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Olin [163]
3 years ago
7

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

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

  • 1.5 × 10⁻⁹M

Explanation:

<u>1. Equilibrium equation</u>

  • H₂(g) + Br₂(l) ⇄ 2HBr(g)

<u>2. Equilibrium constant</u>

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

<u>3. ICE table.</u>

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

<u>4. Substitute into the expression of the equilibrium constant</u>

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

<u>5. Solve the quadratic equation</u>

  • 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

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