Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 3.
Q35 (2025)
Which of the following properties will change when system containing solution 1 will become solution 2? {{IMG1}}
- Molar heat capacity
- Density
- Concentration
- Gibbs free energy
### Related Formula
$G = H - TS$
Extensive properties depend on mass/amount of matter; Intensive properties are independent of amount.
### Core Logic
Both solution 1 ($10\text{ mol}/10\text{ L} = 1\text{ M}$) and solution 2 ($1\text{ mol}/1\text{ L} = 1\text{ M}$) have identical concentrations and chemical compositions.
Intensive properties (molar heat capacity, density, concentration) depend only on composition and temperature, so they remain unchanged.
Gibbs free energy ($G$) is an extensive property, proportional to the total amount of substance present, so it changes.
### Step 1: Final Conclusion
Gibbs free energy changes because it is an extensive property.
### Pattern Recognition
Same concentration/composition $\rightarrow$ Intensive properties stay equal.
Total amount changes $\rightarrow$ Extensive properties ($G, H, U, S$) change.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q49 (2025)
Given:
$$\Delta H_{\mathrm{sub}}^{\ominus} [\mathrm{C(graphite)}] = 710\mathrm{~kJ~mol^{-1}}$$
$$\Delta_{\mathrm{C-H}} H^{\ominus} = 414\mathrm{~kJ~mol^{-1}}$$
$$\Delta_{\mathrm{H-H}} H^{\ominus} = 436\mathrm{~kJ~mol^{-1}}$$
$$\Delta_{\mathrm{C=C}} H^{\ominus} = 611\mathrm{~kJ~mol^{-1}}$$
The $\Delta H_{\mathrm{f}}^{\ominus}$ for $\mathrm{CH_2 = CH_2}$ is ________ $\mathrm{kJ~mol^{-1}}$ (nearest integer value)
### Related Formula
$$\Delta H_{\mathrm{f}}^{\ominus}(\mathrm{C_2H_4}) = 2 \Delta H_{\mathrm{sub}}^{\ominus}[\mathrm{C(s)}] + 2 \Delta_{\mathrm{H-H}} H^{\ominus} - [1 \Delta_{\mathrm{C=C}} H^{\ominus} + 4 \Delta_{\mathrm{C-H}} H^{\ominus}]$$
### Core Logic
Target formation reaction:
$$2\mathrm{C(graphite, s)} + 2\mathrm{H_2(g)} \rightarrow \mathrm{CH_2=CH_2(g)}$$
1. Sublimation of 2 moles of C(s) to gaseous atoms: $2 \times 710 = 1420\mathrm{~kJ}$
2. Dissociation of 2 moles of $\mathrm{H_2(g)}$ to $4\mathrm{H(g)}$ atoms: $2 \times 436 = 872\mathrm{~kJ}$
3. Formation of $1 \times \mathrm{C=C}$ double bond: $-611\mathrm{~kJ}$
4. Formation of $4 \times \mathrm{C-H}$ single bonds: $-4 \times 414 = -1656\mathrm{~kJ}$
### Step 1: Calculation
$$\Delta H_{\mathrm{f}}^{\ominus} = 1420 + 872 - 611 - 1656$$
$$\Delta H_{\mathrm{f}}^{\ominus} = 2292 - 2267 = 25\mathrm{~kJ~mol^{-1}}$$
### Pattern Recognition
Standard Enthalpy of Formation = Atomization Enthalpies of Reactants $-$ Bond Enthalpies of Products formed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q37 (2025)
Consider the given data :
(a) $\mathrm{HCl(g)} + 10\mathrm{H}_2\mathrm{O(l)}\rightarrow \mathrm{HCl.10H_2O} \quad \Delta \mathrm {H} = - 6 9. 0 1 \mathrm {k J} \mathrm {m o l} ^ {- 1}$
(b) $\mathrm{HCl(g)} + 40\mathrm{H}_2\mathrm{O(l)}\rightarrow \mathrm{HCl.40H_2O} \quad \Delta \mathrm {H} = - 7 2. 7 9 \mathrm {k J} \mathrm {m o l} ^ {- 1}$
Choose the correct statement :
- Dissolution of gas in water is an endothermic process
- The heat of solution depends on the amount of solvent.
- The heat of dilution for the HCl $(\mathrm{HCl}.10\mathrm{H}_2\mathrm{O}$ to $\mathrm{HCl}.40\mathrm{H}_2\mathrm{O})$ is $3.78\mathrm{kJ mol}^{-1}$.
