Question

Two moles of NH3 gas are introduced into a
previously evacuated one litre vessel in which it
partially dissociates at high temperature as
2NH3 (g) = N2 (g) + 3H2 (g). At equilibrium, one
mole of NH3(g) remain. The value of K is
(1) 3
(2) 27/16
(3) 3/2
(4) 27/64​

Answer 1

Answer: 4.  $\frac{27}{64}$

Explanation:-

Initial moles of  $NH_3$ = 2 mole

Volume of container = 1 L

Initial concentration of $NH_3=\frac{moles}{volume}=\frac{2moles}{1L}=2M$  

equilibrium concentration of $NH_3=\frac{moles}{volume}=\frac{1mole}{1L}=1M$ [/tex]

The given balanced equilibrium reaction is,

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

Initial conc.         2 M               0     0

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

The expression for equilibrium constant for this reaction will be,

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

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

we are given : 2 - 2x =  1

1 = 2x

$x=\frac{1}{2}$

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

$K_c=\frac{\frac{1}{2}\times (3\times \frac{1}{2})^3}{(2-2\times \frac{1}{2})^2}$

$K_c=\frac{27}{64}$

Thus the value of the equilibrium constant is $\frac{27}{64}$