Question

Sulfur oxychloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), decomposes to sulfur dioxide and chlorine gases. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ At a certain temperature, the equilibrium partial pressures of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) are \(1.88 \mathrm{~atm}, 0.84 \mathrm{~atm},\) and \(0.27 \mathrm{~atm}\) respectively. (a) What is \(K\) at that temperature? (b) Enough \(\mathrm{Cl}_{2}\) condenses to reduce its partial pressure to 0.68 atm. What are the partial pressures of all gases when equilibrium is reestablished?

Short answer

Answer: The equilibrium constant \(K_{p}\) at the given temperature is approximately 5.88. When the partial pressure of \(\mathrm{Cl}_{2}\) decreases to 0.68 atm, the new equilibrium partial pressures are approximately 1.72 atm for \(\mathrm{SO_{2}}\), 0.52 atm for \(\mathrm{Cl_{2}}\), and 0.43 atm for \(\mathrm{SO_{2}Cl_{2}}\).

Step-by-step solution

Part (a): Determining the Equilibrium Constant K

To find the equilibrium constant \(K_{p}\), we first write the expression for the equilibrium constant in terms of partial pressures: $$ K_{p} = \frac{P_{\mathrm{SO_{2}}} \cdot P_{\mathrm{Cl_{2}}}}{P_{\mathrm{SO_{2}Cl_{2}}}} $$ We have given equilibrium partial pressures of \(\mathrm{SO_{2}}, \mathrm{Cl_{2}},\) and \(\mathrm{SO_{2} \mathrm{Cl}_{2}}\). Now, we can plug the values into the expression to find the value of \(K\). $$ K_{p} = \frac{1.88 \cdot 0.84}{0.27} $$ Calculate the value of \(K_{p}\).

Part (a): Calculating the Value of K

Using the equilibrium partial pressures we found, solve for \(K_{p}\). $$ K_{p} = \frac{1.88 \cdot 0.84}{0.27} \approx 5.88 $$ The equilibrium constant \(K_{p}\) at this temperature is 5.88.

Part (b): Le Chatelier's Principle and the Changes in Partial Pressures

Now, let's see how the equilibrium is affected when the partial pressure of \(\mathrm{Cl}_{2}\) decreases to 0.68 atm. According to Le Chatelier's principle, the equilibrium will shift to oppose the changes. In this case, the equilibrium will shift to the left side, meaning the reaction will consume more \(\mathrm{Cl_{2}}\) and produce more \(\mathrm{SO_{2}Cl_{2}}\). Let the change in partial pressure of \(\mathrm{SO_{2}}\) be denoted as \(x\). New partial pressures will be: 1. Partial pressure of \(\mathrm{SO_{2}} = 1.88-x\) 2. Partial pressure of \(\mathrm{Cl_{2}} = 0.68-x\) 3. Partial pressure of \(\mathrm{SO_{2}Cl_{2}} = 0.27+x\) For these new partial pressures, we need to make sure the equilibrium constant remains the same.

Part (b): Solving for the Change in Partial Pressures (x)

According to Le Chatelier's principle, the value of \(K_{p}\) should not change, so we can set up the following equation using the new partial pressures: $$ K_{p} = \frac{(1.88 - x) \cdot (0.68 - x)}{(0.27 + x)} $$ We know the value of \(K_{p} = 5.88\). Solving for \(x\), we get: $$ 5.88 = \frac{(1.88 - x) \cdot (0.68 - x)}{(0.27 + x)} $$ Solve this equation to find the value of \(x\).

Part (b): Finding the New Equilibrium Partial Pressures

Upon solving the equation for \(x\), we find that \(x \approx 0.16\). Now we can find the new equilibrium partial pressures: 1. Partial pressure of \(\mathrm{SO_{2}} = 1.88-0.16 \approx 1.72~\mathrm{atm}\) 2. Partial pressure of \(\mathrm{Cl_{2}} = 0.68-0.16 \approx 0.52~\mathrm{atm}\) 3. Partial pressure of \(\mathrm{SO_{2}Cl_{2}} = 0.27+0.16 \approx 0.43~\mathrm{atm}\) The new equilibrium partial pressures when the partial pressure of \(\mathrm{Cl}_{2}\) decreases to 0.68 atm are approximately 1.72 atm for \(\mathrm{SO_{2}}\), 0.52 atm for \(\mathrm{Cl_{2}}\), and 0.43 atm for \(\mathrm{SO_{2}Cl_{2}}\).