A 1.0\text{ dm}^3 rigid reaction vessel contains an equilibrium mixture of 0.20\text{ mol}… — JAMB UTME

1.0\text{ dm}^3

rigid reaction vessel contains an equilibrium mixture of

0.20\text{ mol}

of sulfur dioxide (

\text{SO}_2

0.10\text{ mol}

of oxygen (

\text{O}_2

), and

0.40\text{ mol}

of sulfur trioxide (

\text{SO}_3

) at a constant temperature according to the equation:

2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)

What is the numerical value of the equilibrium constant,

K_c

, for this reaction?

Answer: 40 dm³ mol⁻¹

The numerical value of the equilibrium constant

K_c

is 40.

Substituting the given equilibrium concentrations into the stoichiometric expression

K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}

gives

K_c = \frac{(0.40)^2}{(0.20)^2 \times 0.10} = \frac{0.16}{0.04 \times 0.10} = 40

Write the equilibrium constant expression (

K_c

) for the reaction.

K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}

Products are in the numerator and reactants in the denominator, each raised to the power of their respective stoichiometric coefficients.

Convert equilibrium moles into molar concentrations (mol dm⁻³).

[SO₂] = 0.20 mol dm⁻³, [O₂] = 0.10 mol dm⁻³, [SO₃] = 0.40 mol dm⁻³

Concentration equals moles divided by volume (

1.0\text{ dm}^3

Substitute the values into the

K_c

expression and solve.

K_c = \frac{(0.40)^2}{(0.20)^2 \times 0.10} = \frac{0.16}{0.004} = 40

Evaluating the square terms yields 0.16 in the numerator and 0.004 in the denominator, which simplifies to 40.

Calculation of equilibrium constant (Kc) from equilibrium concentrations

Question