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 shifts to counteract the change.

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

- In this reaction, we have 1 mole of gas ($CO_{2}$) on the reactant side and 2 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 to the left, favoring the formation of $C(s)$ and $CO_{2}(g)$, which means the reaction goes in the reverse direction. This supports Option 3 as the correct answer.

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

- In this reaction, there are 2 moles of gas on both sides of the equation (1 mole of $H_{2}$ and 1 mole of $I_{2}$ on the reactant side, and 2 moles of $HI$ on the product side). Increasing the pressure does not favor either side because the number of moles of gas is the same on both sides. Therefore, the reaction does not shift in the reverse direction due to pressure changes.

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

- Here, we have 4 moles of gas on the reactant side (1 mole of $N_{2}$ and 3 moles of $H_{2}$) and 2 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 product side. Thus, the reaction shifts to the right, not the reverse direction.

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

- In this reaction, there are 4 moles of gas on the reactant side (1 mole of $CO$ and 3 moles of $H_{2}$) and 2 moles of gas on the product side (1 mole of $CH_{4}$ and 1 mole of $H_{2}O$). Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side. Therefore, the reaction shifts to the right, not the reverse direction.

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, favoring the formation of reactants.