If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for reaction…

If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{l}) \rightarrow 2 \mathrm{HBr}(\mathrm{g})$ is $-72.8 \mathrm{~kJ}$, the standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^\theta\right)$ of $\mathrm{HBr}(\mathrm{g})$ (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is
  1. -36.4
  2. +36.4
  3. -18.2
  4. -18.2

Solution

$\begin{aligned} & \text { } \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{H}_2 \text { gas }\right)=0.0 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{Br}_2 \text { liquid) }\right. \\ & =0.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{T}}^{\circ}=-72.8 \mathrm{~kJ} \text { for } 2 \text { moles of } \mathrm{HBr} \text {. } \\ & \Rightarrow \Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \text { ( } \mathrm{H} \mathrm{Br} \text { gas }\right)-\left(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{H}_2 \text { gas }\right)\right) \\ & \left.+\left(\Delta \mathrm{H}_{\mathrm{f}}{ }^0\left(\mathrm{Br}_2 \text { liq. }\right)\right)\right] \\ & -72.8=2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas })-(0.0+0.0) \\ & =2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas }) \\ & \Rightarrow \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HBr} \text { gas })=\frac{-72.8}{2}=,-36.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {. } \\ & \end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

Practice more Thermodynamics (C) questions on Aicharya