Question

The equilibrium constant of the reaction A2 +B2 = 2AB at 100 C is 50 . If one litre fladk containing one mole A2 is connectef to a two litre flask containing two moles B2 then how many moles of AB will be formed at 373k ?

Answer 1

Answer : The number of moles of AB will be, 1.966 moles.

Solution : Given,

Volume of $A_2$ = 1 L

Volume of $B_2$ = 2 L

Volume of $AB$ = 1 + 2 = 3 L

Moles of $A_2$ = 1 mole

Moles of $B_2$ = 2 mole

The balanced equilibrium reaction is,

                              $A_2+B_2\rightleftharpoons 2AB$

Initially moles           1     2          0

At eqm.                (1 - x)  (2 - x)     2x

The expression for equilibrium constant of the reaction will be,

$K_{eq}=\frac{[AB]^2}{[A_2][B_2]}$

Now put all the given values in this expression, we get

$50=\frac{(\frac{2x}{3})^2}{(\frac{2-x}{2})\times (\frac{1-x}{2})}$

By solving the terms 'x' we get,

x = 0.983

Now we have to calculate the number of moles of AB.

Number of moles of AB = 2x = 2 × 0.983 = 1.966 moles

Therefore, the number of moles of AB will be, 1.966 moles.