Question 6.9: a. Using data from Table 4.1 (see Appendix A, Data Tables), ...
a. Using data from Table 4.1 (see Appendix A, Data Tables), calculate K_{P} at 298.15 K for the reaction CO\left( g \right)+H_{2}O\left( l \right)\to CO_{2}\left( g \right)+H_{2}\left( g \right).
b. Based on the value that you obtained for part (a), do you expect the mixture to consist mainly of CO_{2}\left( g \right) and H_{2}\left( g \right) or mainly of CO\left( g \right)+H_{2}O\left( l \right) at equilibrium?
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a. \ln K_{P}=-\frac{1}{RT}\Delta G°_{R}
=-\frac{1}{R}\left( ^{\Delta G°_{f}\left( CO_{2,g} \right)+\Delta G°_{f}\left( H_{2,g} \right)-\Delta G°_{f}\left( H_{2} O,l\right)} _{-\Delta G°_{f}\left( CO,g \right)}\right)
=-\frac{1}{8.314 J mol^{-1} K^{-1} ×298.15 K}
×\left( ^{-394.4 ×10^{3}J mol^{-1}+0+237.1}_{×10^{3}J mol^{-1}+137.2 ×10^{3}J mol^{-1}} \right)
=8.1087
k_{P}=3.32 ×10^{3}
b. Because K_{P}\gg 1, the mixture will consist mainly of the products CO_{2}\left( g \right)+H_{2}\left( g \right) at equilibrium.
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