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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.

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    $\begingroup$ Here's a hint: How is the equilibrium constant expressed in terms of the partial pressures? $\endgroup$ Jun 29 at 18:00
  • $\begingroup$ 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 ? $\endgroup$
    – Maurice
    Jun 29 at 20:07
  • $\begingroup$ @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 $\endgroup$ Jun 29 at 20:28
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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 :

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

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

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