Ch 18 Chemical Kinetics and Ch 12 Thermodynamics

Ch 18 Chemical Kinetics and Ch 12 Thermodynamics Reaction Mechanism: When the first step is Rate-Limiting ( i.e., slow); the Rate Law is Integer Order: 1st, 2nd or 3rd Order. When the second step is Rate-Limiting; the Rate Law may or may not be Integer. When neither step is Rate-Limiting, the Steady State Apprx: can be used to find the Rate Law! Catalysis: Increasing the Reaction Rate via a substance not Generated or Consumed in the Rxn, and that Provides a Lower Activation Barrier Rxn Path Equilibrium Constant: K(T)=kf/kr=(Af/Ar)exp{-(Eaf-Ear)} aA+bB  cC + dD Forward Rate= Reverse Rate U= Uprod - Ureact =(Eaf – Ear) Internal Energy Change @ V=const

Elementary Reactions: Can also be activated by light/photons Unimolecular A Products Rate=k[A] Ozone photodissociation by ultra violet Sun Light h+ O3 O2 + O h+ Cis  Trans Isomerization

Let [C2H2(CH3)2] =[A] First order Reaction, Find the rate constant k? [A]0 c c c Rate=-d[A]/dt= k[A] d[A]/[A]=-kdt dln[A]= d[A]/[A]=-kdt ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt} [A]=[A]0/2 -In(2)=-kt1/2 t1/2= ln2/k Half-Life c c t t c c t t t

Easier to fit Liner Plots [A]0 Rate=-d[A]/dt= k[A] d[A]/[A]=-kdt dln[A]= d[A]/[A]=-kdt ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt} ln[A] = ln[A]0 - kt Units! ln[A]

2nd Order reactions Rate = k[A]2 Bimolecular are 2nd order reactions or any 2nd order rxn A + A  Products A + A  Products 1/[A] e.g., 2NO2 2NO + O2 let [NO2]=[A] Rate=-(1/2)d[A]/dt = k[A]2 d[A]/[A]2=- d(1/[A])= - 2kdt 1/[A] = 1/[A]0 + 2kt [A]=[A]0/2 (1/[A]0)= 2kt1/2 t1/2= 1/2k[A]0Depends on [A]0 L s mols-1 t1/2 Slope=1/2k 1/[A]0

Elementary Reactions : Single Step Reactions as opposed to complex multi-step reactions Termolecular reaction: A + B + C  Products Rate = k[A][B][C] Recombination Reaction First Step in the formation of liquids as well 3rd body takes away E ≥ KE Energy free bound KE 3H2O  (H2O)2 + H2O D0 bond energy @ R>>Re where V(R)~0 E=KE=2.4 kJ mol-1 D0= 5.0 kJ mol-1 Collision takes away 3 kJ mol-1 E after collision E=-0.6 kJ mol-1

Reaction Mechanism • Given the overall reaction: • 2NO(g)  N2(g) + O2(g) • Propose a plausible 2 step mechanism • 2NO(g)  N2O(g) + O(g) (2) N2O(g) + O(g)  N2(g) + O2(g) • Assume Step (1) is rate limiting • Rate=k1[NO]22nd Order in NO Then is a plausible Mechanism

Given the overall reaction: 2NO(g)  N2(g) + O2(g) • Propose a plausible 2 step mechanism • 2NO(g)  N2O(g) + O(g) (2) N2O(g) + O(g)  N2(g) + O2(g) Assume Step (2) is rate limiting • Forward Rate=k1[NO]2 and Reverse Rate =k-1[N2O][O] • Forward Rate =k2[N2O][O] Fast Equilibrium Forward Rate = Reverse Rate: k1[NO]2 =k-1[N2O][O] Overall Rate=k2[N2O][O] = k2(k1/k-1) [NO]2 = = k2K1[NO]2 Then this is also a plausible Mechanism where K=k1/k-1

