Given : Δ H sub Theta [ C ( graphite )]=710 kJ mol -1 Δ C - H H Theta =414 kJ mol -1 Δ H - H H Theta =436 kJ mol -1 Δ C = C H Theta =611 kJ mol -1 The Δ H f Theta for CH 2 = CH 2 is \ \ \ \ \ kJ mol -1 (nearest integer value)

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 3 April, Morning Shift — Question 17

Given : $\Delta \mathrm{H}_{\text {sub }}^{\Theta}[\mathrm{C}($ graphite $)]=710 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\Theta}=414 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\Theta}=436 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{C}=\mathrm{C}} \mathrm{H}^{\Theta}=611 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The $\Delta \mathrm{H}_{\mathrm{f}}^{\Theta}$ for $\mathrm{CH}_{2}=\mathrm{CH}_{2}$ is $\_\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

$\Delta H_{r}^{\Theta}=2(710)+2 \times 436-611-4(414)$

$=1420+872-611-1656$

$=25 \mathrm{~kJ} \mathrm{~mole}^{-1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Thermochemistry and Enthalpy Changes