Question

The system CO2(g) + H2(g) ⇀↽ H2O(g) + CO(g) is at equilibrium at some temperature. At equilibrium a 4.00 L vessel contains 1.00 mole CO2, 1.00 mole H2, 2.40 moles H2O, and 2.40 moles CO. How many moles of CO2 must be added to the system to bring the equilibrium CO concentration to 0.727 mol/L?

Answer 1

Number of moles added = 4.6 - 0.25 = 4.35 moles

Step-by-step explanation:

CO2(g) + H2(g) ⇄ H2O(g) + CO(g)

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

Concentration = Number of moles / Volume

[H2O] = 2.40 /4 = 0.6

[CO] = 2.40 / 4 = 0.6

[CO2] = 1 / 4 = 0.25

[H2] = 1 / 4 = 0.25

Kc = 0.6 * 0.6 / (0.25 * 0.25)

Kc = 0.36 / 0.0625

Kc = 5.76

If equilibrium [CO] = 0.727, it means that of [H2O] is also 0.727.

This means there was an addition of 0.727 - 0.6 = 0.127 M

The increase is contributed by H2, so [H2] at equilibrium is given as; 1 - 0.127 = 0.873 M

Kc, remains constant; solving for [CO2]

5.76 = 0.727 * 0.727 / (0.873 * [CO2])

5.76 = 0.529 / 0.873 [CO2]

0.461 [CO2] = 0.529

[CO2] = 0.529 / 0.461 = 1.15

Converting to moles;

Number of moles = Concentration * Volume = 1.15 * 4 = 4.6 moles

Number of moles added = 4.6 - 0.25 = 4.35 moles