Finding the molar ratio at equilibrium

Question

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.

Answer 1

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:

$$0<N_{B_2}<N_{B_1}\implies 0<\frac{N_{B_2}}{N_{B_1}}<1\implies0<\frac{\Theta_2}{\Theta_1}<\sqrt{2}$$

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:

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

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

Answer 2

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 :

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