Question

Two moles of ammonia is introduced in a evacuated $500$ mL vessel at high temperature. This decomposition reaction is?
$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
At the equilibrium $N{H}_3$ becomes $1$ mole then the $K_c$ would be:

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

Answer 1

The correct option is C $1.7$

$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)atequil.2-2xx3x$

Amount of $N{H}_3$ left is $1$ mole

$2-2x=1$

$x=0.5$

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

$=\frac{\left(0.5\right)(1.5{)}^3}{1}=1.7$