Rate Laws and stoichiometry

Rate Laws: basic definitions

Rate laws and stoichiometry What we will see: Order of reaction and molecularity Elementary and nonelementary rate laws Reversible and irreversible reactions Reaction rate constant – Arrhenius factor and activation energy Stoichiometric tables for batch and flow systems

Relative Rates of Reaction aA + bB  cC + dD Which is the relationship among (-rA) and (-rB), (rC) and (rD) ?      

The rate law expression The algebraic equation that relates -rA to the species concentrations is called the kinetic expression or rate law (-rA) = [k Af (T)] * [f´(CA, CB, ..)] Reaction rate constant

Dependence of the reaction rate on the concentrations of the species Reaction: aA + bBcC + dD -rA = k CAa CBb Power Law model a = reaction order with respect to reactant A b = reaction order with respect to reactant B n = a + b = overall order of the reaction If a = a e b = b, the reaction is «elementary» Elementary H2 +I2  2HI -rHI = kCH2 CI2 Non-elementary CO + Cl2  COCl2 -rCO = k CCO CCl2 3/2

UNITS OF k (rA) = k * [fn(conc)] Units of k = units of (rA )/ units of [fn(conc)] Zero-order reaction (rA) = k k in mol/m3 ·s First-order reaction (rA) = kCA k in s-1 Second-order reaction (rA) = kCA2 k in m3/mol ·s

Elementary reactions and molecularity We say that a reaction follows an ELEMENTARY RATE LAW when the reaction orders are identical with the stoichiometric coefficients of the reacting species for the reaction as written The MOLECULARITY of a reaction is the number of atoms, ions, or molecules involved (colliding) in a reaction step The terms unimolecular, bimolecular, and termolecular refer to reactions involving one, two, or three atoms (or molecules) interacting or colliding in any one reaction step

Nonelementary Rate Laws A large number of both homogeneous and heterogeneous reactions do not follow simple rate laws [ = k(T) * f(conc.) ]: Reactions involving radicals Catalytic reactions Enzymatic reactions Biomass reactions

Apparent reaction order Sometimes reactions have complex rate expressions that cannot be separated into solely temperature-dependent and concentration-dependent portions We can only speak of reaction order under certain limiting conditions: at very low concentrations of oxygen, the reaction would be "APPARENT" first order with respect to N2O if the concentration of O2 were large enough, the apparent reaction order would be -I with respect to O2 and first order with respect to N2O giving and overall apparent zero order  

Reversible reactions All reactions are …. reversible!!! Reaction aA + bB  cC + dD At equilibrium, the rate of reaction is identically zero for all species (i.e., - rA= 0 ) All rate laws for reversible reactions must reduce to the thermodynamic relationship relating the reacting species concentrations at equilibrium

Rate laws for rev elementary reactions Reaction: aA + bB cC + dD krev kfor Rate of consumption of A for the forward reaction (-rA_forward) = kfor CAa CBb Rate of formation of A for the forward reaction (rA_forward) = – kfor CAa CBb Rate of formation of A for the reverse reaction (rA_reverse) = krev CCc CDd Net rate of formation of A = (rA_reverse) + (rA_forward)

Rate laws for elementary reactions Net rate of formation of A (rA)net = (rA_reverse) + (rA_forward) = krev CCc CDd – kfor CAa CBb At equilibrium (rA)net MUST be =0 krev CCc CDd – kfor CAa CBb = 0

Temperature dependence of Kc From the integration of the van’t Hoff equation

The reaction rate constant (k) (-rA) = kA * [fn(con)] where kA = f(T) ln(k) 1/T Rxn1: High Ea Arrhenius equation k = A exp(-Ea/RT) A = pre-exponential factor or frequency factor Ea = activation energy, J/mol or cal/mol R = gas constant = 8.314 J/mol · K = 1.987 cal/mol · K T = absolute temperature, K Rxn 2: Low Ea ln( k) = ln A - (Ea/R) x 1/T

Why is there an activation energy? Molecules need energy to distort or stretch their bonds so that they break and form new bonds to overcome the steric and electron repulsive forces as they come close together One way to view the barrier to a reaction is through the use of the reaction coordinates potential energy of the system as a function of the progress along the reaction path (see next slide) Energy of formation of the reactants  literature Energy of formation of the transition state  QM calculation (Gaussian)

Activation energy (Ea) Activation energy: can be thought of as a barrier to energy transfer between reacting molecules A+B C+D Ea HR Reaction coordinate Poetential energy

Example - Determination of the Activation Energy Calculate the activation energy for the decomposition of benzene diazonium chloride to give chlorobenzene and nitrogen Using the following information for a first-order reaction:

T dependence of the reaction rate The larger the Ea, the more temperature-sensitive is the rate of reaction Reference values (order of magnitudine) A: 10 13 s-1 E: 300 kJ/mole Correlations exist to relate the activation energy to the heat of reaction Activation energies can be estimated from first principles (molecular modelling)

Factors affecting the reaction rate Temperature (T) Heat transfer Pressure/Temperature Catalyst Concentration (C) Mass transfer rA = f (T, C)

Effect of P & T on concentration In gas phase EOS for real gas: PV = z n RT con C = n/V = P/zRT It is clear that in gas phase concentration is sensitive to P and T In liquid phase Concentration = moli per volume unit = N/V The density of a liquid DOESN’T change much with T Liquids possess low compressibility – effect of P on density is small Concentrations in liquid hase are «generally» not affected by changes of P and T

WHERE WE ARE….

