Given $H_{2} (R} + {Br}_{2} (E) ⟶ 2 HBr (g), {ΔA}_{1}^{\circ}$ and standard enthalpy of condensation of bromine is ${ΔH}_{2}^{\circ}$ standard enthalpy of formation of HBr at 25° C is:

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$\frac{({ΔH}_{1}^{\circ} - {ΔH}_{2}^{\circ})}{2}$

$\frac{{ΔH}_{1}^{\circ}}{2} - {ΔH}_{2}^{\circ}$

$\begin{aligned} \frac{{ΔH}_{1}^{\circ}}{2} \end{aligned}$

$\begin{aligned} \frac{{ΔH}_{1}^{\circ}}{2} + {ΔH}_{2}^{\circ} \end{aligned}$

answer is D.

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Detailed Solution

In the reaction we have,

${Br}_{2} (l) ⟶ {Br}_{2} (g) - ({ΔH}_{2}^{D})$

$\begin{array}{l} H_{2} (g) + {Br}_{2} (g) ⟶ 2 HBr (g) {ΔH}_{1}^{*} \\ H_{2} (g) + {Br}_{2} (l) ⟶ 2 HBr (g) {ΔH}_{1}^{*} - {ΔH}_{2}^{*} \\ {ΔH}_{f}^{*} (HBr (g)) = \frac{[{ΔH}_{1}^{*} - {ΔH}_{2}^{*}]}{2} \end{array}$

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