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 $CH_4\text{ and }H_2O$

$\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1.50mol}{2.00L}=0.75M$

and,

$\text{Concentration of }H_2O=\frac{\text{Moles of }H_2O}{\text{Volume of solution}}=\frac{2.50mol}{2.00L}=1.25M$

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:

$K_c=\frac{[CH_4][H_2O]}{[CO][H_2]^3}$

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

$7.7\times 10^{-23}=\frac{(0.75-x)\times (1.25-x)}{(x)\times (3x)^3}$

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 = $0M\times 2.00L=0mol$

Moles of $H_2O$ at equilibrium = $0.50M\times 2.00L=1mol$

Moles of $CO$ at equilibrium = $0.75M\times 2.00L=1.5mol$

Moles of $H_2$ at equilibrium = $2.25M\times 2.00L=4.5mol$