At 327 °C, the equilibrium constant for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 4.10 M⁻² - VCE - SSCE Chemistry - Question 8 - 2023 - Paper 1

To determine the equilibrium constant for the reaction 2N₂(g) + 6H₂(g) ⇌ 4NH₃(g), we can use the relationship between the equilibrium constants of the two reactions.

The first reaction is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with an equilibrium constant of $K_1 = 4.10 \text{ M}^{-2}$.

The second reaction can be derived from the first by multiplying all coefficients by 2: $2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g)$ The relationship between the two reactions states that if you multiply the coefficients of a balanced equation by a factor, the new equilibrium constant is equal to the old equilibrium constant raised to that power. Thus, we have:

$K_2 = K_1^2$

Substituting the value of $K_1$: $K_2 = (4.10 ext{ M}^{-2})^2$

Calculating this gives: $K_2 = 16.81 ext{ M}^{-4}$

Therefore, we approximate this value to choose the answer among the given options. The closest value is: D. 16.8 M⁻⁴.