Question 6.12: Using the preceding discussion and the tabulated values of Δ...
Using the preceding discussion and the tabulated values of \Delta G°_{f} and \Delta H°_{f} in Appendix A, calculate the CO_{2}\left( g \right) pressure in equilibrium with a mixture of CaCO_{3}\left( s \right) and CaO\left( s \right) at 1000.,1100.,and 1200. K.
For the reaction CaCo_{3}\left( s \right) \rightleftharpoons CaO\left( s \right)+ O_{2}\left( g \right),\Delta G°_{R}=131.1 KJ mol^{-1} and \Delta H°_{R}=178.5 KJ mol^{-1} at 298.15 K. We use Equation (6.71) to calculate K_{P}\left( 298.15 K \right) and Equation (6.66) to calculate K_{P} at elevated temperatures.
\ln K_{P}=\ln\frac{P_{CO2}}{P°}=-\frac{\Delta G°_{R}}{RT} (6.71)
\ln K_{P} \left( T_{f} \right)=\ln K_{P}\left( 298.15 K \right)-\frac{\Delta H°_{R}}{R}\left( \frac{1}{T_{f}}-\frac{1}{298.15 K} \right) (6.66)
\ln\frac{P_{CO2}}{P°}\left( 1000. K \right) =\ln K_{P}\left( 1000. K \right)
=\ln K_{P}\left( 298.15 K \right)-\frac{\Delta H°_{R,298.15 K}}{R}\left( \frac{1}{1000. K}-\frac{1}{298.15 K} \right)
=\frac{-131.1 × 10^{3} J mol^{-1}}{8.314 J K^{-1} mol^{-1} × 298.15 K}-\frac{178.5 ×10^{3} J mol^{-1} }{8.314 J K^{-1} mol^{-1}}
×\left( \frac{1}{1000. K} -\frac{1}{298.15 K}\right)
=-2.348; P_{CO2}\left( 1000. K \right)=0.0956 bar
The values for P_{CO2} at 1100. K and 1200. K are 0.673 bar and 1.23 bar.