Chemical Reaction Engineering Chapter 3, Part 1: Rate Laws Dr. Khashayar Nasrifar

Rate Law: Objectives Part I Write the relationship between the relative rates of reaction. Write a rate law and define reaction order and activation energy. Part II Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion. Write -rA solely as a function of conversion given the rate law and then entering concentration. Calculate the equilibrium conversion for both gas and liquid phase reactions.

Why a Rate Law? X If we have Then we can size a number of CSTR and PFR reaction systems To find -rA= f(X) Need the rate law, -rA=f(CA, CB) Need the reaction stoichiometry, CA=CA0(1-X)

Rate Law: Definitions Homogeneous reaction: any of a class of chemical reactions that occur in a single phase (gaseous, liquid, or solid). The most important of homogeneous reactions are the reactions between gases (e.g., the combination of common household gas and oxygen to produce a flame) and the reactions between liquids or substances dissolved in liquids (e.g., the reactions between aqueous solutions of acids and bases). Heterogeneous reaction: are those reactions involve more than one phase

Elementary Reaction: a reaction that cannot be broken down into smaller steps. A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state. Non-elementary Reaction

Rate Law: Definitions Reversible reaction: a chemical reaction that can proceed in either direction AB Irreversible reaction: AB

Rate Law: Definitions Molecularity: The number of molecules consumed in a chemical reaction Unimolecular: When a single molecules is consumed 92U238  90Th234 + 2He4 Bimolecular: When two molecules are consumed. Br• + C2H6  HBr + C2H5• Termolecular: when three or more reactants and consumed in single step of the reaction. 2NO +O2  2NO2

Rate Law: Relative rates of reaction Consider the general equation The relative rate of reaction of the various species involved in the reaction can be obtained from the ratio of the stoichiometric coefficients. For every mole of A consumed c/a moles of C appears. Rate of formation of C= (c/a) rate of disappearance of A The relationship can be expressed directly from the stoichiometry of the reaction

Rate Law: Reaction Order 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 = n

Reaction Order You can tell the overall reaction order by the units of k C -r Reaction Order Rate Law k A A (mol/dm 3 ) (mol/dm 3 *s) zero -r = k (mol /dm3*s) A 1st s -1 -r = kC A A 2nd -r = kC 2 (dm3 /mol*s) A A k is the specific reaction rate (constant)

Rate Law Basics 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:

One can also write the specific reaction rate as: Taking the ratio:      

Activation Energy The activation energy can be thought of as a barrier to the reaction. One way to view the barrier to a reaction is through the reaction coordinates. These coordinates denote the energy of the system as a function of progress along the reaction path. For the reaction the reaction coordinate is

For the reaction to occur, the reactants must overcome an energy barrier EB. The energy barrier is related to the activation energy, E. The barrier height is a result of (1) the molecules needing energy to distort or stretch their bonds in order to break them and to thus form new bonds, (2) the reaction molecules come close together they must overcome both steric and electron repulsion forces, and (3) the quantum effects that can in some cases produce a barrier.

Activation Energy The activation energy is a measure of how temperature sensitive the reaction is. Reactions with large activations energies are very temperature sensitive.

Example Consider the following elementary reactions 1)   Which reaction has the higher activation energy? 2)   Which reactions have the same activation energy? 3)   Which reaction is virtually temperature insensitive? 4)   Which reaction will dominate (i.e. take place the fastest) at high temperatures?

Elementary rate law A reaction follows an elementary rate law if and only if the stoichiometric coefficients are the same as the individual reaction order of each species. For the reaction in the previous example, The rate law would be:

Non-Elementary Rate Laws If the rate law for the non-elementary reaction is found to be then the reaction is said to be 2nd order in A, 1st order in B, and order overall is 2+1 =3 (3rd order). Most reactions are non-elementary, especially catalytic reactions. This means the kinetic equation is empirical in nature. A generic kinetic equation has the form of Langmuir-Hinshelwood kinetics.

KC = kforward / kreverse Reversible Reactions aA + bB   cC + dD Reversible reactions consist of 2 reactions (forward and backward or reverse reactions). The overall reaction is the net reaction of 2 reactions. The rate at equilibrium is Zero KC = kforward / kreverse

Examples of Rate Laws First Order Reactions (1) Homogeneous irreversible elementary gas phase reaction (2) Homogeneous reversible elementary reaction

Examples of Rate Laws Second Order Reactions (1) Homogeneous irreversible non-elementary reaction This is first order in ONCB, first order in ammonia and overall second order. At 188˚C

Examples of Rate Laws Second Order Reactions (2) Homogeneous irreversible elementary reaction This reaction is first order in CNBr, first order in CH3NH2 and overall second order.

