Question

For the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$ in a vessel, after the addition of equal number of moles of $N_2$ and $H_2$ equilibrium state is achieved. Which of the following is correct?

• A. $\left[H_2\right]=\left[N_2\right]$
• B. $\left[H_2\right]<\left[N_2\right]$
• C. $\left[H_2\right]>\left[N_2\right]$
• D. $\left[H_2\right]>\left[{NH}_3\right]$

Answer 1

Answer: (B)

$\left[H_2\right]<\left[N_2\right]$

For the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$ in a vessel, after the addition of equal number of moles of $N_2$ and $H_2$ equilibrium state is achieved. Then $\left[H_2\right]<\left[N_2\right]$.
Assume total volume is 1 L. So number of moles is equal to molar concentration.
Assume that 4 moles of $N_2$ and 4 moles of $H_2$ are mixed.
As per reaction stoichiometry, if 1 mole of $N_2$ will react with 3 moles of $H_2$ to reach equilibrium, then
$4-1=3$ moles of $N_2$ and $4-3=1$ mole of $H_2$ will remain unreacted. Hence, $\left[H_2\right]<\left[N_2\right]$.