Question

Consider the following reaction: N₂(g) + 3 H₂(g) ⇔ 2 NH₃(g)

At equilibrium in a 2.00 L container, we find 0.200 moles H2, 0.400 moles N₂, and 0.800 moles of NH₃.
Calculate Kc.

Answer 1

To calculate the equilibrium constant (Kc), molar concentrations of H2, N2, and NH3 were determined and plugged into the Kc expression for the reaction N2(g) + 3 H2(g) ⇔ 2 NH3(g), resulting in a Kc of 800 M^-1.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), we use the concentration of each gas at equilibrium. Given a 2.00 L container, we have 0.200 moles of H₂, 0.400 moles of N₂, and 0.800 moles of NH₃. First, we need to find the molar concentrations by dividing the number of moles by the volume of the container. This results in:

• [H₂] = 0.200 moles / 2.00 L = 0.100 M
• [N₂] = 0.400 moles / 2.00 L = 0.200 M
• [NH₃] = 0.800 moles / 2.00 L = 0.400 M

The expression for Kc is:

Kc = [[NH₃]² / ([N₂][[H₂]³)

Plugging in the concentrations:

Kc = (0.400 M)² / (0.200 M)(0.100 M)³ = 0.16 M² / (0.200 M)(0.001 M³) = 0.16 M² / 0.0002 M⁴ = 800 M-1

Therefore, the equilibrium constant, Kc, for the reaction is 800 M-1.