Chemical Reaction Engineering

Chemical Reaction Engineering

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  Chemical Reaction Engineering July 1 2020 Authoredby: Mujeeb UR Rahman Roll No. 17CH106 Email: 17CH106@students.muet.edu.pk Mehran University of Engineering & Technology Jamshoro, Pakistan Submitted to: Pro. Dr. Imran Nazir Associate Professor
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  1 Class Work Problem 3.4 A10-minute experimental run shows that 75% of liquid reactant is converted to product by a half-order rate. What would be the fraction converted in a half- hour run? Solution: For half order decomposition reaction kinetics, simplify rate expression in term of conversion ‘X’ form is −𝒓 𝑨 = 𝒅𝑪 𝑨 𝒅𝒕 = 𝒌√ 𝑪 𝑨 Integrate both side ∫ − 𝒅𝑪 𝑨 √ 𝑪 𝑨 = ∫ 𝒌𝒅𝒕 𝒕 𝟎 𝑪 𝑨 𝑪 𝑨𝟎 𝟐(√ 𝑪 𝑨𝟎 − √ 𝑪 𝑨) = 𝒌𝒕 We know, CA = CA0(1 – X) Above relative becomes 𝟐√ 𝑪 𝑨𝟎 𝟏 𝒌 (𝟏 − √𝟏 − 𝒙) = 𝒕 Here given X= 0.75 at t = 10min
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  2 𝟐√ 𝑪 𝑨𝟎 𝟏 𝒌 (𝟏 −√𝟏 − 𝟎. 𝟕𝟓) = 𝟏𝟎 …(𝟏) Now, conversion in half hour will be 𝟐√ 𝑪 𝑨𝟎 𝟏 𝒌 (𝟏 − √𝟏 − 𝑿) = 𝟑𝟎 … (𝟐) Solving above equations 1 and 2 we get X is X = 0.75 Result: Conversion in half-hour is 75%