Chemical Reaction Engineering 1 Kinetics of Homogenous Reaction Agung Ari Wibowo, S.T, M.Sc Dept. of Chemical Engineering Politeknik Negeri Malang Homogeneous reactions are chemical reactions in which the reactants and products are in the same phase Batch Reactor Continous Stirred Tank / Mixed Flow Reactor Plug Flow Reactor Reaction rate, the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time 1 πππ πππππ π ππππππ ππ = = π ππ‘ (ππ’πππππ π΄πππ)(π‘πππ) 1 πππ πππππ π ππππππ ππ = = π ππ‘ (π£ππ ππ πππ’ππ)(π‘πππ) Volume Mass 1 πππ πππππ π ππππππ ππ = = π ππ‘ (πππ π ππ π ππππ)(π‘πππ) Area Volume ππ = 1 πππ πππππ π ππππππ = π ππ‘ (π£ππ ππ πππππ‘ππ)(π‘πππ) Consider this reaction ππ΄ + ππ΅ → ππΆ + ππ· The Rate Equation −ππ΄ = − 1 πππ΄ πππππ π΄ ππππ π’πππ = π ππ‘ (π£ππ ππ πππ’ππ)(π‘πππ) The Rate Relation − ππ΄ ππ΅ ππ ππ =− = = π π π π Elementary Reaction π΄+π΅ →πΆ The number of collisions of molecules A with B is proportional to the rate of reaction. But at a given temperature the number of collisions is proportional to the concentration of reactants in the mixture; hence, the rate of disappearance of A is given by -ππ΄ = π πΆπ΄ πΆπ΅ Non- Elementary Reaction π»2 + π΅π2 → 2π»π΅π When there is no direct correspondence between stoichiometry and rate, then we have nonelementary reactions Elementary Reaction π΄+π΅ →πΆ The number of collisions of molecules A with B is proportional to the rate of reaction. But at a given temperature the number of collisions is proportional to the concentration of reactants in the mixture; hence, the rate of disappearance of A is given by -ππ΄ = π πΆπ΄ πΆπ΅ Non- Elementary Reaction π»2 + π΅π2 → 2π»π΅π When there is no direct correspondence between stoichiometry and rate, then we have nonelementary reactions Problem: Levenspiel, Example 1.1 Problem: Levenspiel, Example 1.1 ππ = How to solve? 1 πππ πππππ π ππππππ = π ππ‘ (π£ππ ππ πππππ‘ππ)(π‘πππ) −ππ»2 = 2.829 x 104 πππ/π3 π −ππ2 = 1.415 x 104 πππ/π3 π Negative : reactant being comsumed
