Chemistry - Thermochemistry Question with Solution | TestHub

$\text{CaO}\left(\text{s}\right)+{\text{CO}}_2\left(\text{g}\right)\to {\text{CaCO}}_3\left(\text{s}\right)$

${\Delta }_{\text{r}}{\text{H}}^{\circ }=\Delta {\text{H}}_{\text{f}}^{\circ }\left({\text{CaCO}}_3\right)-\Delta {\text{H}}_{\text{f}}^{\circ }\left(\text{CaO}\right)-\Delta {\text{H}}_{\text{f}}^{\circ }\left({\text{CO}}_2\right)$

$=-1207-\left(-635\right)-\left(-394\right)$

$=-\text{178 kJ/mol}$

$\therefore \Delta {\text{E}}_{\text{r}}^{\circ }=\Delta {\text{H}}_{\text{r}}^{\circ }-\Delta {\text{n}}_{\text{g}}RT$

$\Delta {\text{E}}_{\text{r}}^{\circ }=-178-\frac{\left(-1\right)\times \text{8.3}\times 300}{1000}$

$=-\text{175.51 kJ}$

${\text{n}}_{\text{CaO}}=\frac{224}{56}=4$

$\therefore {\text{q}}_{\text{v}}=\text{n}·{\Delta }_{\text{r}}\text{E}=4\times \left(-\text{175.51}\right)$

$=-\text{702.04 kJ}$