# Finding the molar ratio at equilibrium

Consider the reaction $$\ce{SO_2_{(g)} + \frac{1}{2} O_2_{(g)}\longrightarrow SO_3_{(g)}}$$. What effect is there on the molar ratio $$\frac{n_{SO_3}}{n_{SO_2}}$$ at equilibrium if the pressure is increased by reducing the initial volume by half?

Here's what I did: I wrote $$\ce{SO_2_{(g)} + \frac{1}{2} O_2_{(g)} \longrightarrow SO_3_{(g)}}$$
Before the reaction starts, we have $$1$$ mole of $$\ce{SO_2}$$, $$\frac{1}{2}$$ moles of $$\ce{O_2}$$ and $$0$$ moles of $$\ce{SO_3}$$
During the reaction I think we have $$x$$ moles of $$\ce{SO_2}$$, $$\frac{x}{2}$$ moles of $$\ce{O_2}$$ and $$x$$ moles of $$\ce{SO_3}$$.
At equilibrium I think we have $$1-x$$ moles of $$\ce{SO_2}$$, $$\frac{1-x}{2}$$ moles of $$\ce{O_2}$$ and $$x$$ moles of $$\ce{SO_3}$$.

Now, I am not sure if I put those moles correctly. And I also don't really know how to proceed from here, so I guess that I am stuck. I found this similar question Chemical Equilibrium - Le Chatelier's Principle, Change in Volume, but I don't really know how to obtain the molar ratio from there.

• Here's a hint: How is the equilibrium constant expressed in terms of the partial pressures? Jun 29 at 18:00
• Some information is missing. $1)$ You state that at the beginning there was no $\ce{SO3}$. This is an arbitrary decision. $2)$ You arbitrarily decide that the mixture of $\ce{SO2}$ and $\ce{O2}$ is stoichiometric. Why ? Jun 29 at 20:07
• @Maurice 1) I stated that because I wrote those moles before the reaction took place (so this is why I would have no $SO_3$. 2) I don't know, I thought that this was the only way that I may be able to solve this Jun 29 at 20:28

In the beginning, the equilibrium partial pressures of $$\ce{SO2, O2, SO3}$$ are respectively called $$\ce{p_1, p2, p3}$$, and the total pressure $$\pu{P}$$ is given by $$\ce{P_o = p_1 + p_2 + p_3}$$. If suddenly she total pressure is multiplied by $$2$$ it means that some amount $$2x$$ of $$\ce{SO3}$$ has been produced. Then the final pressures of $$\ce{SO2, O2, SO3}$$ are respectively $$\ce{p_1 - 2x, p2 - x , p3 + 2x}$$. The final pressure is : $$\ce{P_f = p_1 - 2x + p_2 - x + p_3 + 2x = P_o - x = P_o/2}$$. So $$\ce{x = P_o/2 = (p_1 + p_2 + p_3)/2}$$. The final pressures are :

$$\ce{p_f(SO2) = p_1 - 2 x = p_1 - 2(p_1 + p_2 + p_3)/2 = p2 + p3}$$ $$\ce{p_f(O2) = p_2 - x = p_2 - (p_1 + p_2 + p_3)/2 = -p1/2 + p2/2 + p3/2}$$ $$\ce{p_f(SO3) = p_3 + 2 x = p_3 + 2(p_1 + p_2 + p_3)/2 = p1 + p2 + 1.5 p3}$$

• if pressure is doubled by reducing volume to half wouldn't the partial pressure of all components be doubled?
– Lllt
Jun 30 at 0:43
• Initially yes but then they will change until they reach equilibrium.
– M.L
Jun 30 at 3:53
• So it should be 2p(1)-2x
– Lllt
Jun 30 at 4:40