Question 15 of 69intermediate🔧 ApplyNumerical3 marks

At 700 K, equilibrium constant for the reaction:

H2(g)+I2(g)⇌2HI(g)H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g)

is 54.8. If 0.5 mol L−1^{-1} of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g)H_2(g) and I2(g)I_2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Correct Answer

The equilibrium concentrations of H2(g)H_2(g) and I2(g)I_2(g) are both 0.06750.0675 mol L−1^{-1}.

Exercise: EXERCISES | Q: 6.15 | (Page 42)
For More Understanding

Explanation

The reaction begins with only HI(g), which decomposes into H2(g)H_2(g) and I2(g)I_2(g) in a 1:1 molar ratio. Therefore, at equilibrium, the concentrations of H2(g)H_2(g) and I2(g)I_2(g) are equal. Using the given equilibrium constant expression Kc=[HI(g)]eq2[H2(g)]eq[I2(g)]eqK_c = \frac{[HI(g)]^2_{eq}}{[H_2(g)]_{eq} [I_2(g)]_{eq}} and substituting the known values of Kc=54.8K_c = 54.8 and [HI(g)]eq=0.5[HI(g)]_{eq} = 0.5 mol L−1^{-1}, we can solve for the unknown concentrations.

Solution Steps

  1. Step 1: Write the expression for the equilibrium constant KcK_c for the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g).

    Kc=[HI(g)]eq2[H2(g)]eq[I2(g)]eqK_c = \frac{[HI(g)]^2_{eq}}{[H_2(g)]_{eq} [I_2(g)]_{eq}}

  2. Step 2: Since we initially started with only HI(g), the decomposition produces equal amounts of H2(g)H_2(g) and I2(g)I_2(g). Thus, let [H2(g)]eq=[I2(g)]eq=x[H_2(g)]_{eq} = [I_2(g)]_{eq} = x.

  3. Step 3: Substitute the given values (Kc=54.8K_c = 54.8 and [HI(g)]eq=0.5[HI(g)]_{eq} = 0.5 mol L−1^{-1}) into the expression.

    54.8=(0.5)2x×x54.8 = \frac{(0.5)^2}{x \times x} 54.8=0.25x254.8 = \frac{0.25}{x^2}

  4. Step 4: Solve for xx.

    x2=0.2554.8x^2 = \frac{0.25}{54.8} x2=0.004562x^2 = 0.004562

    x=0.004562≈0.0675x = \sqrt{0.004562} \approx 0.0675 mol L−1^{-1}

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