Question

In the decomposition reaction of ammonia:
${2NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
$2$ moles of ${NH}_3$ are introduced in the vessel of $1$ litre. At equilibrium, $1$ mole ${NH}_3$ was left, the value of $K_c$ will be:

• A. $0.75$
• B. $0.70$
• C. $1.75$
• D. $1.70$

Answer 1

The correct option is D $1.70$

In the decomposition reaction of ammonia:
${2NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
$2$ moles of ${NH}_3$ are introduced in the vessel of $1$ litre. At equilibrium,
$1$ mole ${NH}_3$ was left, the value of $K_c$ will be $1.70$.
Since total volume is 1 L, the number of moles is equal to molar concentration.
${2NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
Initial moles $t=0$ $2$ $0$ $0$
Moles at equilibrium $t_{eq.}$ $1$ $1/2$ $3/2$
At equilibrium, $2-1=1$ mole ${NH}_3$ was left. So $1$ mole ${NH}_3$ has reacted to form $1/2$ mole of $N_2$ and $3/2$ moles of $H_2$
$K_c=\frac{\left[N_2\right]{\left[H_2\right]}^3}{{\left[{NH}_3\right]}^2}$
$K_c=\frac{\left[1/2\right]{\left[3/2\right]}^3}{{\left[1\right]}^2}$
$K_c=\frac{\left[1/2\right]{\left[3/2\right]}^3}{{\left[1\right]}^2}$
$K_c=1.687$