For the following three reactions $$a, b$$ and $$c$$, equilibrium constants are given: (a) $$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g);\ K_1$$ (b) $$CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g);\ K_2$$ (c) $$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g);\ K_3$$ Which of the following relations is correct?

First write the equilibrium-constant expressions for the three reactions.

Reaction (a): $$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$$
$$K_1=\frac{[CO_2][H_2]}{[CO][H_2O]}$$

Reaction (b): $$CH_4(g)+H_2O(g)\rightleftharpoons CO(g)+3H_2(g)$$
$$K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]}$$

Add reactions (a) and (b) algebraically:

Left-hand side: $$CO+H_2O+CH_4+H_2O=CH_4+2H_2O+CO$$
Right-hand side: $$CO_2+H_2+CO+3H_2=CO_2+4H_2+CO$$
The species $$CO$$ appears on both sides and cancels.

Net reaction obtained is

$$CH_4(g)+2H_2O(g)\rightleftharpoons CO_2(g)+4H_2(g)$$

This is exactly reaction (c), whose equilibrium constant is $$K_3$$.

Rule for combining equilibria: when two reactions are added, their equilibrium constants are multiplied. Therefore

$$K_3 = K_1 \times K_2$$

Among the given choices, this matches Option C.

Final Answer: Option C which is: $$K_3 = K_1 K_2$$