91Ó°ÊÓ

Equilibrium refers to a situation in which the concentrations of reactants and products do not change significantly. Although there appears to be no change in equilibrium, this does not imply that all chemical reactions have stopped.

$\mathrm{Δ³Ò}=-\mathrm{R}×\mathrm{T}×\mathrm{lnK}$

As a result, the reaction has reached equilibrium. Calculate the ${\mathrm{Δ\pm á}}^{\mathrm{o}}$and ${\mathrm{Δ³\S }}^{\mathrm{o}}$for the reaction, assuming they are unaffected by temperature changes, to obtain the equilibrium temperature.

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)→2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

${\mathrm{Δ³\S }}^{\mathrm{o}}=\underset{}{\overset{}{∑}}{\mathrm{Δ³\S }}_{\mathrm{products}}^{\mathrm{o}}-\underset{}{\overset{}{∑}}{\mathrm{Δ³\S }}_{\mathrm{reactants}}^{\mathrm{o}}$

$$\begin{gathered} {\mathrm{Δ³\S }}^{\mathrm{o}}=\left[{2\mathrm{mol}×\mathrm{Δ³\S }}_{{\mathrm{NH}}_3\left(\mathrm{g}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}×\mathrm{Δ³\S }}_{{\mathrm{N}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+3\mathrm{mol}×\mathrm{Δ³\S }}_{{\mathrm{H}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right] \\ {\mathrm{Δ³\S }}^{\mathrm{o}}=\left[2\mathrm{mol}×193.00\frac{\mathrm{J}}{\mathrm{mol}×\mathrm{K}}\right]-\left[3\mathrm{mol}×130.60\frac{\mathrm{J}}{\mathrm{mol}×\mathrm{K}}+1\mathrm{mol}×191.50\frac{\mathrm{J}}{\mathrm{mol}×\mathrm{K}}\right] \\ {\mathrm{Δ³\S }}^{\mathrm{o}}=-197.30\frac{\mathrm{J}}{\mathrm{K}} \end{gathered}$$

$$\begin{gathered} {\mathrm{Δ\pm á}}^{\mathrm{o}}=\left[{2\mathrm{mol}×\mathrm{Δ\pm á}}_{{\mathrm{NH}}_3\left(\mathrm{g}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}×\mathrm{Δ\pm á}}_{{\mathrm{N}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+3\mathrm{mol}×\mathrm{Δ\pm á}}_{{\mathrm{H}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right] \\ {\mathrm{Δ\pm á}}^{\mathrm{o}}=[2\mathrm{mol}×(-45.9\mathrm{kJ}/\mathrm{mol}\left)\right]-[1\mathrm{mol}×(0.00\mathrm{kJ}/\mathrm{mol})+3\mathrm{mol}×(0.00\mathrm{kJ}/\mathrm{mol}\left)\right] \\ {\mathrm{Δ\pm á}}^{\mathrm{o}}=-91.8\mathrm{kJ} \end{gathered}$$

$$\begin{gathered} {\mathrm{Δ³Ò}}^{\mathrm{o}}={\mathrm{Δ\pm á}}^{\mathrm{o}}-\mathrm{T}×{\mathrm{Δ³\S }}^{\mathrm{o}}=0 \\ \mathrm{T}=\frac{{\mathrm{Δ\pm á}}^{\mathrm{o}}}{{\mathrm{Δ³\S }}^{\mathrm{o}}} \\ \mathrm{T}=\frac{-91.8×{10}^3\mathrm{J}}{-197.30\frac{\mathrm{J}}{\mathrm{K}}} \\ \mathrm{T}=465.28\mathrm{K} \end{gathered}$$

At a temperature of $465.28\mathrm{K}$, ${\mathrm{K}}_{\mathrm{p}}=1$.

(b)

Assume$\mathrm{T}=400{}^{\mathrm{o}}\mathrm{C}=673\mathrm{K}$.

$$\begin{gathered} \mathrm{Δ³Ò}=\left[-91.8×{10}^3\mathrm{J}\right]-673\mathrm{K}×\left[-197.30\frac{\mathrm{J}}{\mathrm{K}}\right] \\ \mathrm{Δ³Ò}=40982.9\mathrm{J}=40.983\mathrm{kJ} \end{gathered}$$

$\begin{array}{rcl}\mathrm{Δ³Ò}& =& -\mathrm{R}×\mathrm{T}×{\mathrm{lnK}}_{\mathrm{p}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -\frac{\mathrm{Δ³Ò}}{\mathrm{R}×\mathrm{T}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -\frac{40982.9\mathrm{J}}{8.314\frac{\mathrm{J}}{\mathrm{mol}×\mathrm{K}}×673\mathrm{K}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -7.32449\\ {\mathrm{K}}_{\mathrm{p}}& =& {\mathrm{e}}^{{\mathrm{lnK}}_{\mathrm{p}}}\\ {\mathrm{K}}_{\mathrm{p}}& =& 6.592×{10}^{-4}\end{array}$

At ${400}^{\mathrm{o}}\mathrm{C}$, the equilibrium constant is $6.592×{10}^{-4}$.

(c)

A lower ${\mathrm{K}}_{\mathrm{p}}$indicates that product formation is less favorable, resulting in a lower yield.

The process is faster at higher temperatures because the kinetic energy of molecules and entropy are both larger.

They strive to manufacture fewer products in a shorter amount of time in the industry.