Answer to Question #111108 in Chemistry for Jshacona Abbott

Question

When a 2:1 mixture of SO2 and O2 at a total initial pressure of 5 atm is passed over a catalyst at 425 degrees Celsius the partial pressure of SO3 at equilibrium is found to be 3 atm. Calculate Kp.

Answer 1

Let's write the balanced reaction equation first:

2SO₂ + O₂ $\rightarrow$ 2SO₃.

As you see, 2 mol of SO₂ react with 1 mol of O₂ and 2 mol of SO₃ is produced. Therefore, the expression for K_p is:

$K_P = \frac{p_{SO_3}^2}{p_{SO_2}^{2} p_{O_2}}$ .

Note that $p_i$ are the partial pressures of SO₂,O₂ and SO₃ at the equilibrium.

The total initial pressure of the SO₂/O₂ 2:1 mixture is 5 atm. Therefore, the initial partial pressures of SO₂ and O₂ are 2·5/3 atm and 1·5/3 atm, respectively.

We know that the partial pressure of SO₃ at the equilibrium is 3 atm. Therefore, the partial pressure of SO₂ decreased by 3 atm and the partial pressure of O₂ decreased by 3/2 atm. The equilibrium pressures of SO₂ and O₂ are : (2·5/3 - 3) atm and (1·5/3 - 3/2) atm. Therefore, the pressure constant is:

$K_P = \frac{(3^2)}{(2·5/3-3)^2(1·5/3-3/2)} = 9·9·6 = 486 \text{ atm}^{-1}$

Answer: 486 atm^-1