Questions: Practice Exercise The KP for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) is 4.3 × 10^-4 at 375°C. In one experiment, the initial pressures are: PH2 = 0.40 atm, PN1 = 0.86 atm, and PNH3 = 4.4 × 10^-3 atm. Calculate ΔG for the reaction and predict the direction of the net reaction.

Solution Steps

The equilibrium constant \( K_P \) for the reaction is given by:

\[ K_P = \frac{(P_{\mathrm{NH}_3})^2}{(P_{\mathrm{N}_2})(P_{\mathrm{H}_2})^3} \]

The reaction quotient \( Q_P \) is calculated using the initial pressures:

\[ Q_P = \frac{(P_{\mathrm{NH}_3})^2}{(P_{\mathrm{N}_2})(P_{\mathrm{H}_2})^3} \]

Substitute the given initial pressures:

\[ Q_P = \frac{(4.4 \times 10^{-3})^2}{(0.86)(0.40)^3} \]

Calculate the numerator and the denominator separately:

\[ \text{Numerator} = (4.4 \times 10^{-3})^2 = 1.936 \times 10^{-5} \]

\[ \text{Denominator} = (0.86)(0.40)^3 = (0.86)(0.064) = 0.05504 \]

Now, calculate \( Q_P \):

\[ Q_P = \frac{1.936 \times 10^{-5}}{0.05504} \approx 3.517 \times 10^{-4} \]

Given \( K_P = 4.3 \times 10^{-4} \) and \( Q_P \approx 3.517 \times 10^{-4} \):

Since \( Q_P < K_P \), the reaction will proceed in the forward direction to reach equilibrium.

The change in Gibbs free energy \(\Delta G\) is given by:

\[ \Delta G = \Delta G^\circ + RT \ln Q_P \]

At equilibrium, \(\Delta G^\circ = -RT \ln K_P\). Therefore,

\[ \Delta G = -RT \ln K_P + RT \ln Q_P = RT \ln \left( \frac{Q_P}{K_P} \right) \]

Substitute the values (\(R = 0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1}\), \(T = 375 + 273 = 648 \, \text{K}\)):

\[ \Delta G = (0.0821)(648) \ln \left( \frac{3.517 \times 10^{-4}}{4.3 \times 10^{-4}} \right) \]

Calculate the ratio inside the logarithm:

\[ \frac{3.517 \times 10^{-4}}{4.3 \times 10^{-4}} \approx 0.817 \]

Now, calculate the natural logarithm:

\[ \ln(0.817) \approx -0.2027 \]

Finally, calculate \(\Delta G\):

\[ \Delta G = (0.0821)(648)(-0.2027) \approx -10.77 \, \text{J/mol} \]

Final Answer