Question

Consider the reaction between nitrogen and steam: $$ 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) $$ At a certain temperature, \(K=28.6 .\) Calculate the equilibrium partial pressure of steam if \(P_{\mathrm{NH}_{3}}=1.75\) atm, \(P_{\mathrm{O}_{2}}=0.963 \mathrm{~atm},\) and \(P_{\mathrm{N}_{2}}=0.996\) atm at equilibrium.

Short answer

The equilibrium partial pressure of steam is approximately 1.610 atm.

Step-by-step solution

Write the expression for the equilibrium constant (K) in terms of partial pressures

The equilibrium constant K can be expressed as: $$ K = \frac{(P_{\mathrm{NH}_{3}})^4(P_{\mathrm{O}_{2}})^3}{(P_{\mathrm{N}_{2}})^2(P_{\mathrm{H}_{2}\mathrm{O}})^6} $$ where each partial pressure is raised to the power of the coefficient of the respective substance in the balanced chemical equation.

Substitute the equilibrium constant and given partial pressures into the equation

Plug in the given values for K, \(P_{\mathrm{NH}_{3}}\), \(P_{\mathrm{O}_{2}}\), and \(P_{\mathrm{N}_{2}}\): $$ 28.6 = \frac{(1.75)^4(0.963)^3}{(0.996)^2(P_{\mathrm{H}_{2}\mathrm{O}})^6} $$

Solve for \(P_{\mathrm{H}_{2}\mathrm{O}}\)

Rearrange the equation to isolate \(P_{\mathrm{H}_{2}\mathrm{O}}\) on one side: $$ P_{\mathrm{H}_{2}\mathrm{O}}^6 = \frac{(1.75)^4(0.963)^3}{(0.996)^2 \times 28.6} $$ Now, calculate the value of the right-hand side and take the sixth root to find the equilibrium partial pressure of steam: $$ P_{\mathrm{H}_{2}\mathrm{O}} = \sqrt[6]{\frac{(1.75)^4(0.963)^3}{(0.996)^2 \times 28.6}} \approx 1.610\, \mathrm{atm} $$ Therefore, the equilibrium partial pressure of steam is approximately \(1.610\) atm.