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In a equilibrium reaction $\ce{CO {(g)} + H2O {(g)} <=> CO2 {(g)} + H2 {(g)}}$, initial concentrations of $\ce{CO}$ and H2O are equal and are 0.3 mol/dm3. What is the equilibrium constant of the reaction if the equilibrium concentrations of CO2 and H2 are equal and are 0.1 mol/dm3?

I tried to solve it this way:

$$\begin{split} \ce{K_r} & = \ce{\frac{[CO2][H2]}{[CO][H2O]} \\ & = \frac{(0.1)(0.1)}{(0.3)(0.3)} \\ & = \frac{(0.1)^2}{(0.3)^2} \\ & = \frac{0.01}{0.09} \\ & \approx 0.11 mol/dm^3} \end{split}$$

But the correct answer is 0.25 mol/dm3

What did I do wrong? What is the right way of solving problems like this one?

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0.1 is the equilibrium concentration of $\ce{CO2}$ and $\ce{H2}$. 0.3 is the initial concentration of $\ce{CO}$ and $\ce{H2O}$. Let's plug 0.1 into an ICE-box as $x$.

\begin{array}{|c|c|c|c|c|} \hline \text{Initial}:& 0.3 & 0.3 & 0 & 0 \\ \hline & \ce{CO} & \ce{H2O} & \ce{CO2} & \ce{H2}\\ \hline \text{Change}: & -x & -x & +x & +x \\ \hline \text{Equilibrium}: & 0.3-0.1 & 0.3-0.1 & 0+0.1 & 0+0.1\\ \hline \end{array}

Mass action expression:

$$K_\mathrm{c} = \frac{(0+0.1)^2}{(0.3-0.1)^2} = 0.25$$

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