Question

The $\Delta H_f^0$ of $CaC{O}_3(s),CaO(s)$ and $C{O}_2(g)$ are -1206.9 -635.1 and $-393.5kJ/mol$ respectively. Their $S^{\circ }$ values are 92.9,38.2 and $213.7J/K$ respectively at $1000^{\circ }C$ and 1 atm. The partial pressure of $C{O}_2$ in the reaction
$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$ is approximately:

• A. 1.25 atm
• B. 9.75 atm
• C. 0.03 atm
• D. 0.98 atm

Answer 1

Answer: (B)

$CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$

$k_p={\left(p_{C{O}_2}\right)}^1$

${\Delta }_f^{0}H=\Sigma {\Delta }_f^0{H}_{\text{(Products) }}-\Sigma {\Delta }_f^0{H}_{\text{(Reactants) }}$

$=(-635.1)+((-393.5))-(-1206.9)$

$=(-635.1-393.5)+1206.9$

${\Delta }_f^{0}H=+178.3kJ/mol$

${\Delta }_f^{0}S=\Sigma S_P^0-\Sigma S_R^0$

$=(38.2+213.7)-(92.9)$

${\Delta }_f^{0}S=159J/mol=0.159kJ/mol$

Apply, ${\Delta }^{\circ }G={\Delta }^{\circ }H-T\Delta S^{\circ }$

$=178.3-1273\times 0.159$

$=178.3-202.407$

$=-24.107kJ/mol$

Also, ${\Delta }_f^{0}G=-2.303RT\log K$

$-24.107=-2.303\times \frac{8.314}{1000}\times 1273\log Kp$

log $K_p=\frac{24.107}{24.374}$

$\log K_p=0.98$

$K_p=9.75atm$