Question

In an experiment, 0.35 mol of co and 0.40 mol of h2o were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.22 mol of co remaining. keq at the temperature of the experiment is ________.

Answer 1

The equilibrium constant of reaction at given temperature is 0.28.

Step-by-step explanation:

`CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)`

Initially

0.35 mol 0.40 mol

At equilibrium

(0.35 mol-x) (0.40 mol-x) x x

Moles of CO left at equilibrium = 0.22 mol

0.22 mol = (0.35 mol-x)

x = 0.35 mol - 0.22 mol = 0.13 mol

Moles of
`H_2O`left at equilibrium = 0.40 mol - 0.13 mol = 0.27 mol

Concentration of CO at equilibrium =
`[CO]=(0.22 mol)/(1 L)`

Concentration of
`H_2O` at equilibrium =
`[H_2O]=(0.27 mol)/(1 L)`

Concentration of
`CO_2` at equilibrium =
`[CO_2]=(0.13 mol)/(1 L)`

Concentration of
`H_2` at equilibrium =
`[H_2]=(0.13 mol)/(1 L)`

`K=([CO_2][H_2])/([CO][H_2])`

`=((0.13 mol)/(1 L)* (0.13 mol)/(1 L))/((0.22 mol)/(1 L)* (0.27 mol)/(1 L))=0.28`

The equilibrium constant of reaction at given temperature is 0.28.

Answer 2

We need to set up an I.C.E chart as follows:

CO H20 CO2 H2
I 0.35 0.40 0 0
C .35-x 0.40-x x x
E 0.19 0.40-x x x

We have 0.35 mol of CO at the start and 0.19 remaining. This means that x=0.35-0.19 >> 0.16
Now substitute in 0.16 in your Ke equation.
Just in case you're having trouble with that:
Kb= [Products]/[Reactants] = ([CO][H2O])/([CO2][H2])

Therefore, the constant of equilibrium would be equal to 0.56.