Question

Consider the following reaction in a $1\, L$ closed vessel.
$N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}$
If all the species; $N _{2}, H _{2}$ and $NH _{3}$ are in $1\, mol$ in the beginning of the reaction and equilibrium is attained after unreacted $N _{2}$ is $0.7 \,mol$. 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, $1-x=0.7 \,mol$

$\therefore \, x=0.3\, mol$

Therefore, concentration of $N _{2}, \,H _{2}$ and $NH _{3}$ at equilibrium will be

$\left[ N _{2}\right] =[0.7] $

$\left[ H _{2}\right] =1-(3 \times 0.3)=[0.1] $

$\left[ NH _{3}\right] =1+2 x=1+(2 \times 0.3)=[1.6]$

According to law of equilibrium constant $\left(K_{C}\right)$

$K_{C}=\frac{\left[ NH _{3}\right]^{2}}{\left[ N ^{2}\right],\left[ H _{2}\right]^{3}}=\frac{[1.6]^{2}}{[0.7][0.1]^{3}}$

$K_{C}=\frac{(256)}{0.007}=3657.14$

Hence, equilibrium constant $\left(K_{C}\right)=3657.14$