Chemical Thermodynamics — JEE Main practice

53 questions

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Sample questions with solutions

Q1 · 2026

For the reaction, N2O4⇌2NO2\mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2\mathrm{NO}_2, graph is plotted as shown below. Identify correct statements.

  • A.

    C and E only

  • B.

    D and E only

  • C.

    A and D only

  • D.

    B and C only

Answer: B
Q2 · 2026

Which of the following graphs between pressure 'p' versus volume 'V' represents the maximum work done?

  • A.
  • B.
  • C.
  • D.
Answer: A

  1. Recall the geometric meaning of work in a P-V diagram. Since the area under the curve on a P-V graph represents the magnitude of work done during a process, comparing these areas across the four given graphs tells us which process involves the most work.

  2. Rule out the zero-work option. Given option B shows a path where the enclosed area under the curve is zero, no work is done in that case.

  3. Compare the remaining options. Since all other options show expansion (area lies below the axis, giving negative work by convention), we compare only their magnitudes.

  4. Estimate the magnitudes numerically. Given the areas work out approximately to W1=(44.8)×ln⁡2≈31.04,W3<31.04,W4<31.04W_1 = (44.8)\times \ln 2 \approx 31.04,\qquad W_3 < 31.04,\qquad W_4 < 31.04

Hence, option A has the largest magnitude of work — the answer is option A.

Q3 · 2026

Consider the following data:

ΔfH∘\Delta_f H^{\circ} (methane, g) = −X kJ mol−1-X\ \mathrm{kJ\ mol}^{-1}

Enthalpy of sublimation of graphite = Y kJ mol−1Y\ \mathrm{kJ\ mol}^{-1}

Dissociation enthalpy of H2=Z kJ mol−1H_2 = Z\ \mathrm{kJ\ mol}^{-1}

The bond enthalpy of C–H bond is given by:

  • A.

    X+Y+4Z2\dfrac{X + Y + 4Z}{2}

  • B.

    −X+Y+Z4\dfrac{-X + Y + Z}{4}

  • C.

    X+Y+2Z4\dfrac{X + Y + 2Z}{4}

  • D.

    X+Y+ZX + Y + Z

Answer: C
  1. Write the formation reaction for methane. Since methane forms from solid graphite carbon and hydrogen gas,
C(s)+2H2(g)→CH4(g)\mathrm{C(s)} + 2\mathrm{H}_2\mathrm{(g)} \rightarrow \mathrm{CH}_4\mathrm{(g)}

with enthalpy −X-X.

  1. Break this into elementary bond-related steps using Hess's Law. Given the overall enthalpy equals the energy needed to atomize carbon and hydrogen minus the energy released forming C–H bonds, −X=(ΔHsub of carbon)+2×(B.E. of H–H)−4×(B.E. of C–H)-X = (\Delta H_{sub}\ \text{of carbon}) + 2\times(\text{B.E. of H–H}) - 4\times(\text{B.E. of C–H})

  2. Substitute the given symbols. Since sublimation enthalpy is YY and H–H bond dissociation is ZZ, −X=Y+2Z−4(B.E. of C-H)-X = Y + 2Z - 4(\text{B.E. of C-H})

  3. Solve for the C–H bond enthalpy:

B.E. of C-H=Y+2Z+X4\text{B.E. of C-H} = \frac{Y + 2Z + X}{4}

Hence, the answer is option C.

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