At a constant temperature, 1.0\text{ mole} of carbon monoxide, \text{CO}(g), and 1.0\text{… — JAMB UTME

At a constant temperature,

1.0\text{ mole}

of carbon monoxide,

\text{CO}(g)

, and

1.0\text{ mole}

of steam,

\text{H}_2\text{O}(g)

, are introduced into a

2.0\text{ dm}^3

sealed vessel and allowed to reach equilibrium according to the equation:

\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)

0.6\text{ mole}

of carbon dioxide, \text{CO}_2(g), is present at equilibrium, what is the numerical value of the equilibrium constant,

K_c

, at this temperature?

Answer: 2.25

The numerical value of the equilibrium constant,

K_c

, is 2.25.

At equilibrium,

0.6\text{ mol}

\text{CO}_2

is present, which implies

0.6\text{ mol}

\text{H}_2

is also formed. The remaining amounts of reactants are

1.0 - 0.6 = 0.4\text{ mol}

for both

\text{CO}

and

\text{H}_2\text{O}

. Dividing by the vessel volume (

2.0\text{ dm}^3

) yields concentrations of

0.30\text{ mol dm}^{-3}

for products and

0.20\text{ mol dm}^{-3}

for reactants. Substituting these values into

K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}

gives

K_c = \frac{0.30 \times 0.30}{0.20 \times 0.20} = 2.25

Determine the equilibrium moles for all reactants and products

Equilibrium moles are:

\text{CO} = 0.4\text{ mol}

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

\text{CO}_2 = 0.6\text{ mol}

, and

\text{H}_2 = 0.6\text{ mol}

Since 1 mole of

\text{CO}_2

is produced per mole of

\text{CO}

consumed, forming

0.6\text{ mol}

\text{CO}_2

consumes

0.6\text{ mol}

\text{CO}

and

0.6\text{ mol}

\text{H}_2\text{O}

, while producing

0.6\text{ mol}

\text{H}_2

Calculate equilibrium concentrations by dividing moles by volume (

2.0\text{ dm}^3

)

[\text{CO}] = 0.20\text{ mol dm}^{-3}

[\text{H}_2\text{O}] = 0.20\text{ mol dm}^{-3}

[\text{CO}_2] = 0.30\text{ mol dm}^{-3}

[\text{H}_2] = 0.30\text{ mol dm}^{-3}

Concentration is given by

C = \frac{n}{V}

where volume

V = 2.0\text{ dm}^3

Evaluate the equilibrium expression

K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}

K_c = \frac{0.30 \times 0.30}{0.20 \times 0.20} = 2.25

Plugging in the calculated equilibrium concentrations into the law of mass action expression gives the value of

K_c

Calculating Equilibrium Constant (

K_c

) from Equilibrium Amounts

Question