Question

If Kc = 1.7E8 at 600 K and reacts according to the equation 2SO2(g) + O2(g) <=> 2SO3(g), and at equilibrium, the partial pressure of SO3 is 200 atm and the partial pressure of O2 is 170 atm, what is the partial pressure of SO2?

I tried making an ICE table, and I got that x is equal to 100, but I can't apply it to my SO2 column because I don't know the initial partial pressure. I even tried to allow the initial SO2 pressure to be a variable and had my equilibrium = A - 2x where A is the initial pressure. But that didn't turn out right. How do I do this?

Answer 1

An ICE table works when you're dealing with a system that IS NOT in equilibrium! Remember, the "C" stands for change - this is the change each reactant/product must undergo to reach an equilibrium concentration/partial pressure.

The question tells you that you're system is ALREADY in equilibrium. That means you can just set up your Kc equation, including the partial pressures already given, and solve for the unknown! Hope that helps!

Answer 2

Be careful in your problem you have the value of Kc (that is defined based in the equilibrium concentrations) and Kp is defined with the equilibrium partial pressures.
Besides as @matt101 mention your system it is at the Equilibrium, the value of Kc given in your problem is going to be different that the Kp that you need to solve the problem because the change in number of moles is different than 0.

Answer 3

You're right @Cuanchi - that's what happens when I read the question too fast!

@kyeong the way to solve the problem is the same, just convert Kc into Kp using the following equation: Kp = Kc (RT)^Δn

Does that equation look familiar?

Answer 4

Yeah, it does! Thank you so much!