Same overall reaction: assume Steady-State Apprx Neither Step is rate limiting 2NO(g)+ M  N2O2(g) +M Propose a plausible 2 step mechanism (1) 2NO(g)+ M  N2O2(g) + M (2) N2O2(g)  N2(g) + O2(g) Assume No Step is rate limiting and [N2O2]~constant d/dt [N2O2]= k1[NO]2[M] - k-1[N2O2][M] - k2[N2O2]=0 Overall Rate =k2[N2O2 ] [N2O2]= k1[NO]2[M]/{ k-1[M] +k2} Overall Rate =k2k1[NO]2[M]/{ k-1[M] +k2} Then is a plausible Mechanism

Overall Rate =k2k1[NO]2[M]/{ k-1[M] +k2} Rate [M](Buffer concentration) Fig. 18-19, p. 780

Max Rate Rate Max Rate= k2 (k1/k-1)[NO]2 High PM k-1[M]>>k2 [M]

Steady State Approximation Rate=d[N2]/dt=d[O2]/dt =k2k1[NO]2[M]/{ k-1[M] +k2} [M]=(nM/V) and PV=nRT Ideal Gas Law Therefore PM=[M]RT when PM is high, High Pressures then [M] is large and k-1[M]>>k2 Rate~ k2 (k1/k-1)[NO]2 Step (2) is rate-limiting, Step (1) is in equilibrium! The Rateis independent [M] for high PM

Steady State Approximation Rate=d[N2]/dt=d[O2]/dt =k2k1[NO]2[M]/{ k-1[M] +k2} [M]=(nM/V) and PV=nMRT Ideal Gas Law Therefore PM=[M]RT when PM0, Low Pressures then [M]0 and k-1[M]<<k2 Rate~k1[NO]2[M] 3rd Order

Max Rate Max Rate =k2 (k1/k-1)[NO]2 High PM Rate Slope= k1[NO]2 Low PM [M]

Catalysis Rxn path Diagram Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

Catalytic Converters: e.g., Pt, Rh, ZrO2 etc Fig. 18-19, p. 780

The hydrogenation reaction has a large activation barrier Ea and is Sterically hindered and must go throw a tight transition state, So k is small and the rate is slow

Consider the gas phase(rxn path A) hydrogenation of ethylene to form ethane. This process requires H-H insertion into a C=C bond: a very tight transition state; high barrier and a low steric factor which mean the reaction is slow! C2H4(g) + H2(g)  C2H6(g) (1) H-H H-H

Consider another path B with a lower activation barrier: heterogeneous hydrogenation on a Pt Catalyst C2H4(g) + H2(g)Pt(surface) C2H6(g) C2H4(g)+ H2(g) + Pt(surface)  C2H4(ad) + H2(ad) (1) C2H4(ad) + H2(ad)  C2H6(ad) (2) C2H6(ad)  Pt(surface) C2H6(g) (3) Pt-Pt-Pt Pt-Pt-Pt

Consider another path B with a lower activation barrier: heterogeneous hydrogenation on a Pt Catalyst C2H4(g) + H2(g)Pt(surface) C2H6(g) C2H4(g)+ H2(g) + Pt(surface)  C2H4(ad) + 2H(ad) (1) C2H4(ad) + 2H(ad) C2H6(ad) (2) C2H6(ad)  Pt(surface) C2H6(g) (3) Pt-Pt-Pt Pt-Pt-Pt

Enzyme Catalyzed Reaction The Enzyme plays the role of the surface and the kinetics can be modeled as a steady-state approximation Since the Enzyme/Substrate complex concentration ids nearly constant throughout the reaction

Kinetics of Approaching Equilibrium Equilibrium: Evaporation rate equal the condensation Rate  Equilibrium C2H4(l)  C2H4(g) Rate Law Rate=k1=const, , Order zero • C2H4(l)  C2H4(g) Evap • C2H4(g)  C2H4(l) Cond Rate Law Rate=k2[C2H4(g)] The order n =1 Fig. 10-16, p. 459

Average Rates and Instantaneous Rates [C2H4] Fig. 18-3, p. 838

Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

Ch 18 Chemical Kinetics and Ch 12 Thermodynamics