Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 7.
Q90 (2024)
If $5\text{ moles}$ of an ideal gas expands from $10\text{ L}$ to a volume of $100\text{ L}$ at $300\text{ K}$ under isothermal and reversible condition then work $w$, is $-x\text{ J}$. The value of $x$ is ________
(Given $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$)
### Related Formula
$$W = -2.303 \, nRT \log \left( \frac{V_2}{V_1} \right)$$
### Core Logic
For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention.
Given:
$n = 5\text{ moles}$
$R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$
$T = 300\text{ K}$
$V_1 = 10\text{ L}$
$V_2 = 100\text{ L}$
### Step 1: Calculating Work Done
$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log\left( \frac{100}{10} \right)$$
$$W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$$
$$W = -2.303 \times 12471 \times 1$$
$$W = -28720.713\text{ J}$$
### Step 2: Final Formatting
The question asks for work $w = -x\text{ J}$.
So $x = 28720.713$, which rounds to $28721$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q88 (2024)
Consider the following reaction at 298 K.
$$\frac{3}{2}O_{2(g)} \rightleftharpoons O_{3(g)}. \quad K_p = 2.47 \times 10^{-29}$$
$\Delta_rG^{\ominus}$ for the reaction is ________ kJ.
(Given $R = 8.314 \text{ J } K^{-1} mol^{-1}$)
### Related Formula
$$\Delta_rG^{\ominus} = -RT \ln K_p$$
### Step 1: Calculation
$$\Delta_rG^{\ominus} = -8.314 \times 10^{-3} \text{ kJ } K^{-1} mol^{-1} \times 298 \text{ K} \times \ln(2.47 \times 10^{-29})$$
$$= -8.314 \times 10^{-3} \times 298 \times (-65.87)$$
$$= 163.19 \text{ kJ}$$
### Step 2: Nearest Integer
Rounding $163.19$ to the nearest integer gives $163$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q68 (2026)
Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done?
- $\text{Option 1}$
- $\text{Option 2}$
- $\text{Option 3}$
- $\text{Option 4}$
### Core Logic
The magnitude of work done by or on a gas is given by the area under the P-V curve projected onto the volume axis.
Graph 1: Cyclic process forming a triangle. Area is bounded, represents net work.
Graph 2: Isochoric drop (vertical line at $V=22.4L$). Area = 0, so work done is zero.
Graph 3: Expansion process forming a cycle. Area is enclosed in a convex shape.
Graph 4: Direct expansion from $V=22.4$ to $V=44.8$ at pressure $P=1$ up to $P=2$ (a rectangle combined with a triangle). The total area under the upper curve from $V=22.4$ to $V=44.8$ covers the entire shaded region under the path down to the V-axis.
Option (4) provides the largest total area under the curve extending down to the horizontal axis (maximum magnitude of work done).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q69 (2026)
For the reaction, $N_{2}O_{4} \rightleftharpoons 2NO_{2}$, graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is $-5.40\text{ kJ mol}^{-1}$.
B. As $\Delta G^{\ominus}$ in graph is positive, $N_{2}O_{4}$ will not dissociate into $NO_{2}$ at all.
C. Reverse reaction will go to completion.
D. When 1 mole of $N_{2}O_{4}$ changes into equilibrium mixture, value of $\Delta G = -0.84\text{ kJ mol}^{-1}$.
E. When 2 mole of $NO_{2}$ changes into equilibrium mixture, $\Delta G$ for equilibrium mixture is $-6.24\text{ kJ mol}^{-1}$.
{{IMG1}}
Choose the correct answer from the options given below :
- $\text{D and E only}$
- $\text{C and E only}$
- $\text{A and D only}$
- $\text{B and C only}$
### Core Logic
Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.
{{SOL_IMG1}}
A. Standard free energy change ($\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): $\Delta_r G^{\circ} = G_B^{\circ} - G_A^{\circ}$. Since B is higher than A, $\Delta_r G^{\circ}$ is positive, not $-5.40\text{ kJ mol}^{-1}$. Statement A is false.
B. Even if $\Delta_r G^{\circ}$ is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false.
C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false.
D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is $0.84\text{ kJ mol}^{-1}$. Thus $\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true.
E. The difference from pure products (point B, equivalent to 2 moles $NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is $5.40$, and A to E is $0.84$. So drop from B to E is $- (5.40 + 0.84) = -6.24\text{ kJ mol}^{-1}$. Statement E is true.
### Step 1: Final Conclusion
Only statements D and E are correct.
### Pattern Recognition
The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
Q73 (2026)
Use the following data :
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: center; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">Substance</th><th style="border: 1px solid #888; padding: 8px;">$\frac{\Delta_f H^{\ominus}(500K)}{kJ mol^{-1}}$</th><th style="border: 1px solid #888; padding: 8px;">$\frac{S^{\ominus}(500K)}{J K^{-1} mol^{-1}}$</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">AB(g)</td><td style="border: 1px solid #888; padding: 8px;">32</td><td style="border: 1px solid #888; padding: 8px;">222</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">$A_2(g)$</td><td style="border: 1px solid #888; padding: 8px;">6</td><td style="border: 1px solid #888; padding: 8px;">146</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">$B_2(g)$</td><td style="border: 1px solid #888; padding: 8px;">x</td><td style="border: 1px solid #888; padding: 8px;">280</td></tr></tbody></table></div>
One mole each of $A_{2}(g)$ and $B_{2}(g)$ are taken in a 1L closed flask and allowed to establish the equilibrium at 500K.
$$A_{2}(g) + B_{2}(g) \rightleftharpoons 2AB(g)$$
The value of x (in $\text{kJ mol}^{-1}$) is .... (Nearest integer)
(Given: $\log K=2.2, R=8.314\text{ J K}^{-1}\text{ mol}^{-1}$)
### Related Formula
$$\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$$
$$\Delta_r S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$$
$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$
$$\Delta_r G^{\circ} = -2.303 RT \log K$$
### Core Logic
For the reaction $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$:
1. Enthalpy change ($\Delta_r H^{\circ}$):
$$\Delta_r H^{\circ} = [2 \times \Delta_f H^{\circ}(AB)] - [\Delta_f H^{\circ}(A_2) + \Delta_f H^{\circ}(B_2)]$$
$$\Delta_r H^{\circ} = (2 \times 32) - (6 + x) = (64 - 6 - x) = (58 - x)\text{ kJ mol}^{-1}$$
2. Entropy change ($\Delta_r S^{\circ}$):
$$\Delta_r S^{\circ} = [2 \times S^{\circ}(AB)] - [S^{\circ}(A_2) + S^{\circ}(B_2)]$$
$$\Delta_r S^{\circ} = (2 \times 222) - (146 + 280) = 444 - 426 = 18\text{ J K}^{-1}\text{ mol}^{-1}$$
3. Standard Gibbs Free Energy ($\Delta_r G^{\circ}$) via Equilibrium Constant:
$$\Delta_r G^{\circ} = -2.303 \times R \times T \log K$$
$$\Delta_r G^{\circ} = -2.303 \times 8.314 \times 500 \times 2.2 = -21063.8\text{ J mol}^{-1} = -21.06\text{ kJ mol}^{-1}$$
4. Using Gibbs Equation:
$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$$
$$-21.06 = (58 - x) - 500 \times \left(\frac{18}{1000}\right)$$
$$-21.06 = 58 - x - 9$$
$-21.06 = 49 - x$
$$x = 49 + 21.06 = 70.06\text{ kJ mol}^{-1}$$
Rounding off to nearest integer, $x = 70$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
