Answer to Question #152123 in Chemistry for Alexis

Question

When 1.50 mol of CO2 and 1.50 mol H2 are placed in a 0.750-L container at 395ºC, the following

equilibrium is achieved: CO2(g) + H2(g) ↔ CO(g) + H2O(g).

If Kc = 0.802, what are the concentrations of each substance in the equilibrium mixture?

USE 3 SIG FIGS for each answer?

CO2 & H2 =

CO & H2O =

Answer 1

Solution:

The balanced chemical equation:

CO₂(g) + H₂(g) ↔ CO(g) + H₂O(g)

The equilibrium expression for the reaction system:

The initial concentration (C_o) of each gas is:

C_o(CO₂) = n(CO₂) / V_container = 1.50 mol / 0.750 L = 2.00 mol/L = 2.00 M

C_o(H₂) = n(H₂) / V_container = 1.50 mol / 0.750 L = 2.00 mol/L = 2.00 M

C_o(CO) = C_o(H₂O) = 0 (at the beginning of the reaction).

The reaction table is:

The equilibrium concentration of each gas is:

[CO₂] = [H₂] = 2.00 - x

[CO] = [H₂O] = x

Hence,

or

x = 0.8955 * (2.00 - x)

1.11664x = 2.00 - x

2.11664x = 2.00

x = 0.9449 = 0.945

Thus:

[CO] = [H₂O] = x = 0.945 M

[CO₂] = [H₂] = 2.00 - 0.945 = 1.055 M = 1.06 M

Answer:

[CO] = [H₂O] = 0.945 M;

[CO₂] = [H₂] = 1.06 M.