At a particular temperature, a \(3.0-\mathrm{L}\) flask contains 2.4 moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\) , and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$

We are given the initial moles of all three gases and the volume of the flask, which is 3.0 L. To calculate the initial concentrations, divide the moles by the volume: \([Cl_2] = \frac{2.4 \,\text{moles}}{3.0\,\text{L}} = 0.8\,\text{M}\) \([NOCl] = \frac{1.0 \,\text{moles}}{3.0\,\text{L}} = 0.333\,\text{M}\) \([NO] = \frac{4.5 \times 10^{-3} \,\text{moles}}{3.0\,\text{L}} = 1.5 \times 10^{-3} \,\text{M}\)

Write the expression for Q using the initial concentrations: \(Q = \frac{[NO]^2[Cl_2]}{[NOCl]^2}\) Plug in the initial concentrations for Q: \(Q = \frac{(1.5 \times 10^{-3})^2(0.8)}{(0.333)^2} = 1.80 \times 10^{-5}\)

Since we are asked to calculate the equilibrium constant K, we need to find the direction in which the reaction shifts to reach equilibrium. If Q < K, the reaction will shift to the right (towards products). If Q > K, the reaction will shift to the left (towards reactants). We don't know the value of K yet, but we can still set up the shift expressions for both cases: - If the reaction shifts to the right: \(\Delta NOCl = -2x\), \(\Delta NO = +2x\), and \(\Delta Cl_2 = +x\) - If the reaction shifts to the left: \(\Delta NOCl = +2x\), \(\Delta NO = -2x\), and \(\Delta Cl_2 = -x\)

Set up the expressions for final concentrations using the shift expressions from step 3: - If the reaction shifts to the right: \([NOCl] = 0.333 - 2x\), \([NO] = 1.5 \times 10^{-3} + 2x\), and \([Cl_2] = 0.8 + x\) - If the reaction shifts to the left: \([NOCl] = 0.333 + 2x\), \([NO] = 1.5 \times 10^{-3} - 2x\), and \([Cl_2] = 0.8 - x\)

To find K, rewrite the expressions for final concentrations in terms of Q: - If the reaction shifts to the right: \(K = \frac{[(1.5 \times 10^{-3} + 2x)^2(0.8 + x)]}{(0.333 - 2x)^2} = 1.80 \times 10^{-5}\) - If the reaction shifts to the left: \(K = \frac{[(1.5 \times 10^{-3} - 2x)^2(0.8 - x)]}{(0.333 + 2x)^2} = 1.80 \times 10^{-5}\) Since all concentrations are positive, the reaction must shift to the right. Solving the first equation for x: \(K = \frac{[(1.5 \times 10^{-3} + 2x)^2(0.8 + x)]}{(0.333 - 2x)^2} = 1.80 \times 10^{-5}\) Solving for x gives \(x \approx 1.26 \times 10^{-3}\). Now we can calculate the equilibrium constant K using the final concentrations: \(K = \frac{[(1.5 \times 10^{-3} + 2(1.26 \times 10^{-3}))^2(0.8 + 1.26 \times 10^{-3})]}{(0.333 - 2(1.26 \times 10^{-3}))^2} \approx 5.74 \times 10^{-3}\) Therefore, the equilibrium constant K at this temperature for the given reaction is approximately \(5.74 \times 10^{-3}\).