Chemical Reaction Engineering Asynchronous Video Series Chapter 3, Part 1: Rate Laws H. Scott Fogler, Ph.D. Why a Rate Law? • If we have FA 0 rA X • Then we can size a number of CSTR and PFR reaction systems Summary Why a Rate Law? • If we have FA 0 rA X • Then we can size a number of CSTR and PFR reaction systems 2 CSTRs CSTR PFR FA 0 rA FA 0 rA X FA 0 rA X X Why a Rate Law? • If we have FA 0 rA X • Then we can size a number of CSTR and PFR reaction systems FA 0 rA FA 0 rA X • 2 CSTRs CSTR PFR FA 0 rA X To find -rA= f(X) 1) Need the rate law, -rA=f(CA, CB) 2) Need the reaction stoichiometry, CA=CA0(1-X) X Rate Law Basics • A rate law describes the behavior of a reaction. The rate of a reaction is a function of temperature (through the rate constant) and concentration. Rate Law Basics • A rate law describes the behavior of a reaction. The rate of a reaction is a function of temperature (through the rate constant) and concentration. • Power Law Model k is the specific reaction rate (constant) Rate Law Basics • A rate law describes the behavior of a reaction. The rate of a reaction is a function of temperature (through the rate constant) and concentration. • Power Law Model k is the specific reaction rate (constant) • k is given by the Arrhenius Equation: Where: E = activation energy (cal/mol) R = gas constant (cal/mol*K) T = temperature (K) A = frequency factor (units of A, and k, depend on overall reaction order) Rate Law Basics • To find the activation energy, plot k as a function of temperature: Where: Reaction Order • You can tell the overall reaction order by the units of k CA -rA (mol/dm3) (mol/dm3*s) Reaction Order Rate Law k zero -rA = k (mol /dm3*s) Reaction Order • You can tell the overall reaction order by the units of k CA -rA (mol/dm3) (mol/dm3*s) Reaction Order Rate Law k zero -rA = k (mol /dm3*s) 1st -rA = kC A s -1 Reaction Order • You can tell the overall reaction order by the units of k CA -rA (mol/dm3) (mol/dm3*s) Reaction Order Rate Law k zero -rA = k (mol /dm3*s) 1st -rA = kC A s -1 2nd -rA = kC A2 (dm3/mol*s) Examples of Rate Laws • First Order Reactions (1) Homogeneous irreversible elementary gas phase reaction C2 H 6 C2 H 4 H 2 rA kCC2H6 with 82kcal 1 1 1 mol 1000 T k 0.072s e Examples of Rate Laws • First Order Reactions (1) Homogeneous irreversible elementary gas phase reaction C2 H 6 C2 H 4 H 2 with rA kCC2H6 82kcal 1 1 1 mol 1000 T k 0.072s e (2) Homogeneous reversible elementary reaction n C4 H10 i C4 H10 rn k C nC4 CiC4 K C with T 360 k 31.1exp 7906 360T and T 333 K C 3.03exp 830.3 333T Examples of Rate Laws • First Order Reactions (1) Homogeneous irreversible elementary gas phase reaction C2 H 6 C2 H 4 H 2 82kcal 1 1 1 mol 1000 T k 0.072s e with rA kCC2H6 (2) Homogeneous reversible elementary reaction n C4 H10 i C4 H10 rn k C nC4 CiC4 K C with T 360 k 31.1exp 7906 360T and T 333 K C 3.03exp 830.3 333T • Second Order Reactions (1) Homogeneous irreversible non-elementary reaction with m3 k 0.0017 kmol.min and E 11273 cal At 188˚C mol rA kCONCBC NH3 This is first order in ONCB, first order in ammonia and overall second order. Examples of Rate Laws • Second Order Reactions (2) Homogeneous irreversible elementary reaction CNBr CH3 NH2 CH3Br NCNH2 rA kCCNBr CCH 3 NH2 with 2.2dm 3 k s.mol Examples of Rate Laws • Second Order Reactions (2) Homogeneous irreversible elementary reaction CNBr CH3 NH2 CH3Br NCNH2 rA kCCNBr CCH 3 NH2 with 2.2dm 3 k s.mol This reaction is first order in CNBr, first order in CH3NH2 and overall second order. (3) Heterogeneous catalytic reaction: The following reaction takes place over a solid catalyst: