Question

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sample of NH3(g) is introduced into an evacuated vessel at 650 K, the equilibrium concentration of H2(g) is found to be 0.729 M. Calculate the concentration of NH3 in the equilibrium mixture. M

Answer 1

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

                 $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

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

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M