Question

Two moles of ammonia is introduced in a evacuated 500 ml vessel at high temperature. The decomposition reaction is:

${2NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+{3H}_2\left(g\right)$

At the equilibrium ${NH}_3$ becomes 1 mole then the $K_c$ would be:

• A. $0.42$
• B. $108$
• C. $1.7$
• D. $1.5$

Answer 1

Answer: (B)

$108$

Given that,

Volume $=500ml=0.5L$

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

$2$ $0$ $0$ --- Initial mole

$1$ $1$ $3$ ---- At equilibrium mole

$2$ $2$ $6$ ---- At equilibrium conc.

Now,

$K_c=\frac{\left[N_2\right][H_2{]}^3}{[N{H}_3{]}^2}=\frac{2\times (6{)}^3}{(2{)}^2}=108$

$K_c=108$