Question

Two moles of ammonia is introduced in evacuated 500 mL vessel at high temperature . The decomposition reaction is :
$N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
At the equilibrium $N{H}_3$ becomes 1 mole then the KI would be :-

• A. $0.42$
• B. $6.75$
• C. $1.7$
• D. $1.5$

Answer 1

The correct option is A $0.42$

Answer
$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_{2}g$
initial 2 0 0
equi. 2-2x x 3x
amount of $N{H}_3$ left is 1 mole

$⟹2-2x=1$
$x=0.5$
$K_c=\frac{\left[N_2\right][x_2{]}^3}{[N{H}_3{]}^2}$ $=\frac{\left(0.5\right)(1.5{)}^3}{(2-2\times 0.5{)}^2}=\frac{\left(0.5\right)(1.5{)}^3}{1}$ $=1.7$

Hence, Option "C" is the correct answer.