Question

the equlibrium constant of the reaction a2 + b2 = 2ab at 100 degree celsius is 16.
Initially equal moles of a2 and b2 are taken in 2 litre container then find mole % of a2 at equilibrium.​

Answer 1

chemical reaction is...

a2 + b2 ⇔ 2ab

at t = 0, 1 1 0

at eql. 1 - α 1 - α 2α

so, equilibrium constant, K = [ab]²/[a2][b2]

⇒16 = (2α)²/(1 - α)(1 -α)

⇒16 = 4α²/(1 - α)²

⇒4 = 2α/(1 - α)

⇒4 - 4α = 2α

⇒4 = 6α

⇒α = 2/3

so, total moles at equilibrium = 1 - α + 1 - α + 2α = 2

no of moles of a2 at equilibrium = 1 - α

= 1 - 2/3 = 1/3

so, % mole of a2 at equilibrium = no of moles of a2/total moles at equilibrium × 100

= (1/3)/2 × 100

= 100/6

= 16.67 %

Answer 2

Answer:100/6

Explanation:

chemical reaction is...

a2 + b2 ⇔ 2ab

at t = 0, 1 1 0

at eql. 1 - α 1 - α 2α

so, equilibrium constant, K = [ab]²/[a2][b2]

⇒16 = (2α)²/(1 - α)(1 -α)

⇒16 = 4α²/(1 - α)²

⇒4 = 2α/(1 - α)

⇒4 - 4α = 2α

⇒4 = 6α

⇒α = 2/3

so, total moles at equilibrium = 1 - α + 1 - α + 2α = 2

no of moles of a2 at equilibrium = 1 - α

= 1 - 2/3 = 1/3

so, % mole of a2 at equilibrium = no of moles of a2/total moles at equilibrium × 100

= (1/3)/2 × 100

= 100/6

= 16.67 %