As shown in Table \(15.2, K_{p}\) for the equilibrium $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ is \(4.39 \times 10^{-9}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at $450^{\circ} \mathrm{C}$. If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) $9.93 \mathrm{MPa} \mathrm{NH}_{3}, 4.56 \mathrm{MPa} \mathrm{N}_{2}, 5.57 \mathrm{MPa} \mathrm{H}_{2}$ (b) \(5.78 \mathrm{MPa} \mathrm{NH}_{3}, 14.49 \mathrm{MPa} \mathrm{N}_{2},\) no \(\mathrm{H}_{2}\) (c) $1.32 \mathrm{MPa} \mathrm{NH}_{3}, 2.74 \mathrm{MPa} \mathrm{N}_{2}, 8.31 \mathrm{Mpa} \mathrm{H}_{2}$

The reaction quotient, Qp, is given by the following expression for the given reaction: \[ Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \] Calculate Qp for each given mixture using the pressure (P) values provided: (a) For the mixture \( 9.93 \mathrm{MPa} \mathrm{NH}_{3}, 4.56 \mathrm{MPa} \mathrm{N}_{2}, 5.57 \mathrm{MPa} \mathrm{H}_{2} \): \[ Q_{p(a)} = \frac{ (9.93)^2 }{ (4.56)(5.57)^3 } \] (b) For the mixture \( 5.78 \mathrm{MPa} \mathrm{NH}_{3}, 14.49 \mathrm{MPa} \mathrm{N}_{2}, \) no H2: Since there is no H2 in this mixture, the denominator of the Qp expression will be zero, resulting in an undefined value for Qp. We can conclude that this mixture is not at equilibrium since it requires some amount of H2 for the reaction to take place. (c) For the mixture \( 1.32 \mathrm{MPa} \mathrm{NH}_{3}, 2.74 \mathrm{MPa} \mathrm{N}_{2}, 8.31 \mathrm{Mpa} \mathrm{H}_{2} \): \[ Q_{p(c)} = \frac{ (1.32)^2 }{ (2.74)(8.31)^3 } \]

Now, compare the Qp values we obtained in Step 1 with the given Kp value: (a) \( Q_{p(a)} = 0.224 \) (approx.) Since \( Q_{p(a)} > K_p \), the mixture must shift towards the reactants (N2 and H2). (b) The mixture is not at equilibrium since there is no H2 present. Adding H2 will cause the reaction to proceed towards the formation of NH3. (c) \( Q_{p(c)} = 4.00 \times 10^{-9} \) (approx.) Since \( Q_{p(c)} < K_p \), the mixture must shift towards the products (NH3).

To summarize, all three mixtures are not at equilibrium at \( 450^{\circ}C \). Mixture (a) will shift towards the reactants (N2 and H2), mixture (b) will shift towards the products (NH3) after addition of H2, and mixture (c) will also shift towards the products (NH3).