JEE Chemistry 2025 Question with Solution

Given: The equilibrium is

$$\text{CO(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$$

and the pressure is increased twofold at constant temperature.

Find: Which statements remain correct after increasing pressure.

According to Le Chatelier's principle, increasing the pressure on a gaseous system favors the side with fewer moles of gas.

On the reactant side, total gaseous moles are:

$$1 + 3 = 4$$

On the product side, total gaseous moles are:

$$1 + 1 = 2$$

So the equilibrium shifts toward the right, that is, in the forward direction.

Hence, formation of $\text{CH}_4$ and $\text{H}_2\text{O}$ increases, while reactant concentrations decrease.

The solution states that statements (A) and (B) are correct, and concludes that the correct option is A.

Therefore, the correct option is A.

Given:

$$\text{CO(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}$$

Pressure is doubled at constant temperature.

Find: The correct choice among the listed statement combinations.

First calculate the change in total moles of gas:

$$\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$$ $$\Delta n = (1 + 1) - (1 + 3) = 2 - 4 = -2$$

Since $\Delta n < 0$, the forward reaction reduces the number of moles of gas.

By Le Chatelier's principle, when pressure increases, equilibrium moves in the direction that reduces gaseous moles. Therefore, equilibrium shifts to the product side.

So:

• formation of $\text{CH}_4$ and $\text{H}_2\text{O}$ increases,
• the forward direction is favored.

The extracted solution explicitly concludes:

• (A) is correct,
• (B) is correct,
• final answer is (A) and (B) only.

Therefore, the correct option is A.