Question

$28gms$ of $N_2$ and $6gms$ of $H_2$ were heated in a closed $1litre$ vessel. At equilibrium, $25.5gms$ of $N{H}_3$ is present. The approximate value of $K_c$ is ?

• A. $5.55$
• B. $21.33$
• C. $2.54$
• D. $3.16$

Answer 1

Answer: (B)

$21.33$

$N_2+3H_2\rightleftharpoons 2N{H}_3$

Initial no. of moles of $H_2$ = 6/2 = 3moles; $N_2$ = 28/28 = 1 mole.

At equilibrium no. of moles of $N{H}_3$ = 25.5/17 = 1.5

2x = 1.5; x= 0.75.

at equilibrium no. of N_2 = 1-0.75 = 0.25; H_2 = 3(1-0.25) = 0.75.

$K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right]\ast [H_2{]}^3}$

$K_c=\frac{(1.5{)}^2}{\left(0.25\right)(0.75{)}^3}$

$K_c=\frac{\left(2.25\right)}{0.25\ast 0.421)}=21.33$