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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

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