Question

5 moles of SO2 and 5 moles of O2 are allowed to react. At equilibrium, it was found that 60% of SO2 is used up. lf the partial pressure of the equilibrium mixture is one atmosphere. the partial pressure of O2 is

(A) 0.21 atm
(B) 0.41 atm
(C) 0.82 atm
(D) 0.52 atm

Answer 1

The reaction is depicted as:
SO2+½ O2 <-------> SO3.
(all species are in gaseous state )
We have initially :
SO2= 5 mole
O2 = 5 mole
SO3= 0 mole.
At equilibrium :
SO2= 5- 5(0.6) = 2 mole
O2 = 5- 5 (0.3) = 3.5 mole
SO3 = 5(0.6) = 3 mole.
Now , total moles at equilibrium = 8.5 mole.
Hence mole fraction of O2 = 3.5/8.5.
Total pressure at equilibrium = 1 atm.
Hence Partial pressure due to O2 at equilibrium will be = (1atm)×( 3.5/8.5)
= 0.41 atm.
Hence option B) is correct.

Answer 2

Hola mate

option b is correct