In which of the following reactions, the reaction goes in the reverse direction when pressure is increased:

Question

Accepted Answer

To determine which reaction goes in the reverse direction when pressure is increased, we need to consider Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change.

1. **Option 3: $C(s) + CO_{2}(g) \rightleftharpoons 2CO(g)$**

- In this reaction, we have one mole of gas ($CO_{2}$) on the reactant side and two moles of gas ($2CO$) on the product side. When pressure is increased, the equilibrium will shift towards the side with fewer moles of gas to reduce the pressure. Therefore, the reaction will shift in the reverse direction, favoring the formation of $CO_{2}$ from $2CO$. This is why Option 3 is correct.

2. **Option 1: $H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$**

- In this reaction, there are two moles of gas on the reactant side ($H_{2}$ and $I_{2}$) and two moles of gas on the product side ($2HI$). Since the number of moles of gas is the same on both sides, increasing the pressure does not favor either direction. Therefore, the reaction does not shift in the reverse direction when pressure is increased.

3. **Option 2: $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$**

- Here, we have four moles of gas on the reactant side ($N_{2}$ and $3H_{2}$) and two moles of gas on the product side ($2NH_{3}$). Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the forward direction, favoring the formation of $NH_{3}$. Thus, the reaction does not go in the reverse direction when pressure is increased.

4. **Option 4: $CO(g) + 3H_{2}(g) \rightleftharpoons CH_{4}(g) + H_{2}O(g)$**

- In this reaction, there are four moles of gas on the reactant side ($CO$ and $3H_{2}$) and two moles of gas on the product side ($CH_{4}$ and $H_{2}O$). Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the forward direction, favoring the formation of $CH_{4}$ and $H_{2}O$. Therefore, the reaction does not shift in the reverse direction when pressure is increased.

In conclusion, Option 3 is the correct answer because it is the only reaction where increasing the pressure causes the equilibrium to shift in the reverse direction, according to Le Chatelier's principle.