Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key st

Question

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 125. L tank with 60. mol of sulfur dioxide gas and 12-mol of oxygen gas at 30.C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 12. mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits. ulo 010

Explanation / Answer

2SO2 + 1O2 ---> 2SO3

Initially SO2 moles = 60 , O2 moles = 12 , SO3 moles = 0

after equilibrium SO2 moles = 60-2X , O2 moles = 12-X , SO3 moles = 2X

given SO3 moles = 2X = 12 mol , hence X = 6

hence equilibrium SO2 moles = 60-(2 x 6) = 48 , O2 moles = 12-6 = 6 mol

[SO2] = moles / volume = ( 48/125) = 0.384 M ,

[O2] = 6/125 = 0.048M ,  

[SO3] = 12/125 = 0.096

Kc = [SO3]^2 / [SO2]^2 [O2]

= ( 0.096^2) / ( 0.384^2 x 0.048)

= 1.3

( note this is Kc for reaction 2SO2 + O2 = 2SO3 , if we want for SO2 + 1/2O2 = SO3 we get Kc = ( 1.3)^1/2 = 1.14)

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