carbon disulfide and oxygen react to form carbon dioxide and sulfur dioxide, like this:
cs₂(g)+3o₂(g)→co₂(g)+2so₂(g)
suppose a mixture of cs₂, o₂, co₂ and so₂ has come to equilibrium in a closed reaction vessel. predict what change, if any, the perturbations in the table below will cause in the composition of the mixture in the vessel. also decide whether the equilibrium shifts to the right or left.

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# Explanation:
## Step1: Le Chatelier's Principle for removing $SO_2$
When some $SO_2$ is removed, according to Le - Chatelier's principle, the system will try to counteract this change. The reaction will shift in the direction that produces more $SO_2$. The forward reaction $CS_{2}(g)+3O_{2}(g)\to CO_{2}(g)+2SO_{2}(g)$ produces $SO_2$. So the equilibrium shifts to the right.
As the reaction shifts to the right, $CS_2$ is consumed. So the pressure of $CS_2$ will decrease. And since the reaction is proceeding forward, $O_2$ is consumed. But wait, no, actually, when the reaction shifts right, $CS_2$ and $O_2$ are reactants. As the reaction shifts right, $CS_2$ is used up (so its pressure decreases) and $O_2$ is also used up (but we removed $SO_2$, the system tries to make more $SO_2$ by consuming $CS_2$ and $O_2$). Wait, no:
The reaction is $CS_{2}(g)+3O_{2}(g)
ightleftharpoons CO_{2}(g)+2SO_{2}(g)$. When $SO_2$ is removed, $Q=\frac{P_{CO_2}\times P_{SO_2}^2}{P_{CS_2}\times P_{O_2}^3}<K$ (where $K$ is the equilibrium constant). So the reaction shifts right. As $CS_2$ reacts (to form products), $P_{CS_2}$ decreases. As $O_2$ reacts (to form products), but we removed $SO_2$, the system will consume $CS_2$ and $O_2$ to make more $SO_2$. So $P_{O_2}$ decreases.

## Step2: Le Chatelier's Principle for adding $O_2$
When some $O_2$ is added, $Q=\frac{P_{CO_2}\times P_{SO_2}^2}{P_{CS_2}\times P_{O_2}^3}<K$ (because $P_{O_2}$ increases, making the denominator larger. But actually, when we add $O_2$, the system will try to reduce the amount of $O_2$ by shifting the reaction to the right.
As the reaction shifts right, $CS_2$ is consumed (so $P_{CS_2}$ decreases). And since the reaction is shifting right, $CO_2$ is produced (so $P_{CO_2}$ increases)

# Answer:
| perturbation | change in composition | shift in equilibrium |
|---|---|---|
| Some $SO_2$ is removed | The pressure of $CS_2$ will decrease; The pressure of $O_2$ will decrease | to the right |
| Some $O_2$ is added | The pressure of $CS_2$ will decrease; The pressure of $CO_2$ will increase | to the right |

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Explanation:

Step1: Le Chatelier's Principle for removing \(SO_2\)

Step2: Le Chatelier's Principle for adding \(O_2\)

Answer: