Question #111108
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.
1
Expert's answer
2020-04-22T09:19:33-0400

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 Kp is:

KP=pSO32pSO22pO2K_P = \frac{p_{SO_3}^2}{p_{SO_2}^{2} p_{O_2}} .

Note that pip_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:

KP=(32)(25/33)2(15/33/2)=996=486 atm1K_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



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