Chapter 4: Chemical Thermodynamics Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
- Use of logarithmic tables is allowed. Calculators are not permitted.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
Which of the following is an intensive property?
(A) Volume(B) Enthalpy(C) Temperature(D) Internal energy
-
The mathematical expression of the first law of thermodynamics for an isothermal process is:
(A) $\Delta U = q$(B) $q = -W$(C) $\Delta U = W$(D) $W = 0$
-
A chemical reaction is spontaneous at all temperatures if:
(A) $\Delta H > 0$ and $\Delta S > 0$(B) $\Delta H < 0$ and $\Delta S < 0$(C) $\Delta H > 0$ and $\Delta S < 0$(D) $\Delta H < 0$ and $\Delta S > 0$
-
The work done when a gas expands into a vacuum (free expansion) is:
(A) Infinite(B) Zero(C) Maximum(D) Cannot be determined
Q2. Answer the following questions in one sentence: [3 Marks]
- State Hess's law of constant heat summation.
- Define: Enthalpy of sublimation.
- Write the expression for maximum work ($W_{max}$) done in an isothermal reversible expansion.
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Distinguish between State function and Path function. Give one example of each.
- Distinguish between Isothermal process and Adiabatic process.
- Calculate the work done when 2 moles of an ideal gas expand from a volume of 2 L to 5 L against a constant external pressure of 1.2 bar. ($1 \text{ L bar} = 100 \text{ J}$).
- Define Standard Enthalpy of Formation ($\Delta_f H^\circ$). Give one example.
- What is entropy? Write its SI unit.
SECTION C
Attempt any TWO of the following: [6 Marks]
- Derive the expression for the maximum work ($W_{max}$) done by the system during an isothermal and reversible expansion of an ideal gas.
- Derive the relationship between $\Delta H$ and $\Delta U$ for a chemical reaction involving gases.
- For a certain reaction, $\Delta H = -224 \text{ kJ}$ and $\Delta S = -153 \text{ J K}^{-1}$. Calculate $\Delta G$ at 298 K. Predict whether the reaction is spontaneous or not.
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) Calculate the work done in the oxidation of 4 moles of $SO_2(g)$ at 298 K if:
$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ ($R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$). [2 Marks]
(b) What is an isolated system? Give an example. [2 Marks] - (a) Calculate the standard enthalpy of formation of $CH_4(g)$ if standard enthalpies of combustion of Carbon (graphite), $H_2(g)$, and $CH_4(g)$ are -393.5 kJ/mol, -285.8 kJ/mol, and -890.3 kJ/mol respectively. [3 Marks]
(b) What is the sign convention for heat absorbed by a system? [1 Mark]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (C) Temperature [1 Mark for correct option]
2. (B) $q = -W$ [1 Mark. For isothermal, $\Delta U = 0$, so $0 = q + W$]
3. (D) $\Delta H < 0$ and $\Delta S > 0$ [1 Mark. $\Delta G = \Delta H - T\Delta S$. Both conditions make $\Delta G$ negative]
4. (B) Zero [1 Mark. $P_{ext} = 0$ in vacuum, so $W = -P_{ext}\Delta V = 0$]
Q2. Very Short Answers:
1. Hess's Law:
It 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 multiple steps. [1 Mark for correct statement]
2. Enthalpy of Sublimation:
It is the enthalpy change that accompanies the conversion of exactly one mole of a solid directly into its vapor at constant temperature and pressure. [1 Mark for correct definition]
3. Expression for $W_{max}$:
$W_{max} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$ or $-2.303 nRT \log_{10}\left(\frac{P_1}{P_2}\right)$ [1 Mark for correct formula]
SECTION B [8 Marks]
Q3. State vs Path Function:
State Function: A thermodynamic property whose value depends only on the current state (initial and final states) of the system and is independent of the path followed. Example: Enthalpy ($H$), Entropy ($S$). [1 Mark]
Path Function: A property whose value depends on the path or mechanism followed to go from the initial to the final state. Example: Work ($W$), Heat ($q$). [1 Mark]
Q4. Isothermal vs Adiabatic Process:
| Isothermal Process | Adiabatic Process |
|---|---|
| Temperature of the system remains constant ($\Delta T = 0$). | Temperature of the system changes. |
| Heat is exchanged with surroundings ($q \neq 0$). | No heat is exchanged with surroundings ($q = 0$). |
[1 Mark for each point of distinction = 2 Marks]
Q5. Work done numerical:
Given: $V_1 = 2 \text{ L}$, $V_2 = 5 \text{ L}$, $P_{ext} = 1.2 \text{ bar}$. [1/2 Mark]
Formula: $W = -P_{ext}\Delta V = -P_{ext}(V_2 - V_1)$ [1/2 Mark]
$W = -1.2 \times (5 - 2) = -1.2 \times 3 = -3.6 \text{ L bar}$ [1/2 Mark]
Convert to Joules: $W = -3.6 \times 100 = -360 \text{ J}$. [1/2 Mark for correct answer]
Q6. Standard Enthalpy of Formation ($\Delta_f H^\circ$):
Definition: 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 Mark]
Example: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta_f H^\circ = -285.8 \text{ kJ/mol}$ [1 Mark]
Q7. Entropy and its Unit:
Entropy ($S$): It is a thermodynamic state function that measures the degree of randomness or disorder in a system. The greater the disorder, the higher the entropy. [1 Mark]
SI Unit: Joules per Kelvin per mole ($\text{J K}^{-1}\text{mol}^{-1}$). [1 Mark]
SECTION C [6 Marks]
Q8. Derivation of Maximum Work ($W_{max}$):
