Question

$3$ moles of $N{H_3}$ are allowed to dissociate in a $5$ litre vessel and the equilibrium concentration of ${N_2}$is $0.2\frac{{mole}}{{lit}}$. Then the total number of moles at equilibrium is:
(A) $2.5$
(B) $5$
(C) $1.5$
(D) $7.5$

Answer 1

Hint: One mole of an object is equal to $6.022 \times {10^{23}}$ units of that object. The number $6.022 \times {10^{23}}$ is known as the Avogadro or Avogadro fixed number. The concept of mole can be used to convert between mass and number of particles.

Complete Step by Step Solution:
The reaction when $N{H_3}$ dissociates into ${N_2}$ and ${H_2}$
$N{H_3} \to {N_2} + 3{H_2}$

With an equal number of moles on both sides of the reaction, the increase in volume will not affect the balance and thus there are no fluctuations in orientation. Similarly, if you lower the volume there is no effect on the measurement. To find the number of moles of an object in the scale you take the number of moles at the beginning, and then add the change in the number of moles by going to $x$.

Initially the number of moles of $N{H_3}$is $3$is for ${N_2}$and ${H_2}$is $0$. Then, the equilibrium number for $N{H_3}$ is $3 - 2x$and for ${N_2}$and ${H_2}$ is $x$ and $3x$respectively.
So, the total number of moles are
$ \Rightarrow 3 - 2x + x + 3x$
We know the number of moles for ${N_2}$is $0.2$
So, the value for $x$is $1$
So, total number of moles are:
$ \Rightarrow 3 - 2x + x + 3x$
$ = 5$
Therefore, the correct answer is B) $5$.

Note: When there is an equal number of gas molecules on both sides of an arrow, volume changes have the same effect on the concentration of reactors and products. It has an equal impact on the ups and downs, and the system remains in equilibrium.