Enthalpy change for the reaction $4 H_{(g)} \to 2 H_{2 (g)}$ is $- 869.6 k J$ . The dissociation energy of $H - H$ bond is ?

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Solution

$4 H (g) ⟶ 2 H_{2} (g); Δ H = - 869.6 k J$

$2 H_{2} (g) ⟶ 4 H (g); Δ H = 869.6 k J$

$H_{2} (g) ⟶ 2 H (g); Δ H = 869.6 / 2 = 434.8 k J$

Hence the answer is 434.8kJ

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Similar questions

4 H_{(g)} \to 2 H_{2 (g)}

- 869.6 k J

H - H

{4 H}_{g} ⟶ {2 H}_{2 (g)}

The enthalpy of atomisation for the reaction $C H_{4} (g) \to C (g) + 4 H (g) is 1665 k J m o l^{- 1}$ . What is the bond energy of C—H bond?

C (s) + 2 H_{2} (g) ⟶ C H_{4} (g); Δ H = - X k c a l

C (g) + 4 H (g) ⟶ C H_{4} (g); Δ H = - X_{1} k c a l

C H_{4} (g) ⟶ C H_{3} (g) + H (g); Δ H = + Y k c a l

C - H

H_{2} (g) + C_{2} H_{4} (g) \to C_{2} H_{6} (g)

H - H = 103, C - H = 99, C - C = 80

C = C = 145

K c a l m o l^{- 1}

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