At 500°C the equilibrium constant,K_P ,is 4.00 ×10^-4 for the equilibrium: 2HCN(g) H₂(g) + C₂N₂(g)
What is K_p for the following reaction?
H₂(g) + C₂ N₂(g) 2HCN(g)

Question

At 500°C the equilibrium constant,K_P ,is 4.00 ×10^-4 for the equilibrium: 2HCN(g) H₂(g)+ C₂N₂(g) What is K_p for the following reaction? H₂(g)+ C₂ N₂(g) 2HCN(g)

Quizplus · Accepted Answer

The given equilibrium equation can be written as follows: K_P = [H₂][C₂N₂]/[HCN]^2 From this equation, we can rearrange to get: [HCN]^2 = [H₂][C₂N₂]/K_P Now, we can substitute the given values at 500°C: K_P = 4.00 ×10^-4 H₂ = C₂N₂ = 1 (since they are all in equal amounts) Substituting these values, we get: [HCN]^2 = 1/4.00 ×10^-4 [HCN] = 1/2(4.00 ×10^-4)³ [HCN] = 2.50 × 10³ Therefore, the equilibrium constant for the reverse reaction is: K_p = 1/K_P = 1/4.00 ×10^-4 K_p = 2.50 × 10³. Hence, the answer is D.

At 500°C the Equilibrium Constant,KP ,Is 4

At 500°C the Equilibrium Constant,K_P ,Is 4