Finding initial pressure in a chemical equilibrium

Consider the equilibrium below. If $K = 4$ and initial pressures of $\ce{CO}$ and $\ce{H2O}$ are equal, giving a total pressure of $1.5~\mathrm{bar}$ at equilibrium. What was the initial pressure of $\ce{CO}$ or $\ce{H2O}$?


$$\ce{CO(g) + H2O (g) <=> CO2 (g) + H2 (g)}$$


I don't know how to get the initial pressures in this situation. Do I have to set up an ICE chart? I'm not sure how to start, any help would be appreciated. The answer is apparently $0.75~\mathrm{bar}$.




If the initial pressure of $\ce{CO}$ and $\ce{H\_2O}$ are equal, then let $p\_{\ce{H\_2O}}=p\_{\ce{CO}}=x$.


Since their total is 1.5 bar, i.e. $\ce{p\_{H\_2O} + p\_{CO}}=1.5$


Can you solve this equation?




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Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.


It is given that $K=4$. From the equation for the equilibrium constant, $K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$.


It is also given that initially, $\ce{p\_{H2O} = p\_{\ce{CO}}}$; from the ideal gas law,


$PV=nRT \implies P=\frac{nRT}{V}=kC \tag{1}$


where $k$ is some constant and C is concentration.


So initially, $[\ce{H2O}]=[\ce{CO}]=y$, where $y$ is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.


At equilibrium, $\frac{\ce{[CO2][H\_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$.


Solving this equation gives $x=\frac{2y}{3}$.


Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.



> 
>  Now, from $(1)$, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: $[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$, therefore $[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$.
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So then $p\_{\ce{H2}}= p\_{\ce{CO2}} = 2p\_{\ce{CO}} = 2p\_{\ce{H2O}}$ at equilibrium.


So from substitution into Martin's equation, $\ce{p\_{H\_2O} + p\_{CO} + p\_{\ce{CO2}} + p\_{\ce{H2}}}=1.5$, we find that $p\_{\ce{H2O}} = p\_{\ce{CO}} = \pu{0.25 bar}$.


But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is $\pu{0.75 bar}$ each.


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Here's what I found using an ICE table. We let the initial partial pressures of $\ce{CO}$ and $\ce{H2O}$ be $x$, and assign an arbitrary value $y$ to the change.


$\begin{array}{|c|c|c|c|c|}
\hline
 &\ce{CO} & \ce{H2O} & \ce{CO2} & \ce{H2} \\
\hline
 p\_0 & x & x & 0 & 0 \\
\hline
 \text{change} & -y & -y & +y & +y \\
\hline
 p\_{\text{eq}} & x-y & x-y & y & y \\
\hline
\end{array}$


So we find that $2(x-y+y)=1.5 \Rightarrow x = 0.75$ bar, as required.


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