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$$\ce{CO + H2O <=> CO2 + H2}$$ A stoichiometric mixture of $\ce{CO(g)}$ and $\ce{H2O(g)}$ is allowed to reach equilibrium. $x$ is the mole fraction of $\ce{H2(g)}$ present.

I am meant to find the result that $$K_p=\frac{4x^2}{(1-2x)^2}$$

However, I only get: $K_p=\frac{x^2}{(1-x)^2}$

Here's what I am thinking:

$$K_p=\frac{x_{\ce{CO_2}}x_{\ce{H_2O}}}{x_{\ce{CO}}x_{\ce{H_2}}}$$

But: $x(\ce{CO})=1-\alpha=x(\ce{H_2O})$ and $x(\ce{CO_2})=\alpha=x(\ce{H_2})$.

Then substituting this in gives me my result. What am I doing wrong?

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The initial mole fraction of $\ce{CO}$ present must be $y=0.5$. Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0.

After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.

Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$

I'm sure you can figure out the rest.

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