Question

H2(g) + Br2(l) ⇄ 2HBr(g) Kc = 4.8 × 108

Assume initial conditions of 0.400 M H2(g) and excess Br2(l). What is the equilibrium concentration of H2(g)?

Answer 1

Answer:

• 1.5 × 10⁻⁹M

Explanation:

1. Equilibrium equation

• H₂(g) + Br₂(l) ⇄ 2HBr(g)

2. Equilibrium constant

The liquid substances do not appear in the expression of the equilibrium constant.

    $k_c=\dfrac{[HBr(g)]^2}{[H_2]}=4.8\times 10^8M$

3. ICE table.

Write the initial, change, equilibrium table:

Molar concentrations:

         H₂(g) + Br₂(l) ⇄ 2HBr(g)

I          0.400                   0

C           - x                      +2x

E         0.400 - x              2x

4. Substitute into the expression of the equilibrium constant

     $4.8\times 10^8=\dfrac{(2x)^2}{0.400-x}$

5. Solve the quadratic equation

• 192,000,000 - 480,000,000x = 4x²

• x² + 120,000,000x - 48,000,000 = 0

Use the quadratic formula:

       

$x=\dfrac{-120,000,00\pm\sqrt{(120,000,000)^2-4(1)(-48,000,000}}{2(1)}$

The only valid solution is x = 0.39999999851M

Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M