If the bond energies of H-H, Br-Br and H-Br are 433,192 and 364 ~kJ ~mol^-1 respectively, then Δ H^° for the reaction : H_2(g)+Br_2(g) arrow 2 HBr(g)

Correct answer: (D) - 103 ~kJ. For the reaction, H_2(g)+Br_2(g) arrow 2 HBr(g) Δ H^° reaction =Σ Δ H_bonds booken (reactants) -Σ Δ H_bonds formed (products) = (433+192-2 × 364 ~kJ ~mol^-1 ) =-103 ~kJ ~mol^-1

Q26 / 180 · KCET 2016 · Chemistry

If the bond energies of \( \mathrm{H}-\mathrm{H}, \mathrm{Br}-\mathrm{Br} \) and \( \mathrm{H}-\mathrm{Br} \) are \( 433,192 \) and \( 364 \mathrm{~kJ} \mathrm{~mol}^{-1} \) respectively, then
\( \Delta \mathrm{H}^{\circ} \) for the reaction :
\( \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{HBr}(g) \)

A
- \( 261 \) kJ
B
+103 kJ
C
+261kJ
Correct answer
D
\(- 103 \mathrm{~kJ} \)

Solution

For the reaction, \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{HBr}(g)\) \(\Delta \mathrm{H}^{\circ}\) reaction \(=\Sigma \Delta \mathrm{H}_{\text {bonds booken (reactants) }}-\Sigma \Delta \mathrm{H}_{\text {bonds formed (products) }}\) \(=\left(433+192-2 \times 364 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) \(=-103 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Hence, the correct answer is option: 'D'.

Solution

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