# How to shift SO3/SO2 equilibrium?

I am struggling with a chemical equilibrium question and would appreciate some help. The question involves a closed system where the equilibrium $$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$ has been established. The forward reaction is exothermic.

My question is, how can I decrease the amount of $$\ce{SO3}$$ in the mixture? The options given are to change the temperature, volume, or amount of $$\ce{O2}$$ present.

So far, I have tried to recall the principles of Le Chatelier's principle, which states that when a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will shift its equilibrium position to counteract the change. In this case, if I want to decrease the amount of $$\ce{SO3}$$, I need to shift the equilibrium to the left, meaning that I need to decrease the concentration of $$\ce{SO3}$$.

Based on my understanding of Le Chatelier's principle, changing the temperature will have the opposite effect on the equilibrium because the reaction is exothermic. Decreasing the volume of the container will also have no effect since the reaction involves gases. Therefore, the only option left is to change the amount of $$\ce{O2}$$ present. If I decrease the amount of $$\ce{O2}$$, the equilibrium will shift to the left, reducing the concentration of $$\ce{SO3}$$.

Am I on the right track with my answer? I would appreciate any feedback or additional information that could help me better understand this concept.

• Be sure to review available info about the principle in offline and online sources, including this site, to avoid creation yet another redundant info. Mar 26 at 13:59
• "Decreasing the volume of the container will also have no effect since the reaction involves gases." ?!? If you change the volume of a closed system, you obviously change the pressure.
– Karl
Mar 27 at 14:46

In an exothermic reaction like this one, heating the mixture will push the system to the left hand side of the equation. So it will decrease the amount of $$\ce{SO3}$$ and increase the proportion of $$\ce{SO2}$$ and $$\ce{O2}$$. This is what we are looking for.
As a consequence, both increases (temperature and volume) will have the same effect : destroy $$\ce{SO3}$$ and produce more $$\ce{SO2}$$ and more $$\ce{O2}$$
• What about the question the OP asks? Are they on the right track about $\ce{O2}$? Mar 26 at 14:59
• Of course decreasing the relative amount of $\ce{O2}$ will push the equation to the left-hand-side. Mar 26 at 15:33