How does the partial pressure of gas C change with the change in volume of the container?

Question

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?

Answer 1

The equilibrium constant $K_P$ is independent of pressure but does depend on temperature which we assume to be constant. The equilibrium constant $K_P$ is the ratio of partial pressures $p_i$ of the gaseous substances raised to the stoichiometric powers, $K_P= p_C^4/p_B^4$ and is defined at a standard state of $1$ atm. pressure.

Writing the equilibrium constant for the reaction $\ce{mB <=>lC}$ in terms of the mole fraction $x_i=n_i/n$ for n total moles gives $K_x= x_C^l/x_B^m$ where l and m are constants, both four in your question. The mole fraction i of a species behaving as a perfect gas can also be written $x_i=p_i/P$ where $p_i$ is the partial pressure and P the total pressure.

Rewriting the equilibrium constant in terms of partial pressures of B and C gives $$K_x=\frac{p_C^l}{p_B^m}P^{m-l}= K_PP^{\Delta n}$$ where $\Delta n$ is the change in the number of moles of the gases. Thus unless the change in the number of moles is zero, which it is in your example, the mole fractions of the components will change with change in total pressure even though $K_P$ is constant. If $\Delta n > 0$ the pressure will reduce the mole fraction of products.

Answer 2

If you half the volume then pressure will double. Now by Le Chatelier's principle the equilibrium will shift in order to lessen the pressure, which means the equilibrium must shift to that side where number of moles of gaseous species are less.But in your reaction number of gaseous moles on both sides of the equation are same.Hence there will be no change in equilibrium.