If the bond energies of H – H, Br – Br and H – Br are 433, 192 and 364 kJ m o l − 1 respectively, the Δ H ∘ for the reaction ; H 2 ( g ) + B r 2 ( g ) → 2 H B r ( g ) is Options: A . &nbsp– 261 kJ B . &nbsp+ 103 kJ C . &nbsp+ 261 kJ D . &nbsp– 103 kJ

Answer: Option D : D Δ H ∘ = ∑ B . E ( r e a c t a n t s ) − ∑ B . E ( p r o d u c t s ) = [ B . E ( H − H ) + B . E ( B r − B r ) ] − [ 2 B . E ( H − B r ) ] = ( 433 + 192 ) − ( 2 × 364 ) = 625 − 728 = − 103 k J

If the bond energies of H – H, Br – Br and H

Question

If the bond energies of H – H, Br – Br and H – Br are 433, 192 and 364 kJ

m o l^{- 1}

respectively, the

Δ H^{\circ}

for the reaction ;

H_{2} (g) + B r_{2} (g) \to 2 H B r (g)

Options:

A . – 261 kJ

B . + 103 kJ

C . + 261 kJ

D . – 103 kJ

Answer: Option D
:
D

Δ H^{\circ} = \sum B . E (r e a c t a n t s) - \sum B . E (p r o d u c t s) = [B . E (H - H) + B . E (B r - B r)] - [2 B . E (H - B r)] = (433 + 192) - (2 \times 364) = 625 - 728 = - 103 k J

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