Exemplar Problems

Problem 1 : For the reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) The equilibrium constant Kc is 2.5 x 10⁻⁶ at a certain temperature. Calculate the equilibrium concentrations of N₂, H₂, and NH₃ if initially, 0.2 moles of N₂ and 0.6 moles of H₂ are placed in a 1-liter container.

Solution :

Let’s denote the initial concentrations of N₂, H₂, and NH₃ as [N₂]₀, [H₂]₀, and [NH₃]₀, respectively. We are given: [N₂]₀ = 0.2 moles/L [H₂]₀ = 0.6 moles/L Kc = 2.5 x 10⁻⁶

At equilibrium, the concentrations will be denoted as [N₂], [H₂], and [NH₃].

Using the balanced equation, we know that at equilibrium: [N₂] = [N₂]₀ - x [H₂] = [H₂]₀ - 3x [NH₃] = [NH₃]₀ + 2x

Now, we can write the expression for Kc: Kc = [NH₃]² / ([N₂] * [H₂]³)

Substitute the expressions for [N₂], [H₂], and [NH₃]: Kc = ([NH₃]₀ + 2x)² / (([N₂]₀ - x) * ([H₂]₀ - 3x)³)

Now, we can plug in the given values: 2.5 x 10⁻⁶ = ([NH₃]₀ + 2x)² / ((0.2 - x) * (0.6 - 3x)³)

Since Kc is very small, we can make the assumption that x is small compared to [NH₃]₀, [N₂]₀, and [H₂]₀. Therefore, we can simplify: 2.5 x 10⁻⁶ ≈ (2x)² / ((0.2) * (0.6)³)

Now, solve for x: 2.5 x 10⁻⁶ ≈ (4x²) / (0.432)

4x² ≈ 2.5 x 10⁻⁶ * 0.432 x² ≈ (2.5 x 10⁻⁶ * 0.432) / 4 x² ≈ 2.7 x 10⁻⁷ x ≈ √(2.7 x 10⁻⁷) x ≈ 5.2 x 10⁻⁴

Now, we can find the equilibrium concentrations: [NH₃] = [NH₃]₀ + 2x [NH₃] = 0.6 + 2(5.2 x 10⁻⁴) = 0.60052 moles/L

[N₂] = [N₂]₀ - x [N₂] = 0.2 - 5.2 x 10⁻⁴ = 0.19948 moles/L

[H₂] = [H₂]₀ - 3x [H₂] = 0.6 - 3(5.2 x 10⁻⁴) = 0.59848 moles/L

So, the equilibrium concentrations are approximately: [N₂] ≈ 0.19948 moles/L [H₂] ≈ 0.59848 moles/L [NH₃] ≈ 0.60052 moles/L