Question

A mixture containing $2$ moles of $N_2$ and $3$ moles of $H_2$ gas is brought to equilibrium in a $3L$ vessel. If the molar concentration of $N{H}_3$ at equilibrium is $1/3mol/L$, calculate the $K_c$ for the formation of $N{H}_3$.
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$.

Answer 1

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

At $2$ $3$ $0$

$t=0$

At $2(1-\alpha )$ $3(1-\alpha )$ $2\alpha$

$t=t_{eq}$

Volume of container $=3L$

concentration $=\frac{mol}{volinlitre}$

$\left[{NH}_3\right]=\frac{2\alpha }{3}=\frac{1}{3}$ ($\because$ given)

$\alpha =1/2$

$\left[N_2\right]$ at equilibrium $=\frac{2(1-\alpha )}{3}=\frac{1}{3}$

$\left[H_2\right]$ at equilibrium $=\frac{3(1-\alpha )}{3}=1/2$

$K_C=\frac{{\left(\frac{1}{3}\right)}^2}{{\left(\frac{1}{2}\right)}^3\left(\frac{1}{3}\right)}=8/3$