K_p = 1.5 × 10³ at 400°C for the reaction 2 NH₃(g)⇌ N₂(g)+ 3 H₂(g).What is the value of K_p for the reaction: 2 N₂(g)+ 6 H₂(g)⇌ 4 NH₃(g)?

Question

Quizplus · Accepted Answer

The equilibrium constant for the reverse reaction is the reciprocal of the original, and for the reaction multiplied by a factor (in this case 2), the equilibrium constant is raised to the power of that factor. Given $K_p = 1.5 \times 10^{13}$ for the forward reaction, the reverse reaction's $K_p$ would be $1/(1.5 \times 10^{13})$, and when the reaction is doubled, $K_p$ becomes $(1/(1.5 \times 10^{13}))^2 = 4.4 \times 10^{-27}$.

Kp = 1

K_p = 1