A 4.0\text{ mol} sample of hydrogen iodide, \text{HI}(g), is placed in a 2.0\text{ dm}^3 r… — JAMB UTME

4.0\text{ mol}

sample of hydrogen iodide,

\text{HI}(g)

, is placed in a

2.0\text{ dm}^3

rigid vessel and allowed to dissociate according to the equation:

2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)

20\%

of the

\text{HI}(g)

decomposes at equilibrium, what is the value of the equilibrium constant,

K_c

0.0156

To calculate

K_c

, first find the equilibrium amounts of all species. Starting with

4.0\text{ mol}

\text{HI}

, a

20\%

decomposition means

0.8\text{ mol}

decomposes, leaving

3.2\text{ mol}

\text{HI}

at equilibrium. Based on the balanced reaction stoichiometry

2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)

, decomposing

0.8\text{ mol}

\text{HI}

forms

0.4\text{ mol}

\text{H}_2

and

0.4\text{ mol}

\text{I}_2

. Converting to concentrations in a

2.0\text{ dm}^3

vessel yields

[\text{HI}] = 1.6\text{ mol dm}^{-3}

[\text{H}_2] = 0.2\text{ mol dm}^{-3}

, and

[\text{I}_2] = 0.2\text{ mol dm}^{-3}

. Substituting into

K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

gives

\frac{(0.2)(0.2)}{(1.6)^2} = 0.0156

Determine initial moles and amount decomposed at equilibrium

Initial

\text{HI} = 4.0\text{ mol}

. Amount decomposed

= 0.20 \times 4.0\text{ mol} = 0.8\text{ mol}

The reaction states that 20% of the initial HI dissociates at equilibrium.

Calculate equilibrium mole quantities

Equilibrium

\text{HI} = 4.0 - 0.8 = 3.2\text{ mol}

. According to

2\text{HI} \rightarrow \text{H}_2 + \text{I}_2

, equilibrium

\text{H}_2 = 0.4\text{ mol}

and

\text{I}_2 = 0.4\text{ mol}

Every 2 moles of HI decomposed produces 1 mole of H2 and 1 mole of I2.

Calculate equilibrium concentrations and solve for

K_c

[\text{HI}] = \frac{3.2}{2.0} = 1.6\text{ mol dm}^{-3}

[\text{H}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}

[\text{I}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}

. Thus,

K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac{(0.2)(0.2)}{(1.6)^2} = \frac{0.04}{2.56} = 0.015625 \approx 0.0156

Substitute equilibrium concentrations into the equilibrium constant expression.

Equilibrium Constant (

K_c

) Calculation from Percent Dissociation

Soru