Question

For the reaction $A\left(g\right)+3B\left(g\right)\rightleftharpoons 2C\left(g\right)+D(g,)$ initial mole of $A$ is twice that of $B$, if at equilibrium, mole of $B$ is half of moles of $C$ then percent of $B$ reacted is

• A. $75$
• B. $45$
• C. $91$
• D. $32$

Answer 1

The correct option is A $75$

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

Initial concentration a a/2 0 0

At equilibrium (a-x) (a/2-3x) 2x x

We have been given that at equilibrium moles of B is half of mole of C

i.e. $\frac{a}{2}-3x=\frac{1}{2}\left(2x\right)$

$x=\frac{a}{8}$

Therefore, the maount of B reacted $=3x=\frac{3a}{8}$

Initial amount of B $=\frac{a}{2}$

So, % of B reacted $=\frac{AmountofBreacted}{InitialamountofB}\times 100$

% of B reacted $=\frac{\frac{3a}{8}}{\frac{a}{2}}\times 100=75\%$