Question

In the reaction $2P\left(g\right)+Q\left(g\right)\rightleftharpoons 3R\left(g\right)+S\left(g\right)$. If 2 mol each of P and Q taken initially in a 1 L 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

Answer: (A)

$\left[P\right]<\left[Q\right]$

According to the problem:

Given reaction:

$2P\left(g\right)+Q\left(g\right)\leftrightarrow 3R\left(g\right)+S\left(g\right)$

Initial $2mol2mol00$

At equilibrium $2-2x2-x3xx$

Form the above table,

$\left[P\right]<\left[Q\right]$

Hence the correct answer is $\left(A\right)$