For the reaction equilibrium is established at 472°C when [H₂] = 0.0400 M, [N₂] = 0.0200 M, and [NH₃] = 1.074 × 10^-8M. What is the equilibrium constant for the reverse reaction?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

Kc = [NHS1U1B13S1U1B0]/([HS1U1B12S1U1B0][NS1U1B12S1U1B0])

At 472°C, the equilibrium concentrations are given as:

[HS1U1B12S1U1B0] = 0.0400 M
[NS1U1B12S1U1B0] = 0.0200 M
[NHS1U1B13S1U1B0] = 1.074 × 10S1U1P1-8S1S1P0M

Substituting these values in the expression, we get:

Kc = (1.074 × 10S1U1P1-8S1S1P0M)/((0.0400 M)(0.0200 M))
Kc = 1.109 × 10S1U1P111S1S1P0

Since the given equation is a reversible reaction, the equilibrium constant for the reverse reaction is the reciprocal of Kc for the forward reaction:

Kc(reverse) = 1/Kc(forward)
Kc(reverse) = 1/(1.109 × 10S1U1P111S1S1P0)
Kc(reverse) ≈ 9.01 × 10S1U1P-12S1S1P0

So, the correct choice is (B).

For the Reaction Equilibrium Is Established at 472°C When