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Question

For the following three reactions (i), (ii) and (iii), equilibrium constants are given:
(i) $	ext{CO}(g) + 	ext{H}_2	ext{O}(g) ightleftharpoons 	ext{CO}_2(g) + 	ext{H}_2(g); K_1$
(ii) $	ext{CH}_4(g) + 	ext{H}_2	ext{O}(g) ightleftharpoons 	ext{CO}(g) + 3	ext{H}_2(g); K_2$
(iii) $	ext{CH}_4(g) + 2	ext{H}_2	ext{O}(g) ightleftharpoons 	ext{CO}_2(g) + 4	ext{H}_2(g); K_3$
Which of the following is correct? A). $K_1\sqrt{K_2} = K_3$ B). $K_2 K_3 = K_1$ C). $K_3 = K_1 K_2$ D). $K_3 \cdot K_2^3 = K_1^2$

Edvaya · Accepted Answer

The correct answer is $K_3 = K_1 K_2$. By observing the chemical equations, if we add reaction (i) and reaction (ii):
Left side: $(	ext{CO} + 	ext{H}_2	ext{O}) + (	ext{CH}_4 + 	ext{H}_2	ext{O}) ightarrow 	ext{CH}_4 + 2	ext{H}_2	ext{O} + 	ext{CO}$
Right side: $(	ext{CO}_2 + 	ext{H}_2) + (	ext{CO} + 3	ext{H}_2) ightarrow 	ext{CO}_2 + 4	ext{H}_2 + 	ext{CO}$
Canceling $	ext{CO}$ from both sides yields:
$	ext{CH}_4 + 2	ext{H}_2	ext{O} ightleftharpoons 	ext{CO}_2 + 4	ext{H}_2$, which is exactly reaction (iii).
When two chemical reactions are added, their equilibrium constants are multiplied.
Therefore, $K_3 = K_1 	imes K_2$.

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