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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

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

Grade:12

1 Answers

AskiitianExpert Pramod-IIT-R
47 Points
11 years ago
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] [po2]=1/Kp=1/3.5 = 0.286 atm Ans

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