Question

5. A study of the system, 4 NH3(g) + 7 O2(g) 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [NH3] = [O2] = 3.60 M as the only components initially. At equilibrium, [N2O4] is 0.60 M. Calculate the value of the equilibrium constant, Kc, for the reaction.

Answer 1

Answer: The value of the equilibrium constant, Kc, for the reaction is 0.013

Explanation:

Initial concentration of $NH_3$ = 3.60 M

Initial concentration of $O_2$ = 3.60 M

The given balanced equilibrium reaction is,

            $4NH_3(g)+7O_2(g)\rightleftharpoons 2N_2O_4(g)+6H_2O(g)$

Initial conc.       3.60 M        3.60 M              0 M        0 M

At eqm. conc.     (3.60-4x) M   (3.60-7x) M   (2x) M      (6x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[N_2O_4]^2\times [H_2O]^6}{[NH_3]^4\times[Cl_2]}$

$[N_2O_4]=0.60M$

2x = 0.60 M

x= 0.30 M

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

$K_c=\frac{(2\times 0.30)^2\times (6\times 0.30)^6}{(3.60-3\times 0.30)^4\times (3.60-7\times 0.30)^7}$

$K_c=0.013$