Question

The water-gas shift reaction plays a central role in the chemical methods of obtaining clean fuels from coal.

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

A study was made in which equilibrium was reached with [CO]=[H2O]=[H2]=0.100M and [CO2]=0.400M. After 0.800 mol of H2 is added to the 2.00L container and equilibrium is reestablished, what are the new concentrations of all the components?

Answer 1

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only

If Keq > 1, this favours products, since this relates to a higher amount of C + D

If Keq < 1, this favours reactants, since this relates to a higher amount of A + B

If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios

Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.

get Keq

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

Keq = (0.4)(0.1)/(0.1*0.1)

Keq = 0.4/0.1 = 4

initially:

[CO2] = 0.4

[H2] = 0.1 + 0.8/2 = 0.5

[CO] = 0.1

[H2O] = 0.1

find equilbirium

[CO2] = 0.4 - x

[H2] = 0.5 - x

[CO] = 0.1 + x

[H2O] = 0.1 + x

reaction goes backwards, since products were added

Keq = 1/4 = 0.25

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

solve

0.25 = (0.1+x)(0.1+x) / (0.4 - x)(0.5-x)

0.25 (0.4*0.5 -0.9x + x^2) = 0.1^2 + 0.2x + x^2

0.25 (0.4*0.5) -0.9*0.25x + 0.25^x2 =  0.1^2 + 0.2x + x^2

x^2 ( 1-0.25) + x(0.2 +0.225) + (0.1^2 - 0.05) = 0

0.75x^2 + 0.425x -0.04 = 0

x = 0.08219

[CO2] = 0.4 - 0.08219 = 0.31781 M

[H2] = 0.5 - 0.08219 = 0.41781  M

[CO] = 0.1 + 0.08219 = 0.18219 M

[H2O] = 0.1 + 0.08219 = 0.18219 M