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				   The equlibrium constant kp for the Rxn ,

2So2(g)+O2(g) <----->2So3(g) at 1000k is 3.5/atm
what would be the partial pressure of oxygen gas if the equlibrium is found to have equal moles of So2 and So3

6 years ago


Answers : (1)

										2So2(g)+O2(g) <----->2So3(g)

Kp= [p(so3)2]/[po2].[pso2]2

moles of so2 = moles of so3 [given]

because p proportion to moles

partial pressure of so2 = partial pressure of so3
pso2= pso3

Kp= [p(so3)2]/[po2].[pso2]2

Kp= [p(so3)2]/[po2].[pso3]2

Kp= 1/[po2]

= 0.286 atm Ans
6 years ago

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