Question 17.p.4: The information given includes the equilibrium constant, ini......
The information given includes the equilibrium constant, initial pressures of both reactants and equilibrium pressure for one reactant. First, set up a reaction table showing initial partial pressures for reactants and 0 for product. The change to get to equilibrium is to react some of reactants to form some product. The mole ratio between \text{NO:O}_2\text{:NO}_2\text{ is }2: 1 :2. Use the equilibrium quantity for \text{O}_2 and the expression for \text{O}_2 at equilibrium to solve for the change. From the change find the equilibrium partial pressure for \text{NO and NO}_2.\text{ Calculate }K_\text{p} using the equilibrium values.
Use the equilibrium pressures to calculate K_\text{p}.
K_\text{p}=\frac{\text{P}^2_{\text{NO}_2}}{\text{P}^2_\text{NO}\text{P}_{\text{O}_2}} =\frac{(0.988)^2}{(0.012)^2(0.506)} =1.339679\times 10^4=1.3\times 10^4Check: One way to check is to plug the equilibrium expressions for \text{NO and NO}_2 into the equilibrium expression along with 0.506 atm for the equilibrium pressure of \text{O}_2. Then solve for x and make sure the value of x by this calculation is the same as the 0.494 atm calculated above.
1.3\times 10^4=\frac{(2\text{x})^2(0.506)}{(1.000-2\text{x})^2} \\ (1.3\times 10^4)/0.506=2.569\times 10^4=\frac{(2\text{x})^2}{(1.000-2\text{x})^2} \\ \sqrt{2.569\times 10^4} =\frac{(2\text{x})}{(1.000-2\text{x})} \\ \text{x}=0.497The two values for x agree to give K_\text{p} = 1.3\times 10^4.