Question

For the reaction $A+3B\rightleftharpoons 2C+D$, initial mole of $A$ is twice that of $B$. If at equilibrium moles of $B$ are equal, then percent of $B$ reacted is:

• A. 10%
• B. 20%
• C. 40%
• D. 60%

Answer 1

The correct option is D 60%

Given reaction is

$A+3B\rightleftharpoons 2C+D$

$t=0$, $a$ $a/2$ $0$ $0$

$t=eqm$, $a-x$ $a/2-3x$ $2x$ $x$

At equilibrium,

$\frac{a}{2}-3x=2x$ (Given)

$\frac{a}{2}=5x\Rightarrow x=\frac{a}{10}$

$\therefore$ Amount of $B$ reacted= $3x=\frac{3a}{10}$

Initial amount of $B=a/2$

So, % of $B$ reacted=$\left(\frac{3a}{10}/\frac{a}{2}\right)\times 100$

$=\frac{6}{10}\times 100=60$%