Question

For the reaction $A+3B\rightleftharpoons 2C$, initial mole of A is twice that of B. If at equilibrium moles of B and C are equal, then percent of B reacted is:
A. 10%
B. 20%
C. 40%
D. 60%

Answer 1

Hint: Since the number of moles of B and C is equal at equilibrium, the amount of unreacted B is equal to the amount of C formed at equilibrium. Equate both of them to get an expression. Solving it will give the percentage of B that reacted.

Complete answer:
Formula Used: $Percent\quad Reacted=\dfrac { Reacted\quad Amount }{ Initial\quad Amount } \times 100$
Let at the initial stage, before the reaction started, the number of moles of A taken is ‘a’.
It is given that the initial mole of A taken is twice that of B. So, the number of moles of B will be ‘$\dfrac { a }{ 2 }$’. Also, in the initial stage, the number of moles of C and D will be 0, as the reaction has not started yet.
Given in the reaction $A+3B\rightleftharpoons 2C+D$, 3 moles of B react with one mole of A and at equilibrium, 2 moles of C are formed along with 1 mole of D. Let the number of moles be ‘x’
So, the number of moles of A reacting will be ‘x’, and the number of moles of B reacting with ‘x’ moles of A will be ‘3x’. Number of moles of C forming at equilibrium will be ‘2x’ and the number of moles of D will be ‘x’.
Since ‘x’ moles of A have reacted, the number of moles of A left at equilibrium will be ‘$a-x$’.
Since ‘3x’ moles of B have reacted, the number of moles of B left at equilibrium will be ‘$\dfrac { a }{ 2 } -3x$’.
$A+3B\rightleftharpoons 2C+D$

Number of Moles of:	A	B	C	D
Initial:	a	$\dfrac { a }{ 2 }$	0	0
Equilibrium:	$a-x$	$\dfrac { a }{ 2 } -3x$	2x	x

It is given in the question that, at equilibrium, the number of moles of B and C is equal.
So, $\dfrac { a }{ 2 } -3x=2x$
$\Rightarrow \dfrac { a }{ 2 } =2x+3x=5x$
$\Rightarrow \dfrac { a }{ 2\times 5 } =x$
$\Rightarrow x=\dfrac { a }{ 10 }$ ……………..(i)
Now, the amount of B reacted = 3x
Putting the value of x from eq (i), we get;
$3x=\dfrac { 3a }{ 10 }$
Now, the initial amount of B = $\dfrac { a }{ 2 }$
Amount of B reacted $= 3x=\dfrac { 3a }{ 10 }$
So, percentage of B reacted $= \dfrac { Reacted\quad Amount }{ Initial\quad Amount } \times 100$
$=\dfrac { \dfrac { 3a }{ 10 } }{ \dfrac { a }{ 2 } } \times 100$
$=\dfrac { 3a\times 2 }{ 10\times a } \times 100$
$=60$

So, the percentage of B reacted is 60%. Thus option D is the correct one.

Note: Always remember that in the case of finding the percentage reacted, the amount reacted should be on the numerator while the initial amount should be on the denominator. It should not be the other way round.