Consider the equilibrium $$CO(g)+3H_{2}(g)\rightleftharpoons CH_{4}(g)+H_{2}O(g)$$ If the pressure applied over the system
increases by two fold at constant temperature then (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below :

The reaction: $$CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)$$

Moles of gas: Reactants = 4, Products = 2. So $$\Delta n_g = -2$$.

When pressure is doubled at constant temperature:

(A) Concentration of reactants and products increases. Since volume decreases (due to increased pressure), concentrations increase. TRUE.

(B) Equilibrium will shift in forward direction. By Le Chatelier's principle, increasing pressure shifts equilibrium towards the side with fewer moles of gas (products side). TRUE.

(C) Equilibrium constant increases. $$K_c$$ depends only on temperature. It does NOT change with pressure. FALSE.

(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. $$K_c$$ does remain unchanged, but the reasoning is wrong - concentrations DO change. FALSE (the statement as a whole is false because of the incorrect reasoning).

Only statements A and B are correct.

The correct answer is Option 2: (A) and (B) only.