Atkins’ Physical Chemistry Eighth Edition Chapter 22 – Lecture 2 The Rates of Chemical Reactions Copyright © 2006 by Peter Atkins and Julio de Paula Peter Atkins Julio de Paula

Integrated Rate Laws Relation Between Concentration and Time Rate law provides rate as a function of concentration Need relationship between concentration and time : i.e., How do we determine the concentration of a reactant at some specific time?

Fig Exponential decay of the reactant, 2 nd -order Plot of [A]/[A] o vs. t [A] o Rearranges to:

Summary of the Kinetics of Reactions OrderRate Law Concentration-Time Equation Half-Life rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] o - kt 1 [A] = 1 [A] o + kt [A] = [A] o - kt t½t½ ln 2 k = t ½ = [A] o 2k2k t ½ = 1 k[A] o rate = k [A][B] t ½ = 1 k[A] o 2

 Generally, as T increases, so does the reaction rate  k is temperature dependent Temperature affects the rate of chemiluminescence in light sticks

Dependence of rate constant on temperature

 In a chemical rxn, bonds are broken and new bonds are formed  Molecules can only react if they collide with each other  Furthermore, molecules must collide with the correct orientation and with enough energy Cl + NOCl Cl 2 + NO

Activation Energy An energy barrier Reactant Activated complex Product Transition state

Activation energy (E a ) ≡ minimum amount of energy required to initiate a chemical reaction Transition state ≡ very short-lived and cannot be removed from reaction mixture methyl isonitrile acetonitrile

Rank the following series of reactions from slowest to fastest The lower the E a the faster the reaction Slowest to fastest: (2) < (3) < (1) Exothermic Endothermic

How does a molecule acquire sufficient energy to overcome the activation barrier? Effect of temperature on distribution of kinetic energies Fraction of molecules with E ≥ E a R = J/(mol ∙ K) T = kelvin temperature g ≡ degeneracy of state Maxwell–Boltzmann Distributions

Fraction of molecules with E ≥ E a e.g., suppose E a = 100 kJ/mol at T = 300 K: = 3.9 x (implies very few energetic molecules) at T = 310 K:= 1.4 x (about 3.6 times more molecules) R = J/(mol ∙ K) T = kelvin temperature g ≡ degeneracy of state

Temperature Dependence of the Rate Constant k = A exp(− E a /RT ) E a ≡ activation energy (J/mol) R ≡ gas constant (8.314 J/Kmol) T ≡ kelvin temperature A ≡ frequency factor ln k = −E a R 1 T + ln A (Arrhenius equation) y = mx +b

Fig Arrhenius equation plot of ln k versus 1/T Rearranges to:

k 1 = A exp(−E a /RT 1 ) The Arrehenius equation can be used to relate rate constants k 1 and k 2 at temperatures T 1 and T 2. k 2 = A exp(−E a /RT 2 ) combine to give: Better as an estimate, since slope depends only on two points

Fig Potential energy profile for exothermic reaction Note: this rate constant gives the rate of successful collisions Transition state