Examples of Rate Laws (3) Heterogeneous catalytic reaction: The following reaction takes place over a solid catalyst:

Part II Stoichiometry

Stoichiometry If the rate law depends on more than one species we must relate the concentrations of different species to each other This relationship is most easily established with the aid of a stoichiometric table The stoichiometric table presents the stoichiometric relationship between reacting molecules for a single reaction

Reaction Stoichiometry The relative rate of reaction of the various species involved in the reaction can be obtained from the ratio of the stoichiometric coefficients. For every mole of A consumed c/a moles of C appears. Rate of formation of C= (c/a) × rate of disappearance of A The relationship can be expressed directly from the stoichiometry of the reaction

Batch Stoichiometric Table NA0 NB0 NC0 ND0 NI0 t=t NA NB NC ND NI

Batch Stoichiometric Table Species Symbols Initial Change Remaining A NA0 -NA0X NA=NA0(1–X) B NB0=NA0ΘB -(b/a)NA0X NB=NA0(ΘB –(b/a)X) C NC0=NA0ΘC (c/a)NA0X NC=NA0(ΘC+(c/a)X) D ND0=NA0ΘD (d/a)NA0X ND=NA0(ΘD+(d/a)X) Inert I NI=NA0ΘI NI = NA0ΘI NT0 δNA0X NT=NT0+δNA0X

Concentration: Batch Systems Constant Volume Batch: Note:  if the reaction occurs in the liquid phase or if a gas phase reaction occurs in a rigid (e.g., steel) batch reactor Then

Example

Species Symbols Initial Change Remaining NaOH A NA0 -NA0X NA=NA0(1–X) C17H35COO)3-C3H5 B NB0=NA0ΘB -(1/3)NA0X NB=NA0(ΘB –(1/3)X) C17H35COONa C NC0=NA0ΘC NA0X NC=NA0(ΘC+X) C3H5(OH)3 D ND0=NA0ΘD (1/3)NA0X ND=NA0(ΘD+(1/3)X) Water (Inert) I NI=NA0ΘI NI = NA0ΘI NT0 NT=NT0+δNA0X

Flow System Stoichiometric Table

Flow System Stoichiometric Table Species Symbols Initial Change Remaining A FA0 -FA0X FA=FA0(1–X) B FB0=FA0ΘB -(b/a)FA0X FB=FA0(ΘB –(b/a)X) C FC0=FA0ΘC (c/a)FA0X FC=FA0(ΘC+(c/a)X) D FD0=FA0ΘD (d/a)FA0X FD=FA0(ΘD+(d/a)X) Inert I FI=FA0ΘI FI = FA0ΘI FT0 FT=FT0+δFA0X

Concentration: Flow System Liquid Phase Flow System: This gives us -rA = f(X) If the rate of reaction were: then we would have:

Concentration: Gas Flow System For gaseous reactants and assuming ideal gases, therefore from ideal gas law we may get From Stoichiometric Table Substitute in v Where:

Concentration: Gas Flow System Combining the compressibility factor equation of state with Z = Z0 with and For example if the gas phase reaction has the rate law then k with

Example Production of Nitric acid Nitric acid is made commercially from nitric oxide. Nitric oxide is produced by the gas-phase oxidation of ammonia. 4NH3 + 5O2  4NO + 6H2O The feed consists of 15 mol% ammonia in air at 8.2 atm and 227°C.

What is the total entering concentration? Assume ideal gas behavior.

(b) What is the entering concentration of ammonia?

(c) Set up a stoichiometric table with ammonia as your basis of calculation. Express Ci for all species as functions of conversion for a constant-volume batch reactor. Express PT as a function of X. Express Pi and Ci for all species as functions of conversion for a flow reactor.

(d) Write the combined mole balance and rate law solely in terms of the molar flow rates and rate law parameters for C1 and C2 above. Assume the reaction is first order in both reactants

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Measures Other Than Conversion Uses: A. Membrane reactors B. Multiple reaction Liquids: Use concentrations, i.e., CA For the elementary liquid phase reaction carried out in a CSTR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry are:

Measures Other Than Conversion For Reactant A: For CSTR

Measures Other Than Conversion For Reactant B: There are two equations, two unknowns, CA and CB

Measures Other Than Conversion Gases: Use molar flow rates, i.e., FT A < === > B

Measures Other Than Conversion If the reaction, A < === > B carried out in the gas phase in a PFR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry yield, for isothermal operation (T=To) and no pressure drop (DP=0) are:

Summary At the start of the chapter we saw we needed -rA=f(X). This result is achieved in two steps. Rate Laws -rA=k f(Ci) 1st order A  B or 1st order 2nd order A+B ==> C Rate laws are found by experiment Stoichiometry Liquid: Gas: A < === > B -rA=kCA -rA=kACACB