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

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

Solution

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

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