Question

The equilibrium constant for the reaction, N$_2$(g) + 3H$_2$ $\rightleftharpoons$ 2NH$_3$(g) and 2N$_2$(g) + 6H$_2$ $\rightleftharpoons$ 4NH$_3$(g) are $K_1$ and $K_2$, respectively. The relationship between $K_1$ and $K_2$ is

• A. $K_2 = K_1^2$
• B. $K_2 = K_1^{-2}$
• C. $K_1 = K_2^2$
• D. $K_2 = \sqrt{K_1}$
• E. $K_1 = \sqrt{K_2}$

Hint:

If you reverse a reaction, the new equilibrium constant is $1/K$. If you multiply a reaction by $n$, the constant becomes $K^n$.

Answer 1

The Correct Option is A

Concept: When a balanced chemical equation is multiplied by a factor $n$, the new equilibrium constant ($K'$) is equal to the original equilibrium constant ($K$) raised to the power of $n$.

Step 1: {Analyze the first reaction ($K_1$).} Reaction 1: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ The expression for $K_1$ is: $$K_1 = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$

Step 2: {Analyze the second reaction ($K_2$).} Reaction 2: $2\text{N}_2(g) + 6\text{H}_2(g) \rightleftharpoons 4\text{NH}_3(g)$ The expression for $K_2$ is: $$K_2 = \frac{[\text{NH}_3]^4}{[\text{N}_2]^2[\text{H}_2]^6}$$

Step 3: {Establish the relationship.} Observe that Reaction 2 is simply Reaction 1 multiplied by a factor of 2. Therefore, the relationship between their constants is: $$K_2 = (K_1)^2$$