Question

Consider the reaction CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) Kc = 7.7×10–23 at 25 °C. 1.50 mol CH4 and 2.50 mol H2O are added to an empty 2.00-L container at 25 °C and allowed to reach equilibrium. What are the equilibrium amounts (in moles) of each species? 1.5

Answer 1

Answer : The moles of
`CH_4`,
`H_2O`,
`CO` and
`H_2` at equilibrium is, 0 mol, 1 mol, 1.5 mol and 4.5 mol respectively.

Explanation :

First we have to calculate the concentration of
`CH_4\text{ and }H_2O`

`\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=(1.50mol)/(2.00L)=0.75M`

and,

`\text{Concentration of }H_2O=\frac{\text{Moles of }H_2O}{\text{Volume of solution}}=(2.50mol)/(2.00L)=1.25M`

The given chemical reaction is:

`CO(g)+3H_2(g)\rightleftharpoons CH_4(g)+H_2O(g)`

Initial conc. 0 0 0.75 1.25

At eqm. x 3x (0.75-x) (1.25-x)

The expression for equilibrium constant is:

`K_c=([CH_4][H_2O])/([CO][H_2]^3)`

Now put all the given values in this expression, we get:

`7.7* 10^(-23)=((0.75-x)* (1.25-x))/((x)* (3x)^3)`

x = -2.19317 × 10¹⁰

x = 0.75

x = 1.25

x = 2.19317 × 10¹⁰

We are accepting value of x = 0.75 while all the values of x are neglecting because equilibrium concentration can not be more than initial concentration.

Equilibrium concentration of
`CH_4` = (0.75-x) = (0.75-0.75) = 0 M

Equilibrium concentration of
`H_2O` = (1.25-x) = (1.25-0.75) = 0.50 M

Equilibrium concentration of
`CO` = x = 0.75 M

Equilibrium concentration of
`H_2` = 3x = 3(0.75) = 2.25 M

Now we have to calculate the moles of each species in terms of mole.

Moles of
`CH_4` at equilibrium =
`0M* 2.00L=0mol`

Moles of
`H_2O` at equilibrium =
`0.50M* 2.00L=1mol`

Moles of
`CO` at equilibrium =
`0.75M* 2.00L=1.5mol`

Moles of
`H_2` at equilibrium =
`2.25M* 2.00L=4.5mol`