Question

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The reaction: CO(g) + H₂O(g) ⇔ CO₂(g) + H₂(g) is used to increase the ratio of hydrogen in synthesis gas (mixtures of CO and H₂. Suppose you start with 1.00 mol each of carbon monoxide and water in a 50.0 L vessel. What is the concentration of each substance in the equilibrium mixture at 1000°C given that K_c = 0.58 at this temperature?

Answer 1

The given chemical equation,

CO(g) + H2O(g) <===> CO2(g) + H2(g)

We can write,

Kc = [CO2][H2]/[CO][H2O]

Given, 1 mole of CO and 1 mole of H2O are reacting,

so initial molarity of [CO] = [H2O] = moles/L = 1/50 = 0.02 M

Let x M change (reacted) in the starting material concentration at equlibrium then,

[CO2] = [H2] = x M

[CO] = [H2O] = 0.02-x

Kc = 0.58 = x^2/(0.02-x)(0.02-x)

0.58 = x^2/(0.0004-0.04x+x^2)

2.32 x 10^-4 - 0.0232x + 0.58x^2 = x^2

0.42x^2 + 0.0232x - 2.32 x 10^-4 = 0

Solve the quadratic equation for x,

x = 8.65 x 10^-3 M

So the equlibrium concentration of each at equlirbium would be,

[CO] = 0.02-8.65 x 10^-3 = 0.01135 M

[H2O] = 0.02-8.65 x 10^-3 = 0.01135 M

[CO2] = 8.65 x 10^-3 M

[H2] = 8.65 x 10^-3 M