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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 $\ce{p_{H_2O}=p_{CO}}=x$.
Since their total is 1.5 bar, i.e. $\ce{p_{H_2O} + p_{CO}}=1.5$
Can you solve this equation?
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, $\frac{\ce{[CO_2][H_2]}}{[CO][H_2O]}$.
It is also given that initially, $\ce{p_{H_2O} = p_{CO}}$; from the ideal gas law, $PV=nRT \implies P=\frac{nRT}{V}=kC \text{ }(1)$, where k is some constant and C is concentration.
So initially, $[\ce{H_2O}]=[\ce{CO}]=y$, y being a constant. Let the concentration consumed by the equilibrium shifting to the right be x.
At equilibrium, $\frac{\ce{[CO_2][H_2]}}{[CO][H_2O]}=\frac{x^2}{(y-x)^2}=4$.
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: $[\ce{CO_2}]=[\ce{H_2}]=\frac{2y}{3}, \\ \therefore [\ce{CO}]=[\ce{H_2O}]=\frac{y}{3}$.
So then $\ce{p_{H_2}= p_{CO_2} = 2p_{CO} = 2p_{H_2O}}$ at equilibrium.
So from substitution into Martin's equation, $\ce{p_{H_2O} + p_{CO} + p_{CO_2} + p_{H_2}}=1.5$, we find that $\ce{p_{H_2O} = p_{CO}} = 0.25\text{ bar}$.
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 $0.75$ bar each.
Here's what I found using an ICE table. We let the initial partial pressures of CO and H[sub]2[/sub]O be $x$, and assign an arbitrary value $y$ to the change.
+----------+-----+------+------+-----+
| | CO | H2O | CO2 | H2 |
+----------+-----+------+------+-----+
| p_0 | x | x | 0 | 0 |
+----------+-----+------+------+-----+
| change | -y | -y | +y | +y |
+----------+-----+------+------+-----+
| p_{eq} | x-y | x-y | y | y |
+----------+-----+------+------+-----+
So we find that $2(x-y+y)=1.5 \Rightarrow x = 0.75$ bar, as required.
