Question

Given reaction:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
What would happen if there is a decrease in volume of container in which reaction occurs?

• A. Less $N{H}_3$ would form
• B. More $N_2$ would form
• C. More $N{H}_3$ would form
• D. More $H_2$ would form

Answer 1

The correct option is C More $N{H}_3$ would form

Calculating the change in no. of gaseous moles i.e. $\Delta n_g=2-4=-2(<0)\Delta n_g<0$

$As,V\downarrow \Rightarrow \frac{n_g}{V}\uparrow$
So according the Le - chatelier's principle reaction will go towards where $n_g$ will decrease.
Since the product side has only two moles of gas, compared to the reactant side which has four gaseous moles.
$n_g$ decreases in forward direction.
$\Rightarrow$ The reaction would shift towards the product side.

Hence more $N{H}_3$, i.e. more product will be formed.