- The heat of formation of HCl solution is represented by both (a) and (b)
### Related Formula
$$\Delta H_{\text{dilution}} = \Delta H_2 - \Delta H_1$$
### Core Logic
Analyzing the thermodynamic statements:
- $\Delta H$ values are negative, so the dissolution of $HCl(g)$ is clearly exothermic, eliminating option (1).
- Since the enthalpy release changes when the moles of water solvent shift from 10 to 40 ($-69.01$ vs $-72.79$), the **heat of solution depends explicitly on the amount of solvent** (Statement 2 is true).
- Let's check Statement 3: By subtracting equation (a) from (b):
$$\mathrm{HCl\cdot10H_2O} + 30\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl\cdot40H_2O}$$
$$\Delta H = -72.79 - (-69.01) = -3.78 \mathrm{~kJ\cdot mol^{-1}}$$
The value is negative, indicating an exothermic process, so calling it $+3.78$ makes option (3) incorrect.
### Pattern Recognition
The standard integral enthalpy of solution varies with solvent concentration until infinite dilution is achieved. Thus, concentration dependence is a core property of partial molar solution variables.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q27 (2025)
Let us consider a reversible reaction at temperature, $T$.
In this reaction, both $\Delta H$ and $\Delta S$ were observed to have positive values. If the equilibrium temperature is $T_e$, then the reaction becomes spontaneous at:
- $T = T_e$
- $T_e > T$
- $T > T_e$
- $T_e = 5T$
### Related Formula
$$\Delta G = \Delta H - T\Delta S$$
### Core Logic
For a reaction to be spontaneous, the change in Gibbs free energy must be negative:
$$\Delta G < 0 \implies \Delta H - T\Delta S < 0$$
Given that both $\Delta H > 0$ and $\Delta S > 0$:
$$\Delta H < T\Delta S \implies T > \frac{\Delta H}{\Delta S}$$
At the equilibrium temperature $T_e$, $\Delta G = 0$, which gives:
$$T_e = \frac{\Delta H}{\Delta S}$$
Substituting this back into the inequality reveals that the reaction is spontaneous when:
$T > T_e$
### Pattern Recognition
When both $\Delta H$ and $\Delta S$ are positive, the reaction is entropy-driven and becomes spontaneous only at higher temperatures ($T > T_e$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q30 (2025)
One mole of an ideal gas expands isothermally and reversibly from $10\mathrm{~dm}^3$ to $20\mathrm{~dm}^3$ at $300\mathrm{~K}$. $\Delta U$, $q$ and work done in the process respectively are:
Given: $R = 8.3\mathrm{~J~K^{-1}~mol^{-1}}$, $\ln 10 = 2.3$, $\log 2 = 0.30$, $\log 3 = 0.48$
- $0, 21.84\mathrm{~kJ}, -1.26\mathrm{~kJ}$
- $0, -17.18\mathrm{~kJ}, 1.718\mathrm{~J}$
- $0, 21.84\mathrm{~kJ}, 21.84\mathrm{~kJ}$
- $0, 1.718\mathrm{~kJ}, -1.718\mathrm{~kJ}$
### Related Formula
$$\Delta U = n C_v \Delta T$$
$$w = -n R T \ln\left(\frac{V_2}{V_1}\right)$$
$$\Delta U = q + w$$
### Core Logic
Since the expansion step is strictly **isothermal** ($\Delta T = 0$):
$\Delta U = 0$
Now compute the work command parameter $w$:
$$w = -n R T \ln\left(\frac{V_2}{V_1}\right) = -1 \cdot 8.3 \cdot 300 \cdot \ln\left(\frac{20}{10}
ight)$$
$$w = -2490 \cdot \ln(2) = -2490 \cdot (2.3 \cdot \log 2)$$
$$w = -2490 \cdot (2.3 \cdot 0.30) = -2490 \cdot 0.69 = -1718.1\mathrm{~J} = -1.718\mathrm{~kJ}$$
Applying the first law equation constraint:
$$q = -w = +1.718\mathrm{~kJ}$$
Hence, $\Delta U = 0$, $q = 1.718\mathrm{~kJ}$, $w = -1.718\mathrm{~kJ}$.
### Pattern Recognition
Isothermal expansion of an ideal gas ALWAYS yields $\Delta U = 0$. Work is negative (done by system) and heat exchange $q$ matches work magnitude inversely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
