Question 19.8: Calculating K from the Standard Free-Energy Change (Molecula...
Calculating K from the Standard Free-Energy Change (Molecular Equation)
Find the value of the equilibrium constant K at 25^{\circ} \mathrm{C}(298 \mathrm{~K}) for the reaction
2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
The standard free-energy change, \Delta G^{\circ}, at 25^{\circ} \mathrm{C} equals -13.6 \mathrm{~kJ}. (We calculated this value of \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} just before Section 19.4.)
PROBLEM STRATEGY
Rearrange the equation \Delta G^{\circ}=-R T \ln K to give
\ln K=\frac{\Delta G^{\circ}}{-R T}
\Delta G^{\circ} and R must be in compatible units. You normally express \Delta G^{\circ} in joules and set R equal to 8.31 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol}).
Substituting numerical values into this equation,
\ln K=\frac{-13.6 \times 10^{3}}{-8.31 \times 298}=5.49
Hence,
K=e^{5.49}=2.42 \times 10^{2}
Note that although the value of K indicates that products predominate at equilibrium, K is only moderately large. You would expect that the composition could be easily shifted toward reactants if you could remove either \mathrm{NH}_{3} or \mathrm{CO}_{2} (according to Le Chatelier’s principle). This is what happens when urea is used as a fertilizer. As \mathrm{NH}_{3} is used up, more \mathrm{NH}_{3} is produced by the decomposition of urea.