Question

In a reversible reaction A + B C Initially
mol of each reactant is taken in 2L container if
at equilibrium 20% of A reacts with B. Then what
will be equilibrium constant K for reaction :-
(1) 0.225
(2) 0.412
(3) 0.625
(4) 0.925​

Answer 1

answer : option (3) 0.625

It has given that in a reversible reaction, A + B ⇔C , initially 1 mole each reactant is taken in 2L container if at equilibrium 20% of A reacts with B.

we have to find equilibrium constant K for reaction.

A + B ⇔ C

at t = 0 1 1 0

at eqⁿ 1 - α 1 - α α

equilibrium constant, K = [C]/[A][B]

= α/V/(1 - α)²/V²

= αV/(1 - α)²

=(0.2 × 2)/(1 - 0.2)²

= 0.4/(0.8)²

= 0.4/0.64

= 1/1.6

= 0.625

therefore equilibrium constant,K = 0.625

Answer 2

Answer:

the equilibrium constant is 0.625

Explanation: