For reaction  A(g) + B(g) → AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB

Q:  For reaction  A(g) + B(g) $ \rightleftharpoons $ AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB

(A) 20%

(B) 40%

(C) 60%

(D) 4%

Solution:  A(g)     +    B(g) $ \rightleftharpoons $  AB(g)

Initial moles     2           2                      0

At equilibrium  2 – x       2 – x              x = 0.8

% of A changed to AB $\large = \frac{0.8}{2} × 100 $

= 40%

Ans: (B)