Chemical Kinetics Chapter 14

Summary of the Kinetics 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 Fig Temperature affects the rate of chemiluminescence In light sticks

Fig 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 Fig Cl + NOCl Cl 2 + NO

Activation Energy Fig An energy barrier Reactant Activated complex or Transition state Product

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

Sample Exercise 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? Fig Effect of temperature on distribution of kinetic energies f = e - E a RT Fraction of molecules with E ≥ E a R = J/(mol ∙ K) T = kelvin temperature Maxwell–Boltzmann Distributions

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

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 14.12

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:

Plot of ln k vs 1/T Slope = −E a /R ln k =− EaEa R 1 T + ln A Fig Graphical determination of activation energy y = m x + b Arrhenius Equation

Sample Exercise The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at K?

Solution Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, E a, and the rate constant, k, at a particular temperature. Plan: We can obtain E a from the slope of a graph of ln k versus 1/T and the rate constant, k, at a particular temperature. Once we know E a, we can use Equation together with the given rate data to calculate the rate constant at K.

Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k.

A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17: slope = − E a /R = − 1.9 x 10 4

We use the value for the molar gas constant R in units of J/mol-K (Table 10.2). We thus obtain

(b) To determine the rate constant, k 1, at T 1 = K, we can use Equation with E a = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as: k 2 = 2.52 × 10 –5 s –1 and T 2 = K: Thus, Note that the units of k 1 are the same as those of k 2.

Practice Exercise The first-order rate constant for the reaction of methyl chloride with water to produce methanol and hydrochloric acid is 3.32 x s -1 at 25 °C. Calculate the rate constant at 40 °C if the activation energy is 116 kJ/mol. k 2 = s -1