The following equilibrium pressures were observed at a certain temperature for the reaction $$ \begin{array}{c} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\ P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{~atm} \\ P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{~atm} \end{array} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{~atm}\), and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

We first need to write the general expression for the equilibrium constant \(K_p\) for the reaction given: \(\textrm{N}_2(g)+3 \textrm{H}_2(g) \rightleftharpoons 2 \textrm{NH}_3(g)\) The equilibrium constant expression for this reaction is as follows: $$K_p = \frac{(P_{\textrm{NH}_3})^2}{(P_{\textrm{N}_2})(P_{\textrm{H}_2})^3}$$

Now we need to plug in the given equilibrium pressures into the expression for \(K_p\) and calculate the value: $$K_p = \frac{(3.1 \times 10^{-2} \textrm{~atm})^2}{(8.5 \times 10^{-1} \textrm{~atm})(3.1 \times 10^{-3} \textrm{~atm})^3}$$ After solving this expression, we get: $$K_p \approx 6.3 \times 10^3$$

Now we need to check if the new pressure values provided represent a system in equilibrium. For that, we need to calculate the reaction quotient \(Q_p\), and compare it with the \(K_p\) we just found. The expression for the reaction quotient is similar to the expression for equilibrium constant: $$Q_p = \frac{(P'_{\textrm{NH}_3})^2}{(P'_{\textrm{N}_2})(P'_{\textrm{H}_2})^3}$$ Substitute the new pressures into the expression and calculate \(Q_p\): $$Q_p = \frac{(0.0167 \textrm{~atm})^2}{(0.525 \textrm{~atm})(0.00761 \textrm{~atm})^3}$$ After solving this expression, we get: $$Q_p \approx 6.1 \times 10^3$$ Now, compare the values of \(Q_p\) and \(K_p\). If \(Q_p = K_p\), the system is in equilibrium. If \(Q_p < K_p\), the system is not in equilibrium, and the reaction will shift to the right (forward) to reach equilibrium. If \(Q_p > K_p\), the system is not in equilibrium, and the reaction will shift to the left (reverse) to reach equilibrium. In our case, \(Q_p \approx 6.1 \times 10^3\) and \(K_p \approx 6.3 \times 10^3\). The values are close but not equal. The reaction is not in equilibrium, but it is only slightly off. The reaction will shift to the right (forward) to reach equilibrium.