Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 8.
Q56 (2026)
Consider the following data:
$\Delta_f H^{\ominus}(\text{methane, g}) = -X \text{ kJ mol}^{-1}$
Enthalpy of sublimation of graphite $= Y \text{ kJ mol}^{-1}$
Dissociation enthalpy of $\text{H}_2 = Z \text{ kJ mol}^{-1}$
The bond enthalpy of $\text{C-H}$ bond is given by:
- $(1) \ \frac{X + Y + 2Z}{4}$
- $(2) \ \frac{X + Y + 4Z}{2}$
- $(3) \ X + Y + Z$
- $(4) \ \frac{-X + Y + Z}{4}$
### Core Logic
Reaction for formation of methane:
$$\text{C}(\text{s}) + 2\text{H}_2(\text{g}) \rightarrow \text{CH}_4(\text{g})$$
$-X = (\Delta H_{\text{sub}} \text{ of carbon}) + 2 \times (\text{B.E. of H-H}) - 4 \times (\text{B.E. of C-H})$
$-X = Y + 2Z - 4(\text{B.E. of C-H})$
### Step 1: Rearranging for Bond Enthalpy
Rearranging the equation for C-H bond enthalpy:
$$\text{B.E. of C-H} = \frac{X + Y + 2Z}{4}$$
### Pattern Recognition
Sees: bond enthalpy derivation from heat of formation, sublimation, and dissociation.
Trap: Sign convention errors when substituting formation enthalpies.
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q60 (2026)
Match the LIST-I with LIST-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I (Thermodynamic Process)</th><th style="border: 1px solid #888; padding: 8px;">List-II (Magnitude in kJ)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Work done in reversible, isothermal expansion of 2 mol of ideal gas from $2\text{ dm}^{3}$ to $20\text{ dm}^{3}$ at 300 K.</td><td style="border: 1px solid #888; padding: 8px;">I. 4</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Work done in irreversible isothermal expansion of 1 mol ideal gas from $1\text{ m}^{3}$ to $3\text{ m}^{3}$ at 300 K against a constant pressure of 3kPa.</td><td style="border: 1px solid #888; padding: 8px;">II. 11.5</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and $\overline{C}_{v} = \frac{3}{2}R$</td><td style="border: 1px solid #888; padding: 8px;">III. 6</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and $\overline{C}_{p} = \frac{5}{2}R$</td><td style="border: 1px solid #888; padding: 8px;">IV. 7</td></tr></tbody></table></div>
Choose the correct answer from the option given below:
- $\text{A-III, B-II, C-IV, D-I}$
- $\text{A-II, B-III, C-I, D-IV}$
- $\text{A-I, B-II, C-III, D-IV}$
- $\text{A-II, B-I, C-III, D-IV}$
### Core Logic
Evaluate each option one by one:
**Option (A)**: Work done in reversible isothermal expansion
$W = -nRT \ln\left(\frac{V_2}{V_1}\right)$
Magnitude $= \frac{2 \times 8.314 \times 300}{1000} \times \ln(10)\text{ kJ}$
$= 4.9884 \times 2.303 \approx 11.5\text{ kJ}$
So, A $\rightarrow$ II.
**Option (B)**: Work done in irreversible isothermal expansion against constant external pressure
$W = -P_{\text{ext}} (V_2 - V_1)$
Magnitude $= 3\text{ kPa} \times (3 - 1)\text{ m}^3 = 3 \times 10^3\text{ Pa} \times 2\text{ m}^3 = 6000\text{ J} = 6\text{ kJ}$
So, B $\rightarrow$ III.
**Option (C)**: Change in internal energy
$\Delta U = n C_v \Delta T$
Magnitude $= 1 \times \left(\frac{3}{2} \times 8.314\right) \times 320 = \frac{12.471 \times 320}{1000}\text{ kJ} \approx 3.99\text{ kJ} \approx 4\text{ kJ}$
So, C $\rightarrow$ I.
**Option (D)**: Change in enthalpy
$\Delta H = n C_p \Delta T$
Magnitude $= 1 \times \left(\frac{5}{2} \times 8.314\right) \times 337 = \frac{20.785 \times 337}{1000}\text{ kJ} \approx 7\text{ kJ}$
So, D $\rightarrow$ IV.
### Step 1: Final Matching
Matches are A-II, B-III, C-I, D-IV.
### Pattern Recognition
Standard match-list question. Calculate A and D first, as they quickly eliminate wrong options.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q75 (2026)
Dissociation of a gas $A_{2}$ takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300K.
$$A_{2}(g) \rightleftharpoons 2A(g)$$
The standard Gibbs energy of formation of the involved substances has been provided below:
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">Substance</th><th style="border: 1px solid #888; padding: 8px;">$\Delta G_{f}^{\circ} / \text{kJ mol}^{-1}$</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">$A_{2}$</td><td style="border: 1px solid #888; padding: 8px;">-100.00</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">A</td><td style="border: 1px solid #888; padding: 8px;">-50.832</td></tr></tbody></table></div>
The degree of dissociation of $A_{2}(g)$ is given by $(x \times 10^{-2})^{1/2}$ where $x = $ ____. (Nearest integer).
