Provide me with a right answer. Thanks Question 13 of 15 Due Date: Points Possib

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Provide me with a right answer. Thanks

Question 13 of 15 Due Date: Points Possible Grade Category: Description: Policies: Map Sapling Learning At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container 2SO At equilibrium, 0.110 mol of O2 is present. Calculate Kc You can check you You can view solut up on any questio Number You can keep tryin you get it right or You lose 5% of the in your question f nswer 95 95 eTextbook O Help With This T O Web Help & Vid O Technical Suppo Previous Give Up & View Solution Q, Check Answer Next Exit 95 Hint 4:13 A Type here to search A ENG

Explanation / Answer

The reaction is 2SO3 (g)<---->2SO2(g)+ O2(g)

2 moles of SO3 decomposes to give 2 moles of SO2 and 1 mole of O2

moles of O2 at equilibrium = 0.11moles

Moles of SO3 required to get 0.11 moles of O2= 2*0.11=0.22 moles moles of SO2 formed= 0.22 moles

Moles of SO3 at equilibrium =0.66-0.22=0.44, SO2=0.22, O2=0.11

volume of reactor= 4.5L

concentrations(M) = moles/L

Concentrations at equilibrium (M) : SO2=0.22/4.5 =0.049, SO3=0.44/4.5=0.098, O2=0.11/4.5 =0.024

Kc= [SO3]2/ [SO2]2[ [O2] = 0.049*0.049/0.098*0.098*0.024)=10.42

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