Problem 92 - Chemistry: An Atoms-Focused Approach

Question

The equilibrium constant \(K_{\mathrm{p}}\) of the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ is 7.69 at \(830^{\circ} \mathrm{C} .\) If a vessel at this temperature initially contains pure \(\mathrm{SO}_{3}\) and if the partial pressure of \(\mathrm{SO}_{3}\) at equilibrium is 0.100 atm, what is the partial pressure of \(\mathrm{O}_{2}\) in the flask at equilibrium?

Step-by-Step Solution

Answer

Answer: The partial pressure of O2 in the flask at equilibrium is approximately 0.0272 atm.

Step 1: Write the expression for Kp

The equilibrium constant Kp for the given reaction can be written as: $$ K_p = \frac{[\mathrm{SO_2}]^2 \cdot [\mathrm{O_2}]}{[\mathrm{SO_3}]^2} $$ where the concentrations represent the partial pressure of each gas at equilibrium.

Step 2: Relate partial pressures at equilibrium

Using stoichiometry, notice that when 2 moles of SO3 dissociate, they produce 2 moles of SO2 and 1 mole of O2. Therefore, if the change in SO3's partial pressure is x, the change in SO2's partial pressure is 2x, and the change in O2's partial pressure is x during the reaction. Since the initial partial pressures of SO2 and O2 are zero, we can write their equilibrium partial pressures as 2x and x, respectively. We're given that the partial pressure of SO3 at equilibrium is 0.100 atm, meaning x = (initial_pSO3 - 0.100 atm).

Step 3: Substitute the pressures into Kp equation

We can now substitute the equilibrium partial pressures in terms of x into the Kp equation: $$ 7.69 = \frac{(2x)^2 \cdot x}{(0.100)^2} $$

Step 4: Solve for x

Now, we can solve this equation for x: $$ 7.69 = \frac{4x^3}{0.01} \\ x^3 = \frac{7.69 \cdot 0.01}{4} \\ x^3 = 0.019225 \\ x = \sqrt[3]{0.019225} \\ x \approx 0.0272 $$

Step 5: Calculate partial pressure of O2

Now that we have x, we can find the partial pressure of O2 by substituting x into the expression for partial pressure of O2 at equilibrium: $$ \mathrm{O_2} = x \\ \mathrm{O_2} \approx 0.0272 \text{ atm} $$ The partial pressure of O2 in the flask at equilibrium is approximately 0.0272 atm.