Given : H sub [ C(graphite) ] = 710 kJ mol - 1 C-H H = 414 kJ mol - 1 H-H H = 436 kJ mol - 1 C=C H = 611 kJ mol - 1 The H f for CH 2 = CH 2 is ____ kJ mol - 1 (nearest integer value). [2025]

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Solution

Q.
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). [2025]

Ans.
(25)

$2 C (graphite) + 2 H_{2} (g) \to C H_{2} = C H_{2} (g), Δ H_{f}^{\circ} (C_{2} H_{4})$

Enthalpy of any reaction = Total bond enthalpy of reactants - total bond enthalpy of products

$Δ H_{f}^{\circ} (C_{2} H_{4}) = [2 \times Δ H_{s u b}^{\circ} (graphite) + 2 \times B.E (H_{2})] - [4 \times B.E (C - H) + B.E (C = C)]$

$Δ H_{f}^{\circ} (C_{2} H_{4}) = [2 \times 710 + 2 \times 436] {kJ mol}^{- 1} - [4 \times 414 + 611] {kJ mol}^{- 1}$

$= 25 {kJ mol}^{- 1}$

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