An industrial chemist puts 1.40 mol each of H2 (g) and CO2 (g) in a 1.00-L container at a certain temperature. When equilibrium is reached, 0.52 mol of CO (g) is in the container. Find Keq at this temperature for the following reaction: H2 (g) + CO2 (g) ↔H2O (g) + CO (g)

H₂ (g) + CO₂ (g) ⇆ H₂O (g) + CO (g) Kc → 0.35

Explanation:

The equilibrium is:

H₂ (g) + CO₂ (g) ⇆ H₂O (g) + CO (g)

Initially we have 1.40 moles of hydrogen and 1.40 mol of carbon dioxide.

During the reaction x amount has reacted.

In the equilibrium we have 0.52 moles of CO. Therefore the scheme is:

H₂ (g) + CO₂ (g) ⇆ H₂O (g) + CO (g)

Initially 1.40 1.40 - -

React x x x x

Eq. 1.40-x 1.40-x 0+x 0+x=0.52

Then x = 0.52. Ratio is 1:1, so If x reacted, x is formed during the equilibrium

1.40 - 0.52 = 0.88 → [H₂] and [CO₂]

0.52 = [CO] and [H₂O]

The expression for Kc is: [CO]. [H₂O] / [H₂]. [CO₂]

Kc = (0.52. 0.52) / (0.88. 0.88) → 0.35