Browse Questions
Thermodynamics

# Estimate the value of $\Delta S^{\circ}_{298}$ for the dissociation of calcium carbonate given that, $\Delta H^{\circ}_{298}$ for the reaction is $178\: kJ$. The equilibrium pressure is one bar at $897^{\circ}C.(R = 8.314\: J/Kmol)$

$\begin {array} {1 1} (A)\;132.66\: J/K & \quad (B)\;-152.15 \: J/K \\ (C)\;-132.66\: J/K & \quad (D)\;152.15 \: J/K \end {array}$

The equilibrium is $CaCO_3(s) \leftrightharpoons CaO(s) + CO_2(g)$
Let the equilibrium constant at $897^{\circ}C\: be\: K_{eq}$
So, $K_{eq} = P_{CO2 } = 1\: bar$
Let equilibrium constant at $298 \: K \: be\: K’_{eq}$
$log\large\frac{ K’_{eq}}{K_{eq}}$$= \large\frac{ \Delta H^{\circ}}{2.303R}$$ \bigg[ \large\frac{(T_2 – T_1)}{T_1T_2} \bigg]$$= \large\frac{178 \times 10^3}{(2.303\times8.314)}$$ \times \bigg[ \large\frac{(298 – 1170)}{(298 \times1170)} \bigg]$
$log\: K’_{eq } - log\: K_{eq} = -23.25$
$\Rightarrow log\: K’_{eq} – log \: 1 = -23.25$
$\Rightarrow log\: K’_{eq} = -23.25 + log\: 1 = -23.25$
Now, $\Delta G^{\circ}_{298} = -2.303 \times RT+log K’_{eq} = -2.303 \times 8.314 \times 298 \times (-23.25)$
$\Rightarrow \Delta G^{\circ}_{298} = \Delta H^{\circ}_{298} - T \Delta S^{\circ}_{298} = 132660.9 \: J = 132.66\: kJ$
$\Rightarrow \Delta S^{\circ}_{298} = \large\frac{\Delta H^{\circ}_{298} - \Delta G^{\circ}_{298}}{ T}$$= \large\frac{(178-132.66)}{298}$$ = 15215\: J/K = 152.15\: J/K$
Ans : (D)