Question

The following reaction takes place in a 1.00 liter container at 750.0 °C.

H2(g) + CO2(g) ⇌ H2O(g) + CO(g)
At equilibrium there are 0.106 moles of H2, 0.106 moles of CO2, 0.094 moles of water and 0.094 moles of CO. What is the Keq for this reaction?

Answer 1

The equilibrium constant (Keq) for a reaction is a measure of the relative concentrations of the products and reactants at equilibrium. The equation for the equilibrium constant (Keq) is:

Keq = [H2O(g)][CO(g)] / [H2(g)][CO2(g)]

In order to find the Keq, we need to substitute the equilibrium concentrations of the species into the equation.

At equilibrium, there are 0.106 moles of H2, 0.106 moles of CO2, 0.094 moles of water and 0.094 moles of CO.

Keq = [0.094 moles H2O][0.094 moles CO] / [0.106 moles H2][0.106 moles CO2]

By canceling out the moles unit and substituting the number we got the value of Keq as:

Keq = 0.0940.094 / 0.1060.106

Keq = 0.87

Therefore, the Keq for this reaction is 0.87.

It is worth to note that the temperature, pressure and volume don't affect the Keq value but they affect the position of equilibrium.