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)?

Explanation:

1. Equilibrium equation

2. Equilibrium constant

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

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

5. Solve the quadratic equation

Use the quadratic formula:

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

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explanation:

3 1/4 %

explanation: