Given : $$\begin{aligned} & \Delta \mathrm{H}_{\text {sub }}^{\ominus}[\mathrm{C}(\text { graphite })]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}=\mathrm{C}} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$ The $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{CH}_{2}=\mathrm{CH}_{2}$ is _____ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Chemical Thermodynamics

Given :

Δ H_{sub}^{⊖} [C (graphite)] = 710 kJ mol^{- 1} Δ_{C - H} H^{⊖} = 414 kJ mol^{- 1} Δ_{H - H} H^{⊖} = 436 kJ mol^{- 1} Δ_{C = C} H^{⊖} = 611 kJ mol^{- 1}

The

Δ H_{f}^{⊖}

for

CH_{2} = CH_{2}

is _____

kJ mol^{- 1}

(nearest integer value)

Answer: 25

Formation of ethene is

2 C(graphite) + 2 H_{2} (g) \to C_{2} H_{4} (g)

. Atomising the reactants absorbs

2 Δ H_{sub} (C) + 2 Δ H_{H-H} = 2 (710) + 2 (436) = 2292 kJ

. Forming the one

C=C

bond and four

C-H

bonds of ethene releases

611 + 4 (414) = 2267 kJ

. The enthalpy of formation is the difference,

Δ H_{f}^{⊖} = 2292 - 2267 = 25 kJ mol^{- 1}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer