In the following hypothetical reaction, A+3 B leftharpoons 2 C+D initial moles of A are twice as that of B. If at equilibrium moles of B and C are equal. Percentage of B reacted is

Question

In the following hypothetical reaction, A+3 B leftharpoons 2 C+D initial moles of A are twice as that of B. If at equilibrium moles of B and C are equal. Percentage of B reacted is (A) 60 % (B) 40 % (C) 10 % (D) 20 %

Tardigrade Admin · Accepted Answer

Given that at equilibrium, x-3 y=2 y, x=2 y+3 y=5 y Moles of B reacted =3 y %=( text Moles reacted / text Total moles ) × 100=(3 y/x) × 100=(3 y/5 y) × 100 =60 %

Q. In the following hypothetical reaction, $A + 3 B ⇌ 2 C + D$ initial moles of $A$ are twice as that of $B$ . If at equilibrium moles of $B$ and $C$ are equal. Percentage of $B$ reacted is

$60%$

$40%$

$10%$

$20%$

Given that at equilibrium,

$x - 3 y = 2 y, x = 2 y + 3 y = 5 y$

Moles of $B$ reacted $= 3 y$

$% = \frac{Moles reacted}{Total moles} \times 100 = \frac{3 y}{x} \times 100 = \frac{3 y}{5 y} \times 100$

$= 60%$

Q. In the following hypothetical reaction, A+3B⇌2C+D initial moles of A are twice as that of B. If at equilibrium moles of B and C are equal. Percentage of B reacted is

60%

40%

10%

20%