[Given: $R = 8\text{ J mol}^{-1}\text{K}^{-1}$, $\log 2 = 0.3010$, $\log 3 = 0.48$]
### Related Formula
$$\Delta G^{\circ}_{reaction} = \sum \Delta G^{\circ}_{f}(\text{products}) - \sum \Delta G^{\circ}_{f}(\text{reactants})$$
$$\Delta G^{\circ} = -RT \ln K_p$$
$$K_p = \frac{4\alpha^2 P_0}{1 - \alpha^2}$$
### Core Logic
First, find standard Gibbs free energy of the reaction:
$$\Delta G^{\circ}_{reaction} = 2 \times \Delta G^{\circ}_{f}(A) - \Delta G^{\circ}_{f}(A_{2})$$
$$\Delta G^{\circ} = 2(-50.832) - (-100.00) = -101.664 + 100.00 = -1.664\text{ kJ mol}^{-1}$$
$$\Delta G^{\circ} = -1664\text{ J mol}^{-1}$$
Use this to find $K_p$:
$$-1664 = -8 \times 300 \times \ln K_p$$
$$1664 = 2400 \ln K_p$$
$$\ln K_p = \frac{1664}{2400} = 0.6933$$
Since $\ln 2 \approx 0.693$, we have:
$K_p = 2$
### Step 1: Calculate Degree of Dissociation
For the reaction $A_2 \rightleftharpoons 2A$, with initial moles 1 and degree of dissociation $\alpha$:
Moles at equilibrium: $1-\alpha$ (for $A_2$) and $2\alpha$ (for A).
Total moles = $1+\alpha$.
$$K_p = \frac{(P_A)^2}{P_{A_2}} = \frac{\left(\frac{2\alpha}{1+\alpha} P_0\right)^2}{\frac{1-\alpha}{1+\alpha} P_0} = \frac{4\alpha^2 P_0}{1 - \alpha^2}$$
Given total pressure $P_0 = 1\text{ bar}$ and $K_p = 2$:
$$2 = \frac{4\alpha^2 (1)}{1 - \alpha^2}$$
$$2 - 2\alpha^2 = 4\alpha^2$$
$$6\alpha^2 = 2 \implies \alpha^2 = \frac{1}{3} \implies \alpha = \frac{1}{\sqrt{3}}$$
### Step 2: Match to Given Format
We are given $\alpha = (x \times 10^{-2})^{1/2}$.
Squaring both sides:
$$\alpha^2 = x \times 10^{-2}$$
$$\frac{1}{3} = x \times 10^{-2} \implies x = \frac{100}{3} = 33.33$$
### Step 3: Rounding
Nearest integer is 33.
### Pattern Recognition
Whenever $\Delta G^{\circ}$ yields an $RT \ln K_p$ around 0.693, $K_p$ is 2. The formula $K_p = 4\alpha^2/(1-\alpha^2)$ for $A_2 \rightleftharpoons 2A$ at $P=1$ is standard and should be memorized.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
Q72 (2026)
If the enthalpy of sublimation of Li is $155\text{ kJ mol}^{-1}$, enthalpy of dissociation of $\text{F}_2$ is $150\text{ kJ mol}^{-1}$, ionization enthalpy of Li is $520\text{ kJ mol}^{-1}$, electron gain enthalpy of F is $-313\text{ kJ mol}^{-1}$, standard enthalpy of formation of LiF is $-594\text{ kJ mol}^{-1}$. The magnitude of lattice enthalpy of LiF is ____ $\text{kJ mol}^{-1}$ (Nearest integer).
### Related Formula
$$\Delta_f H^0 = \Delta_{\text{sub}} H + \text{IE} + \frac{1}{2}\Delta_{\text{bond}} H + \Delta_{eg} H + \text{L.E.}$$
### Core Logic
Step 1: Substitute given thermodynamic cycle values into Born-Haber equation:
$$-594 = 155 + 520 + \frac{150}{2} + (-313) + \text{L.E.}$$
$$-594 = 155 + 520 + 75 - 313 + \text{L.E.}$$
$$-594 = 437 + \text{L.E.}$$
$$\text{L.E.} = -594 - 437 = -1031\text{ kJ mol}^{-1}$$
Step 2: Magnitude of lattice enthalpy is $1031\text{ kJ mol}^{-1}$.
{{SOL_IMG1}}
### Pattern Recognition
Sees: Born-Haber cycle parameters for ionic solid.
Shortcut: Add sublimation, ionization, half-dissociation, and electron gain enthalpies, then subtract from formation enthalpy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q58 (2026)
A cup of water at $5^{\circ}C$ (system) is placed in a microwave oven and the oven is turned on for one minute during which, the water begins to boil. Which of the following option is true?
- $q = +ve, w = 0, \Delta U = -ve$
- $q = +ve, w = -ve, \Delta U = +ve$
- $q = -ve, w = -ve, \Delta U = -ve$
- $q = +ve, w = -ve, \Delta U = -ve$
### Related Formula
$$\Delta U = q + w$$
where:
$\Delta U$ = change in internal energy
$q$ = heat added to system
$w$ = work done on the system
### Core Logic
Analyze the state changes step-by-step applying IUPAC sign conventions for thermodynamics.
{{SOL_IMG1}}
### Step 1: Heat Transfer
Since heat is supplied by the microwave oven to the water (system), the system absorbs heat. Thus, $q = +ve$.
### Step 2: Work Done
As water boils, it converts from liquid to vapor, which implies a massive volume expansion. Work is done by the system against the atmosphere. Thus, work done on the system is negative: $w = -ve$.
### Step 3: Internal Energy Change
The temperature of water increases from $5^\circ C$ to $100^\circ C$, and liquid converts to gas. The internal energy of steam at $100^\circ C$ is much greater than that of liquid water at $5^\circ C$. Hence, $\Delta U = +ve$.
### Pattern Recognition
Heating + Boiling inherently means heat is entering ($q > 0$), expanding volume pushes outward doing work ($w < 0$), and increasing thermal energy raises the internal state function ($\Delta U > 0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
