Question

Consider the following reaction.

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

If the reaction begins in a 10.00 L vessel with 2.5 mol CO and 2.5 mol H2O gas at 588K (Kc= 31.4 at 588 K). Calculate the concentration of CO, H2O, CO2, and H2 at equilibrium.

Answer 1

[CO] = mol of CO / volume

= 2.5 mol / 10.00 L

= 0.25 M

[H2O] = mol of H2O / volume

= 2.5 mol / 10.00 L

= 0.25 M

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

0.25 0.25 0 0 (initial)

0.25-x 0.25-x x x (at equilibrium)

Kp = p(CO2)p(H2) / {p(CO)*p(H2O)}

31.4 = x^2 / (0.25-x)^2

sqrt(31.4) = x/ (0.25-x)

5.60 = x/ (0.25-x)

1.40 - 5.60*x = x

6.60*x = 1.40

x = 0.212 M

[CO] = 0.25 - x = 0.25 - 0.212 = 0.038 M

[H2O] = 0.25 - x = 0.25 - 0.212 = 0.038 M

[CO2] = x = 0.212 M

[H2] = x = 0.212 M