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Consider the following reaction in a $1 \mathrm{~L}$ closed vessel.
$\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$
If all the species; $\mathrm{N}_2, \mathrm{H}_2$ and $\mathrm{NH}_3$ are in $1 \mathrm{~mol}$ in the beginning of the reaction and equilibrium is attained after unreacted $\mathrm{N}_2$ is $0.7 \mathrm{~mol}$. What is the value of equilibrium constant?
Options:
$\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$
If all the species; $\mathrm{N}_2, \mathrm{H}_2$ and $\mathrm{NH}_3$ are in $1 \mathrm{~mol}$ in the beginning of the reaction and equilibrium is attained after unreacted $\mathrm{N}_2$ is $0.7 \mathrm{~mol}$. What is the value of equilibrium constant?
Solution:
2851 Upvotes
Verified Answer
The correct answer is:
3657.14
For the given reaction:
$\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$
$\begin{array}{llll}\text { Initial moles } & 1 & 1 & 1\end{array}$
At equilibrium $1-x \quad 1-3 x \quad 2 x$
Given, $1-x=0.7 \mathrm{~mol}$
$\therefore \quad x=0.3 \mathrm{~mol}$
Therefore, concentration of $\mathrm{N}_2, \mathrm{H}_2$ and $\mathrm{NH}_3$ at equilibrium will be
$\begin{aligned}
{\left[\mathrm{N}_2\right] } & =[0.7] \\
{\left[\mathrm{H}_2\right] } & =1-(3 \times 0.3)=[0.1] \\
{\left[\mathrm{NH}_3\right] } & =1+2 x=1+(2 \times 0.3)=[1.6]
\end{aligned}$
According to law of equilibrium constant $\left(K_C\right)$
$\begin{aligned}
K_C & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}^2\right],\left[\mathrm{H}_2\right]^3}=\frac{[\mathrm{l} .6]^2}{[0.7][0.1]^3} \\
K_C & =\frac{(256)}{0.007}=3657.14
\end{aligned}$
Hence, equilibrium constant $\left(K_C\right)=3657.14$ and (b) is the correct option.
$\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3$
$\begin{array}{llll}\text { Initial moles } & 1 & 1 & 1\end{array}$
At equilibrium $1-x \quad 1-3 x \quad 2 x$
Given, $1-x=0.7 \mathrm{~mol}$
$\therefore \quad x=0.3 \mathrm{~mol}$
Therefore, concentration of $\mathrm{N}_2, \mathrm{H}_2$ and $\mathrm{NH}_3$ at equilibrium will be
$\begin{aligned}
{\left[\mathrm{N}_2\right] } & =[0.7] \\
{\left[\mathrm{H}_2\right] } & =1-(3 \times 0.3)=[0.1] \\
{\left[\mathrm{NH}_3\right] } & =1+2 x=1+(2 \times 0.3)=[1.6]
\end{aligned}$
According to law of equilibrium constant $\left(K_C\right)$
$\begin{aligned}
K_C & =\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}^2\right],\left[\mathrm{H}_2\right]^3}=\frac{[\mathrm{l} .6]^2}{[0.7][0.1]^3} \\
K_C & =\frac{(256)}{0.007}=3657.14
\end{aligned}$
Hence, equilibrium constant $\left(K_C\right)=3657.14$ and (b) is the correct option.
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