To solve this problem, we will use the equilibrium expression and the mole ratio of the components in the reaction:
Reaction: SO2 + ½ O2 ↔ SO3
Step 1: Define the Initial Moles
Let's assume we have 1 mole of the mixture. Since the mole ratio of SO2 to O2 is 2:1, the total number of moles of the mixture is 3 moles (2 moles of SO2 and 1 mole of O2). Therefore, the initial amounts are:
SO2: 2/3 moles
O2: 1/3 moles
SO3: 0 moles (no SO3 initially)
Step 2: Define the Changes in Moles
At equilibrium, the reaction will proceed until equilibrium is reached. Let the change in the number of moles of SO2 be represented by x, where x is the number of moles of SO2 that reacts. Since the reaction shows that 1 mole of SO2 reacts with ½ mole of O2 to form 1 mole of SO3, the changes in moles of the gases will be:
SO2: 2/3 - x
O2: 1/3 - x/2
SO3: x (since 1 mole of SO3 is produced for every mole of SO2 reacted)
Step 3: Set Up the Equilibrium Expression
We are given that at equilibrium, the mixture contains 87% of SO3 by volume. This means that 87% of the total volume of the gas at equilibrium is SO3. Since volume is directly proportional to the number of moles (at constant temperature and pressure), we can use the mole ratio to find the fraction of SO3.
Let’s calculate the total number of moles at equilibrium:
Total moles at equilibrium = (2/3 - x) + (1/3 - x/2) + x = 1 - x/2
Now, we know that 87% of the total moles are SO3. Therefore, the number of moles of SO3 is 0.87 times the total moles at equilibrium:
x = 0.87 × (1 - x/2)
Step 4: Solve for x
Now, solve the equation for x:
x = 0.87 × (1 - x/2)
x = 0.87 - 0.435x
x + 0.435x = 0.87
1.435x = 0.87
x = 0.87 / 1.435
x ≈ 0.605 moles
Step 5: Calculate the Partial Pressures at Equilibrium
Now that we have x ≈ 0.605, we can calculate the equilibrium concentrations of the gases in terms of their partial pressures. Since the total pressure is 1 atm, and we know the mole fractions, the partial pressures will be proportional to the mole fractions.
At equilibrium:
SO2: (2/3 - x) = (2/3 - 0.605) ≈ 0.060 moles
O2: (1/3 - x/2) = (1/3 - 0.605/2) ≈ 0.092 moles
SO3: x ≈ 0.605 moles
The total number of moles at equilibrium is:
Total moles = 0.060 + 0.092 + 0.605 ≈ 0.757 moles
Now, the partial pressures of the gases are:
SO2: (0.060 / 0.757) × 1 atm ≈ 0.079 atm
O2: (0.092 / 0.757) × 1 atm ≈ 0.122 atm
SO3: (0.605 / 0.757) × 1 atm ≈ 0.799 atm
Step 6: Write the Equilibrium Expression for Kp
The equilibrium constant in terms of partial pressures, Kp, for the reaction is:
Kp = (P_SO3) / (P_SO2 × (P_O2)^1/2)
Substitute the values of the partial pressures:
Kp = (0.799) / (0.079 × (0.122)^(1/2))
Step 7: Calculate Kp
First, calculate the square root of 0.122:
(0.122)^(1/2) ≈ 0.349
Now, substitute into the Kp equation:
Kp = 0.799 / (0.079 × 0.349)
Kp ≈ 0.799 / 0.0276
Kp ≈ 28.9 atm
Final Answer:
The equilibrium constant, Kp, is approximately 28.9 atm.