1. For a reversible process, opposing pressure $P_{ex} = P - dP$. [1/2 Mark]
2. Work done for a small volume change: $dW = -P_{ex} \cdot dV = -(P - dP) \cdot dV$.
Since $dP \cdot dV$ is negligible, $dW = -P \cdot dV$. [1 Mark]
3. Total work by integrating from $V_1$ to $V_2$: $W_{max} = -\int_{V_1}^{V_2} P \, dV$.
4. Using ideal gas law $P = \frac{nRT}{V}$: $W_{max} = -\int_{V_1}^{V_2} \frac{nRT}{V} dV$. [1/2 Mark]
5. For isothermal process, $T$ is constant: $W_{max} = -nRT [\ln V]_{V_1}^{V_2} = -nRT \ln\left(\frac{V_2}{V_1}\right)$. [1/2 Mark]
6. Converting to base 10: $W_{max} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$. [1/2 Mark]
Q9. Derivation of $\Delta H = \Delta U + \Delta n_g RT$:
1. By definition, $H = U + PV$. For a change at constant pressure, $\Delta H = \Delta U + P\Delta V = \Delta U + P(V_2 - V_1)$. [1 Mark]
2. Applying ideal gas equation for reactants and products: $PV_1 = n_1RT$ and $PV_2 = n_2RT$. [1 Mark]
3. Substituting: $\Delta H = \Delta U + n_2RT - n_1RT = \Delta U + (n_2 - n_1)RT$.
4. Let $\Delta n_g = n_2 - n_1$. Therefore, $\Delta H = \Delta U + \Delta n_g RT$. [1 Mark]
Q10. Numerical on Spontaneity ($\Delta G$):
Given: $\Delta H = -224 \text{ kJ} = -224,000 \text{ J}$, $\Delta S = -153 \text{ J K}^{-1}$, $T = 298 \text{ K}$. [1/2 Mark]
Formula: $\Delta G = \Delta H - T\Delta S$ [1/2 Mark]
$\Delta G = -224,000 - (298 \times -153) = -224,000 - (-45,594) = -224,000 + 45,594$ [1 Mark]
$\Delta G = -178,406 \text{ J} = -178.406 \text{ kJ}$. [1/2 Mark]
Since $\Delta G < 0$, the reaction is spontaneous. [1/2 Mark]
SECTION D [4 Marks]
Q11. (a) Work in chemical reaction [2 Marks] (b) Isolated System [2 Marks]
(a) Work Calculation:
Reaction: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$
For 2 moles of $SO_2$: $\Delta n_g = n_{prod} - n_{react} = 2 - (2+1) = -1 \text{ mol}$. [1/2 Mark]
For 4 moles of $SO_2$, $\Delta n_g = -2 \text{ mol}$. [1/2 Mark]
$W = -\Delta n_g RT = -(-2) \times 8.314 \times 298 = 2 \times 8.314 \times 298$ [1/2 Mark]
Answer: $W = +4955.14 \text{ J} = +4.955 \text{ kJ}$ [1/2 Mark]
(b) Isolated System: A system which can exchange neither mass (matter) nor energy with its surroundings. [1 Mark] Example: Hot tea in a perfectly insulated thermos flask. [1 Mark]
Q12. (a) Enthalpy of Formation Numerical [3 Marks] (b) Sign Convention [1 Mark]
(a) Enthalpy Calculation:
Target reaction: $C(graphite) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta_f H^\circ = ?$
Given:
(1) $C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \text{ kJ/mol}$
(2) $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H = -285.8 \text{ kJ/mol}$
(3) $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H = -890.3 \text{ kJ/mol}$ [1 Mark for setting up equations]
To get target reaction: Add Eq(1) and $2 \times$ Eq(2), then subtract Eq(3).
$\Delta_f H^\circ = (-393.5) + 2(-285.8) - (-890.3)$ [1 Mark]
$\Delta_f H^\circ = -393.5 - 571.6 + 890.3 = -965.1 + 890.3$
$\Delta_f H^\circ = -74.8 \text{ kJ/mol}$ [1 Mark]
(b) Sign Convention: Heat absorbed by the system is denoted with a positive sign ($+q$). [1 Mark]
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