Question

In the reaction $2P\left( g \right) + Q\left( g \right) \rightleftharpoons 3R\left( g \right) + S\left( g \right)$. If $2$ moles each of $P$ and $Q$ taken initially in a $1litre$ flask. At equilibrium which is true?
A. $[P] < [Q]$
B. $[P] = [Q]$
C. $[Q] < [R]$
D. None of these

Answer 1

Hint: We can take the moles of each element dissociated in terms of any one element. The moles of each element consumed will be related to their respective stoichiometric coefficients. Thus, by writing the equilibrium relations, we can check which component has dissociated more, that is, which component’s concentration has reduced.

Complete step by step answer:
Let the moles of the component $Q$ dissociated be ‘$x$’. Thus, by looking at the balanced equation, we see that the moles of $P$ dissociated will be twice this amount, since the stoichiometric coefficient of $P$ ($2$) is double that of $Q$ ($1$). Similarly, the moles of $R$ formed will be thrice the moles of $Q$ dissociated, that is, $3x$. And finally, moles of $S$ formed will be equal to the moles of $Q$ dissociated, since their stoichiometric coefficients are equal.
And it is given that initially, $2$ moles each of $P$ and $Q$ are taken in the flask.
Thus, at equilibrium, we have the following relations:
\[2P + Q( g) \rightleftharpoons 3R(g) + S(g\]

Time	Moles of $P$	Moles of $Q$	Moles of $R$	Moles of $S$
Initial conditions	$2$	$2$	$0$	$0$
At equilibrium	$2 - 2x$	$2 - x$	$3x$	$x$

Thus, at equilibrium, we see that the number of moles of $P(2 - 2x)$ is lesser than the number of moles of $Q(2 - x)$. Thus, the concentration of $P$ will be lesser than that of $Q$. In other words, $[P] < [Q]$

So, the correct answer is Option A .

Note: The stoichiometric coefficients denote the exact number of moles of each component needed for the reaction to occur. In this case, from the balanced reaction, only one mole of $Q$ needs two moles of $P$ to react, indicating that the $2$ moles of $Q$ present in the flask requires four moles of $P$ for the reaction to be completed. However, since only two moles of $P$ are provided, $P$ is known as the limiting reagent here.