Exam

JEE Main

Year

2025

Date

3 April

Shift

Morning Shift

Subject

Chemistry

Chapter

Thermodynamics (C)

JEE Main 2025 · 3 April · Morning Shift · Chemistry · Thermodynamics (C)

Given : aligned & Δ H_ sub ^ [C( graphite )]=710 kJ mol^ -1 \& Δ_...

Given :\begin{aligned}& \Delta \mathrm{H}_{\text{sub }^{\ominus}[\mathrm{C}(\text{ graphite })]=710 \mathrm{ kJ} \mathrm{ mol}^{-1} \& \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=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}\end{aligned}The \Delta \mathrm{H}_{\mathrm{f}^{\ominus} for $\mathrm{CH}_2=\mathrm{CH}_2$ is _____ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)

Step-by-step solution

$$\begin{aligned} & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{(4)}}=2 \times\left[\Delta \mathrm{H}_{\mathrm{sub}}^{\mathrm{o}}\right]_{\mathrm{C}_{(\mathrm{s})}}+2 \times \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}^{\mathrm{o}}-1 \times \Delta \mathrm{H}_{\mathrm{C}=\mathrm{C}}^{\mathrm{o}}-4 \times \Delta \mathrm{H}_{\mathrm{CH}}^{\mathrm{o}} } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=(2 \times 710)+(2 \times 436)-611-4 \times 414 } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=25 \mathrm{~kJ} / \mathrm{mol} }\end{aligned}$$