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.
Answers
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: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 %