5 moles of so2 and 5 moles of o2 react in a closed vessel at equilibrium 60 percent of the so2 is consumed the total no. of gaseous moles (so2, o2, so3)
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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.
Read more on Brainly.in - https://brainly.in/question/1049362#readmore
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.
Read more on Brainly.in - https://brainly.in/question/1049362#readmore
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