Question

In the following equilibrium ,2A+B =3C+D initial moles of 

B is twice  that of A,When  equilibrium is reached, moles of A and C are equal .Thus , percentage dissociation of B is 
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Answer 1

Answer:

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Answer 2

Given :

2A+B =3C+D

Initial moles of B is twice  that of A.

When equilibrium is reached, moles of A and C are equal.

To find :

Percentage dissociation of B

Solution :

Given reaction is:

                 2A  +  B  =  3C  +  D

t = 0           a       2a      0       0

t = eqm    a-2x  2a-x    3x      x

At equilibrium, moles of A and C are equal

a-2x = 3x

a = 3x + 2x

a = 5x

x = a/5

Therefore, amount of B reacted = x = a/5

Initial amount of B = 2a

So, % of B reacted = $\frac{a/5}{2a} *100$

% of B reacted = 10%

Therefore, 10% of B reacted.