JEE Main 2025 Apr 3 Shift 1, Chemistry Q24: Hess’s Law and Bond Energy

JEE Main2025Apr 3, Shift 1Chemistry

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)

Your answer

Solution

Build ethene from its standard-state elements via a Hess cycle:

Step 1: $2 C(graphite) \to 2 C(g)$ , energy $= 2 \times 710 = 1420$ kJ.

Step 2: $2 H_{2} (g) \to 4 H(g)$ , energy $= 2 \times 436 = 872$ kJ.

Step 3: Assemble 1 C=C bond and 4 C–H bonds. Bond formation releases energy:

$- Δ_{C = C} - 4 Δ_{C - H} = - 611 - 4 (414) = - 611 - 1656 = - 2267 kJ .$

Sum:

$Δ H_{f}^{⊖} (C_{2} H_{4}) = 1420 + 872 - 2267 = 25 kJ mol^{- 1} .$

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.