Question

Given, $H_2\left(g\right) + Br_2\left(g\right) \to 2HBr\left(g\right), \Delta H^{\circ}_{1}$, and standard enthalpy of condensation of bromine is $\Delta H^{\circ}_{2}$, standard enthalpy of formation of $HBr$ at $25^{\circ}C$ is

• A. $\Delta H^{\circ}_1/2$
• B. $\Delta H^{\circ}_1/2+\Delta H^{\circ}2$
• C. $\Delta H^{\circ}_1/2-\Delta H^{\circ}2$
• D. $\left(\Delta H^{\circ}_{1}-\Delta H^{\circ}_{2}\right)/2$

Answer 1

Answer: (D)

$H_{2}\left(g\right)+Br_{2}\left(g\right)\to2HBr\left(g\right);\Delta H=\Delta H^{\circ}_{1} ....\left(i\right)$
$Br_{2}\left(g\right)\to Br_{2}\left(\ell\right);\Delta H=\Delta H^{\circ}_{2} ....\left(ii\right)$
Subtracting equation (ii) from equation (i), we get
$H_{2}\left(g\right)+Br_{2}\left(l\right)\to2HBr\left(g\right);\Delta H=\Delta H^{\circ}_{1}-\Delta H^{\circ}_{2} $
Required equation, $\frac{1}{2}H_{2}\left(g\right)+\frac{1}{2}Br_{2}\left(l\right)\to HBr\left(g\right); $
$\Delta H=\left[\frac{\Delta H^{\circ}_{1}=\Delta H^{\circ}_{2}}{2}\right]$