Calculate the heat produced (in $\mathrm{kJ}$ ) when $224 \mathrm{~g}$ of $\mathrm{CaO}$ is completely…

Calculate the heat produced (in $\mathrm{kJ}$ ) when $224 \mathrm{~g}$ of $\mathrm{CaO}$ is completely converted to $\mathrm{CaCO}_{3}$ by reaction with $\mathrm{CO}_{2}$ at $27^{\circ} \mathrm{C}$ in a container of fixed volume.
Given : $\Delta \mathrm{H}_{f}^{\circ}\left(\mathrm{CaCO}_{3}, \mathrm{~s}ight)=-1207 \mathrm{~kJ} / \mathrm{mol}$
$\Delta \mathrm{H}_{f}^{\circ}(\mathrm{CaO}, \mathrm{s})=-635 \mathrm{~kJ} / \mathrm{mol}, \Delta \mathrm{H}_{f}^{\circ}\left(\mathrm{CO}_{2}, \mathrm{~g}ight)=$
$-394 \mathrm{~kJ} / \mathrm{mol} ; \quad\left[ight.$ Use $\left.\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}ight]$
  1. $702.04 \mathrm{~kJ}$
  2. $721.96 \mathrm{~kJ}$
  3. $712 \mathrm{~kJ}$
  4. $721 \mathrm{~kJ}$

Solution

$\begin{aligned} \text { } & \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})-\mathrm{CaCO}_{3}(\mathrm{~s}) \\ \Delta_{\mathrm{f}} \mathrm{H}^{\circ} &=\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CaCO}_{3}ight)-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{CaO})-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{CaCO}_{2}ight) \\ &=-1207-(-635)-(-394) \\ &=-178 \mathrm{~kJ} / \mathrm{mol} \\ & \therefore \quad \Delta \mathrm{U}=\Delta \mathrm{H}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & \Delta \mathrm{U}=-178-\left(\frac{(-1) \times 8.3 \times 300}{1000}ight) \\ &=-175.51 \mathrm{~kJ} \\ & \mathrm{n}_{\mathrm{CaO}}=\frac{224}{56}=4 \\ & \therefore \quad \mathrm{q}_{\mathrm{v}}=\mathrm{n} \Delta \mathrm{r}=4 \times(-175.51) \\ &=-702.04 \mathrm{~kJ} \end{aligned}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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