Exemplar Problems

Problem 3: Consider the reaction:

2HI(g) ⇌ H₂(g) + I₂(g) At a certain temperature, the equilibrium constant Kc is 54. If the initial concentration of HI is 0.5 M, calculate the equilibrium concentrations of HI, H₂, and I₂.

Solution :

Let’s denote the initial concentration of HI as [HI]₀, and the equilibrium concentrations of HI, H₂, and I₂ as [HI], [H₂], and [I₂], respectively. We are given: [HI]₀ = 0.5 M Kc = 54

Using the balanced equation, we know that at equilibrium: [HI] = [HI]₀ - 2x [H₂] = x [I₂] = x

Now, we can write the expression for Kc: Kc = ([H₂] * [I₂]) / ([HI]²)

Substitute the expressions for [HI], [H₂], and [I₂]: 54 = (x * x) / ((0.5 - 2x)²)

Now, solve for x: 54 = x² / (0.25 - 2x)²

54(0.25 - 2x)² = x²

13.5(0.25 - 2x)² = x²

13.5(0.0625 - 0.5x + 4x²) = x²

1.125 - 9x + 54x² = x²

1.125 - 9x + 54x² - x² = 0

53x² - 9x + 1.125 = 0

Now, use the quadratic formula to solve for x: x = [9 ± √(9² - 4(53)(1.125))] / (2(53))

x = [9 ± √(81 - 238.5)] / 106

x = [9 ± √(-157.5)] / 106

Since the discriminant is negative, there are no real roots for x. This means that the reaction does not reach equilibrium under the given conditions, and the concentrations of HI, H₂, and I₂ remain unchanged from their initial values.

So, at equilibrium: [HI] ≈ 0.5 M [H₂] ≈ 0 M [I₂] ≈ 0 M