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 $1$ litre flask. At equilibrium which is true?

• A. $\left[P\right]<\left[Q\right]$
• B. $\left[P\right]=\left[Q\right]$
• C. $\left[Q\right]=\left[R\right]$
• D. None of these

Answer 1

The correct option is A $\left[P\right]<\left[Q\right]$

$2P_{\left(g\right)}+Q_{\left(g\right)}\rightleftharpoons 3R_{\left(g\right)}+S_{\left(g\right)}$

Initially $2moles$ $2moles$ $0$ $0$

At eqm $2-2\alpha$ $2-\alpha$ $\alpha$ $\alpha$

Here $2$ moles of $P$ reacts with $1$ mole of $Q$. So at equilibrium number of moles of $P$ consumed will be higher than that of $Q$. Therefore, rate of decrease of $\left[P\right]$ is higher than that of $\left[Q\right]$.

$\therefore \left[P\right]<\left[Q\right]$ at equilibrium.