Design equations for isothermal reactors REACTOR DIFF. EQN ALGEBR. EQN INTEGRAL FORM BATCH CSTR PFR

Design equations for isothermal reactors From the previous table we see that if we know the relationship –rA = g(XA) then it is possible to size batch, CSTRs, PFRs (and PBRs) operating at the same conditions under which -rA = g(X) was obtained In general the rate law is known -rA = k [f(Ca, Cb,…)] Thus, if we know Cj =hj(X) then in fact we have -rA as a function of X and this is all that is needed to evaluate the isothermal design equations

stoichiometric tables Batch systems Constant volume reactions Flow systems Variable volume reactions

Stoichiometric tables A stoichiometric table presents the stoichiometric relationships between reacting molecules for a single reaction. t = t NA NB NC ND NI t = 0 NA0 NB0 NC0 ND0 NI0 Batch reactor

Stoichiometric tables for Batch Reactors Reaction: Concentration in batch reactors:

Stoichiometric tables for Batch Reactors From stoichiometric table we have Ni = f(X) The goal is to find the value of V needed to obtain a given concentration For constant volume reactors: V = VR = VO (gas phase reaction) V = VO (liquid phase reaction) For constant volume reactors (gas phase reaction): V has to be expressed as a function of P, T The increment in the number of moles has to be considered

Stoichiometric tables for Batch Reactors VARIABLE VOLUME Ideal gas PV = ZNT RT POVO = ZO NTO RTO Def. “d” – change in the total number of moles per mole of A reacted NT = NTO + d NAO X Def. “e” Change in total number of moles for complete conversion Total moles fed

Stoichiometric tables for Batch Reactors Reaction: Concentration in batch reactor:

Stoichiometric tables for FLOW Reactors Reaction Concentration in flow reactor: The form of the stoichiometric table for a continuous-flow system is virtually identical to that for a batch system except that we replace Nj0 by Fj0 and Nj by Fj

Concentrations for flow reactors From the stoichiometric table, we have Fi = f(X) which value of the volumetric flow rate v, should we use? Liquid phase reactions v = vO Gas phase reactions For isothermal and isobaric reactors, without change in the number of moles v = vo

Concentrations for flow reactors – volume change Per gas ideale PV = ZNT RT POVO = ZO NTO RTO CT = FT /v = P/(ZRT) Def. “d” – change in the total number of moles per mole of A reacted FT = FTO + d FAO X Def. “e” Change in total number of moles for complete conversion Total moles fed

Concentrations for flow reactors – volume change To derive the concentrations in terns of conversion

Concentrations for flow reactors – gas pahse, volume change

Example The following reaction (liquid phase) is irreversible and follows an elementary cinetic law: A+B  C The reaction is conducted in a PFR of volume V1 dm3 at T1 K and in a CSTR of volume V2 dm3 working at T2 K. The feed is obtained mixing two streams, one containing only A, the other only B. The concentration of each reactant in the stream is CO mol/L and the total volumetric flow entering the reactor is vo dm3 /min. Find the relationship needed to calculate which X can be achieved in each reactor. Further information: at T1 K, k = 0.07 dm3/mol-min Ea = 85000 J/mol-K ATTENZIONE, mancano alkcuni dati numerici che sono forniti durante l’esercizio

Steps Draw a scheme containing all important information Write the equations: Rate law as a function of concentration Velocity constant as a function of T and Ea Express concentrations as a function of X Design equation for your reactor Solve for X

Solution A A PFR V1 ; T1 B CAO = 0.5 CO CSTR CBO = 0.5 CO B V2 ; T2 0.5 vO CO A CAO = 0.5 CO CBO = 0.5 CO vO 0.5 vO CO A B PFR V1 ; T1 CSTR V2 ; T2 CAO = 0.5 CO CBO = 0.5 CO vO

Solution A+B  C 1. Rate law (Elementary) (-rA) = k CA CB = (-rB) = (rC) We need k at T1 e T2 ; CA and CB expressed as function of X 2. Velocity constant (k) k = A exp(-Ea/RT) k a T1 = k1 = A exp (-Ea/RT1) = 9.1 L/mol.min 3. Concentration as a function of X (Stoichiometric tables) CA = CAO (1-X) CB = CAO (B –b/aX) = CAO (1-X) [ B = 1]

Solution 4.Reaction rate as a function of X dFA=FA0 dXA= v0 CA0 4.Reaction rate as a function of X (-rA) = k CAO2 (1-X)2 = (-rB) = (rC) 5. Design equation from the mole balance 5.6 182 XPFR = 0.84 XCSTR = 0.928

Expressing concentration as a function of X

Expressing concentration as a function of X

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