The reaction,CaCO_{3(s)} rightleftharpoons Ca | vedclass

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(A) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = \Delta G_{f}^{\circ}(CaO) + \Delta G_{f}^{\circ}(CO_{

Vedclass · Accepted Answer

$1.13 	imes 10^{-23}$

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(A) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = \Delta G_{f}^{\circ}(CaO) + \Delta G_{f}^{\circ}(CO_{2}) - \Delta G_{f}^{\circ}(CaCO_{3})$.Substituting the given values: $\Delta G^{\circ} = (-603.501) + (-394.389) - (-1128.79) = +130.9 \, kJ \, mol^{-1} = 130900 \, J \, mol^{-1}$.Using the relation $\Delta G^{\circ} = -RT \ln K_{p}$,we have $\ln K_{p} = -\frac{\Delta G^{\circ}}{RT}$.$\ln K_{p} = -\frac{130900}{8.314 	imes 298} \approx -52.834$.Converting to base $10$ logarithm: $\log_{10} K_{p} = \frac{-52.834}{2.303} \approx -22.941$.Thus,$K_{p} = 10^{-22.941} \approx 1.13 	imes 10^{-23}$.Since $K_{p} = P_{CO_{2}}$,the partial pressure of $CO_{2(g)}$ is $1.13 	imes 10^{-23} \, atm$.

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