# 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 $0.75~\mathrm{bar}$.

• Are we meant to assume that the system initially has no $\ce{H_2} \text{ or } \ce{CO_2}$? – Sherlock Holmes Oct 23 '14 at 5:57

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.

• I am not so sure, this is the right (generic) way to solve this problem, since it is given as $$p_e(\ce{CO}) + p_e(\ce{H2O}) + p_e(\ce{CO2}) + p_e(\ce{H2}) = 1.5~\mathrm{bar}$$ – Martin - マーチン Oct 23 '14 at 5:52
• @Martin Is it possible to solve this for the case where there is some $\ce{CO_2}$ or $\ce{H_2}$ present in varying amounts? I can't seem to do the algebra.. – Sherlock Holmes Oct 23 '14 at 6:24
• I don't understand what you mean, do you refer to $p(\ce{CO2})\neq p(\ce{H2})$ at equilibrium?|| I highly recommend you use a clearer way of presenting your calculations. I was initially not able to follow your thoughts, until I did the calculation myself. Using $K$ in two instances is never a good idea (substitution of physical properties with a nameless variable $x$ is usually not a good idea.) – Martin - マーチン Oct 23 '14 at 6:58
• Yes, that and my assumption that there was no $\ce{CO_2}$ or $\ce{H_2}$ initially. And thanks for pointing that out, I shall endeavour to clean it up and make it easier to follow in future. – Sherlock Holmes Oct 23 '14 at 7:35

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.