Question

1 mole of n2 and three moles of h2 are placed in 1 l vessel. Find the concentration of nh3 at equilibrium if the kc at 400 k is 4/27

Answer 1

Answer : The concentration of $NH_3$ at equilibrium is 0.762 mole/L.

Solution : Given,

Volume = 1 L

Temperature = 400 K

Equilibrium constant,  $K_c$ = $\frac{4}{27}$

The Net balanced chemical reaction is,

                                          $N_2+3H_2\rightarrow 2NH_3$

Initial concentration           1         2             0

At equilibrium                  (1-x)      (3-3x)      2x

The expression for equilibrium constant is,

$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$

Now put all the values in this expression, we get

$\frac{4}{27}= \frac{(2x)^2}{(1-x)(3-3x)^3}$

By rearranging the terms, we get the value of 'x'.

x = 0.381 mole/L

The concentration of $NH_3$ at equilibrium = 2x = 2 × 0.381 = 0.762 mole/L

Therefore, the concentration of $NH_3$ at equilibrium is 0.762 mole/L.