Question

${K_P}$ for the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$ at ${400^o}C$ is $1.64 \times {10^{ - 4}}$ . Find ${K_C}$ . Also find $\Delta {G^o}$ using ${K_P}$ and ${K_C}$values and interpret the difference.

Answer 1

Hint: he standard Gibbs free energy of formation of a compound is the change in Gibbs free energy that consists of the formation of one mole of that substance from its component elements, at their standard states which is the most stable form of the element at \[25{\text{ }}^\circ C\] and\[100{\text{ }}kPa\] . Its symbol is .

Complete answer:
${K_P}$ for the reaction is the equilibrium constant calculated from the partial pressure of the gases used in the reaction as reactants. As per the question, the reaction given is as follows:
${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$
The value of difference in the number of moles of product and reactants in the reaction will be:
$\Delta {n_g} = {n_p} - {n_r}$
Thus, the value of $\Delta {n_g}$ comes out to be:
$\Delta {n_g} = 2 - 4 = - 2$
${K_C}$ for the reaction is the equilibrium constant calculated from the concentration of the gases used in the reaction as reactants.
The relation between ${K_P}$ and ${K_C}$ is as follows:
${K_P} = {K_C}{(RT)^{\Delta {n_g}}}$
Where, ${K_P}$ for the reaction at ${400^o}C$= $1.64 \times {10^{ - 4}}$
$\Delta {n_g} = 2 - 4 = - 2$
${K_C} = ?$
$R = 0.082Latm{K^{ - 1}}mo{l^{ - 1}}$
$T = {400^o}C = (400 + 273)K = 673K$
Substituting the values and solving, we have:
$1.64 \times {10^{ - 4}} = {K_C}{(0.082 \times 673)^{ - 2}}$
$ \Rightarrow {K_C} = 1.64 \times {10^{ - 4}} \times {(0.082 \times 673)^2} \approx 0.50$
Thus the value of ${K_C}$ is found to be $0.50$.
The relation between equilibrium constant and change in standard Gibbs free energy is:
$\Delta {G^o} = - RT\ln K$
Now, substituting the values of ${K_P}$ and ${K_C}$separately, we get:
$\Delta {G^o} = - RT\ln {K_P}$….(i)
$\Delta {G^o} = - RT\ln {K_C}$….(ii)
Substituting the values in equation (i) and (ii) and on solving, we have:
$\Delta {G^o} = - 8.314 \times 673 \times \ln (1.64 \times {10^{ - 4}}) = - 8.314 \times 673 \times ( - 3.506) = 19617.19J/mol$
The above value is for the change in standard Gibbs free energy, when the equilibrium is taken for the partial pressure of the reactants.
$\Delta {G^o} = - 8.314 \times 673 \times \ln (0.50) = - 8.314 \times 673 \times ( - 0.693) = 3877.55J/mol$
The above value is for the change in standard Gibbs free energy, when the equilibrium is taken for the concentration of the reactants.
The standard free energy change calculated form ${K_P}$ is higher than the standard free energy change calculated from \[{K_C}\] ​as the numerical value of \[{K_C}\]​ is higher than the numerical value of ${K_P}$.

Note:
The value of the universal gas constant should always be decided on the basis of the values of the other variables given in the question and not any random values. In the above reaction of formation of ammonia from nitrogen and hydrogen, the value of ${K_P}$ is less than the value of ${K_C}$.