Question

When H₂(g) is mixed with CO₂(g) at 2,000 K, equilibrium is achieved according to the following equation

CO₂(g) + H₂(g) « H₂O(g) + CO(g)

In one experiment, the following equilibrium concentrations were measured.

[H₂] = 0.20 mol/L

[CO₂] = 0.30 mol/L

[H₂O] = [CO] = 0.55 mol/L

What is the mole fraction of CO(g) in the equilibrium mixture?

Answer 1

The mole fraction of CO(g) in the equilibrium mixture is 0.34375, calculated by dividing the concentration of CO(g) by the total moles of all substances present in the mixture.

Step-by-step explanation:

To calculate the mole fraction of CO(g) in the equilibrium mixture, we need to first determine the total number of moles present in the mixture. This can be done by summing up the equilibrium concentrations of each component:

Total moles = [H₂] + [CO₂] + 2[H₂O] (Since [H₂O] = [CO])

Total moles = 0.20 mol/L + 0.30 mol/L + 2(0.55 mol/L)

Total moles = 1.60 mol/L

Now, we calculate the mole fraction of CO(g) which is the ratio of the moles of CO(g) to the total moles:

Mole fraction of CO(g) = [CO] / Total moles

Mole fraction of CO(g) = 0.55 mol/L / 1.60 mol/L

Mole fraction of CO(g) = 0.34375