Question

A student ran the following reaction in the laboratory at 1100 K: 2SO2(g) O2(g) 2SO3(g) When he introduced SO2(g) and O2(g) into a 1.00 L evacuated container, so that the initial partial pressure of SO2 was 3.43 atm and the initial partial pressure of O2 was 1.61 atm, he found that the equilibrium partial pressure of O2 was 0.809 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.

Answer 1

Kp = 0.949

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction:

`2SO_2(g) +O_2(g) \rightleftharpoons 2SO_3(g)`

It is possible to set up the equilibrium expression as shown below:

`Kp=(p_(SO_3)^2)/(p_(SO_2)^2p_(O_2))`

Whereas the initial pressure of SO2 was 3.43 atm and that of O2 was 1.61 atm. Now, since the partial pressure of O2 decreased to the 0.809 atm, it is possible to calculate the change in the pressure of O2 via:

`x=1.61atm-0.809atm=0.801atm`

Which is actually applied to SO3 and SO2 according to the stoichiometry in the equilibrium expression to calculate Kp:

`Kp=((2*x)^2)/((3.43-2x)^2(0.809))`

Thus, by plugging in x, we obtain:

`Kp=((2*0.801)^2)/((3.43-2*0.801)^2(0.809))\\\\Kp=0.949`

Best regards!