Suppose I have the equilibrium in a closed container: $$\ce{3A(s) + 4B(g) -> 4C(g) + D(s)}$$
What happens to the partial pressure of C if the volume of the container is halved?
I was pondering about this question, because by the gas equilibrium constant $$K_\mathrm{p} = \frac{[C]^4}{[B]^4}$$ the partial pressure of each should remain equal to each other. But using an ice table, $K_\mathrm{p} = \frac{[C]+x}{[B]-x}$, or you can flip it around. Either way, $x$ should equal 0 by this equation for them to remain the same, which means that the partial pressures don't change, which then means that the gases are turning into the solids. But by the chemical equation 1 mole of one gas will produce 1 mole of the other gas, so how does the gas even turn into the solid? So wouldn't the partial pressures of both increase?
Another way to think about it is to imagine a container with only B and C. That means a reaction cannot even occur, so when volume decreases, the partial pressures increase.
This is my thought on the problem. Can someone please explain to me what really happens?
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