Question

Consider the following reaction in a $1L$ closed vessel.
$N_2+3H_2\rightleftharpoons 2N{H}_3$
If all the species; $N_2,H_2$ and $N{H}_3$ are in $1mol$ in the beginning of the reaction and equilibrium is attained after unreacted $N_2$ is $0.7mol$. What is the value of equilibrium constant?

• A. 3600.00
• B. 3657.14
• C. 2657.14
• D. 1828.57

Answer 1

Answer: (B)

For the given reaction:



Given, $1-x=0.7mol$

$\therefore x=0.3mol$

Therefore, concentration of $N_2,H_2$ and $N{H}_3$ at equilibrium will be

$\left[N_2\right]=[0.7]$

$\left[H_2\right]=1-(3\times 0.3)=[0.1]$

$\left[N{H}_3\right]=1+2x=1+(2\times 0.3)=[1.6]$

According to law of equilibrium constant $\left(K_C\right)$

$K_C=\frac{{\left[N{H}_3\right]}^2}{\left[N^2\right],{\left[H_2\right]}^3}=\frac{[1.6{]}^2}{[0.7][0.1{]}^3}$

$K_C=\frac{(256)}{0.007}=3657.14$

Hence, equilibrium constant $\left(K_C\right)=3657.14$