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? Commented Jun 29, 2021 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 ? Commented Jun 29, 2021 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 Commented Jun 29, 2021 at 20:28

Let:

A = $$\ce{SO2}$$

B = $$\ce{O2}$$

C = $$\ce{SO3}$$

The only information given is the change in moles for the reaction and the volume decrease factor:

$$\Delta n=c-a-b=1-1-0.5=-0.5$$

$$\frac{V_2}{V_1}=0.5$$

For simplicity, the molar ratios of interest before and after compression are:

$$\Theta_1=\left(\frac{N_{\ce{SO3}}}{N_{\ce{SO2}}}\right)_1=\frac{N_{C_1}}{N_{A_1}}$$

$$\Theta_2=\left(\frac{N_{\ce{SO3}}}{N_{\ce{SO2}}}\right)_2=\frac{N_{C_2}}{N_{A_2}}$$

The initial (before compression) equilibrium system in terms of moles is given by:

$$K_{N_1}=\frac{N_{C_1}}{N_{A_1}N^{0.5}_{B_1}}=K_C\;V_1^{\Delta n}$$

The final (after compression) equilibrium system in terms of moles is given by:

$$K_{N_2}=\frac{N_{C_2}}{N_{A_2}N^{0.5}_{B_2}}=K_C\;V_2^{\Delta n}$$

Dividing the final state by the initial state:

$$\frac{K_{N_2}}{K_{N_1}}=\frac{\frac{N_{C_2}}{N_{A_2}\;N_{B_2}^{0.5}}}{\frac{N_{C_1}}{N_{A1}\;N_{B_1}^{0.5}}}=\frac{\Theta_2}{\Theta_1}\left(\frac{N_{B_1}}{N_{B_2}}\right)^{0.5}=\frac{\require {cancel} \cancel{K_C}}{\cancel{K_C}}\left(\frac{V_2}{V_1}\right)^{\Delta n}=0.5^{-0.5}=2^{0.5}$$

In other words:

$$\frac{\Theta_2}{\Theta_1}=\left(2\frac{N_{B_2}}{N_{B_1}}\right)^{0.5}$$

It's not possible to proceed without knowing the ratio of moles of B at both equilibrium states:

$$\frac{N_{B_2}}{N_{B_1}}=\;?$$

But we at least know that due to Le Châtelier's Principle, an increase in pressure with $$\Delta n <1$$ leads to the consumption of some A and some B in order to produce more C, which means necessarily:

$$0

If we evaluate the special case in which the ratio of B is 0.5, then:

$$\frac{N_{B_2}}{N_{B_1}}=0.5\implies \frac{\Theta_2}{\Theta_1}=1\implies no\;change$$

Which means we now know the conditions under which the ratio of interest either increases or decreases as a function of the ratio of B:

$$0.5<\frac{N_{B_2}}{N_{B_1}}<1\implies 1<\frac{\Theta_2}{\Theta_1}<\sqrt{2}\implies increase$$

$$0<\frac{N_{B_2}}{N_{B_1}}<0.5\implies 0<\frac{\Theta_2}{\Theta_1}<1\implies decrease$$

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? Commented Jun 30, 2021 at 0:43
• Initially yes but then they will change until they reach equilibrium.
– M.L
Commented Jun 30, 2021 at 3:53
• So it should be 2p(1)-2x Commented Jun 30, 2021 at 4:40