\circ < 0), increasing temperature will:

A. Decrease $K$, shifting the equilibrium toward reactants B. Increase $K$, shifting the equilibrium toward products C. Have no effect on $K$ D. Invert the sign of \\Delta G^\\circ’,

Answer: A

Answer: A — For \\Delta H^\\circ < 0, $d\\ln K/dT < 0$, so $K$ decreases with increasing temperature. By Le Chatelier’s principle, increasing temperature favors the endothermic direction (which absorbs the added heat). For an exothermic reaction, that means shifting toward reactants. The van’t Hoff equation, integrated: \\ln(K_2/K_1) = -\\frac\{\\Delta H^\\circ\}\{R\}(1/T_2 - 1/T_1).

See Also

• Physical Chemistry
• Thermodynamics
• Chemical Kinetics — Practice Problems

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Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

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Prerequisites

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Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.