Question

The $\Delta H _{ f }^{0}$ of $CaCO _{3}( s ), CaO ( s )$ and $CO _{2}( g )$ are -1206.9 -635.1 and $-393.5 \,kJ / mol$ respectively. Their $S ^{\circ}$ values are 92.9,38.2 and $213.7\, J / K$ respectively at $1000^{\circ} C$ and 1 atm. The partial pressure of $CO _{2}$ in the reaction
$CaCO _{3}( s ) \rightleftharpoons CaO ( s )+ CO _{2}( g )$ is approximately:

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

Answer 1

Answer: (B)

$CaCO _{3}( s ) \rightleftharpoons CaO ( s )+ CO _{2}( g )$

$k _{ p }=\left( p _{ CO _{2}}\right)^{1}$

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

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

$=(-635.1-393.5)+1206.9$

$\Delta_{ f }^{0} H =+178.3 kJ / mol$

$\Delta_{ f }^{0} S =\Sigma S _{ P }^{0}-\Sigma S _{ R }^{0}$

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

$\Delta_{ f }^{0} S =159\, J / mol =0.159 \,kJ / mol$

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

$=178.3-1273 \times 0.159 $

$=178.3-202.407$

$ =-24.107 \,kJ / mol$

Also, $\Delta_{ f }^{0} G =-2.303 RT\, \log \,K$

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

log $K _{ p } =\frac{24.107}{24.374}$

$\log\, K _{ p } =0.98$

$K_{ p } =9.75 \,atm$