# Le Chatelier’s principle when pressure of gases increased with the same stoichiometric coefficient

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

What will happen to the equilibrium when the pressure of gases is increased? What will happen to the yield of $\ce{H2}$?

Since pressure increased means the volume is decreased, equilibrium will shift to the direction with the least number of moles. But since the stoichiometric coeffients are all the same, am I right to say that it will not shift?

But I doubt my answer as the question states - "pressure of gases increased". The reactants include solid carbon, meaning I do not count that in. Does this mean that equilibrium shifts to the right?

Your thinking is on the right track. However, what you really need to look out for is number of gas moles.

In order to understand this, think in terms of the equilibrium constant. The equilibrium constant for this reaction is: $$K_\text{p}=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$ Remember, since carbon is a solid, changing the volume/pressure is not going to change anything about it, so it is not in the equilibrium constant.

Since we are introducing a supposed change into the system, we have to write its reaction quotient. The reaction quotient, $Q$, is just a way to determine which way the reaction will have to shift in order to return to equilibrium. We write it in pretty much the same way: $$Q=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$ Even though it looks the same, remember that we are calculating this after introducing a change. So what we really plug into this equation is the following, where $x$ is the factor by which we increase pressure (suppose that $P_{\text{CO}_2}$, $P_{\text{H}_2}$, and $P_{\text{H}_2\text{O}}$ are the equilibrium pressures): $$Q=\frac{(P_{\text{CO}_2}\cdot x)(P_{\text{H}_2}\cdot x)}{(P_{\text{H}_2\text{O}}\cdot x)}$$ $$Q=\frac{x^2P_{\text{CO}_2}P_{\text{H}_2}}{xP_{\text{H}_2\text{O}}}$$ Since we are increasing pressure, $x$ is obviously positive. Thus, $Q>K_\text{p}$.

Due to Le Chatelier's principle, since we introduced a stress into the system, the system will shift away from that stress. $Q$ and $K_\text{p}$ are a good way of mathematically representing this stress. $Q$ is essentially a representation of the system's current state, and $K_\text{p}$ is the representation of the system's ideal state.

So, how do we reach this ideal state? It's simple; we want $Q$ to equal $K_\text{p}$. Obviously this will happen through a change in pressure. Since $Q$ is too big, we need to decrease the numerator in order to make it equal to $K_\text{p}$. The numerator represents the pressure of the products, so the reaction will shift away from the products, using up more products and creating more reactants, until it reaches equilibrium.

Hope that helps.

First of all, How can you practically increase the pressure of the gases individually when they are all present in the reaction vessel in some definite proportion and they are all in the equilibrium state? You can only increase the pressure of the whole system. That's the only thing possible.

Here $\Delta n_\mathrm{g} >0$. According to Le Chatelier's principle, increase in pressure favours the backward reaction. So, equilibrium shifts to the left.

If you have any doubt in the statement "If $\Delta n_\mathrm{g} > 0$, pressure favours the backward reaction. See my answer in the opposite analogous statement of If $\Delta n_\mathrm{g} > 0$, pressure favours the forward reaction.

Pressure dependency in Haber Bosch ammonia synthesis