Question

For the equilibrium reaction A+3B = 2C +D,initial mole of A is twice that of B.If at equilibrium moles of A and B are equal,then the percent of B reacted?

Answer 1

Answer:

ANSWER

Given reaction is

                    A+3B⇌                      2C+D

t=0,               a        a/2                    0     0

t=eqm,    a−x     a/2−3x                2x      x

At equilibrium,

a  /2 −3x=2x (Given)

a /2=5x⇒x=  a/10

∴ Amount of B reacted= 3x=  3a/10

Initial amount of B=a/2

So, % of B reacted=(  3a/10  /  a/2  )×100

                               =  6/10 ×100=60%

Explanation: