Question

Calculate the heat produced (in $\text{kJ}$ ) when $\text{224}\text{gm}$ of $\text{CaO}$ is completely converted to ${\text{CaCO}}_3$ by reaction with ${\text{CO}}_{\text{2}}$ at ${\text{27}}^{\text{o}}\text{C}$ in a container of fixed volume.
Given : $\Delta H_f^{\circ }\left[CaC{O}_3(s)\right]=-1207kJ/mol;\Delta H_f^{\circ }[CaO(s)]=-635kJ/mol$
$\Delta H_f^{\circ }\left(C{O}_2,g\right)=-394kJ/mol;[$ Use $R=8.3K^{-1}mo{l}^{-1}]$

• A. $-702.04kJ$
• B. $721.96kJ$
• C. $712kJ$
• D. $721kJ$

Answer 1

Answer: (A)

$CaO(s)+C{O}_2(g)\to CaC{O}_3(s)$
${\Delta }_r{H}^{\circ }=\Delta H_f^{\circ }\left(CaC{O}_3\right)-\Delta H_f^{\circ }(CaO)-\Delta H_f^{\circ }\left(C{O}_2\right)$
$=-1207-(-635)-(-394)$
$=-178kJ/mol$
$\therefore \Delta E_r^{\circ }=\Delta H_r^{\circ }-\Delta n_{g}RT$
$\Delta E_r^{\circ }=-178-\frac{(-1)\times 8.3\times 300}{1000}$
$=-175.51kJ$
$n_{CaO}=\frac{224}{56}=4$
$\therefore q_v=n\cdot \Delta r_{r}E=4\times (-175.51)$
$=-702.04kJ$