Chemical Kinetics

Chemical Kinetics Chapter 14

Reminders • Assignment 1 due today (end of class) • Assignment 2 up on ACME, due Jan. 29 (in class) • Assignment 3 will be up Mon., Jan 29 and will be due Mon., Feb. 05 • Assignment 4 (Ch. 15) will not be due before Midterm 1, but Ch. 15 will be on the midterm

A product = s-1 k = rate = [A] M/s D[A] - M = k [A] Dt D[A] rate = - Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 First-Order Reactions Rate constant Average rate rate = k [A] Rate law Differential rate law [A] = [A]0exp(-kt) Integrated rate law Integrated rate law ln[A] = ln[A]0 - kt (linear form) 14.3

A product D[A] - = k [A] Dt D[A] rate = - Dt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 First-Order Reactions Average rate rate = k [A] Rate law Differential rate law [A] = [A]0exp(-kt) Integrated rate law Integrated rate law ln[A] = ln[A]0 - kt (linear form) 14.3

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? 0.88 M ln 0.14 M = 2.8 x 10-2 s-1 t = ln ln[A]0 – ln[A] = k k [A]0 [A] [A]0 = 0.88 M ln[A] = ln[A]0 - kt [A] = 0.14 M kt = ln[A]0 – ln[A] = 66 s Recall: 14.3

t [A]0 ln t½ [A]0/2 0.693 = = = = k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? ln t½ ln2 ln2 0.693 = = k k k [A]0 5.7 x 10-4 s-1 [A] First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s-1) 14.3

= What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½ = 1200 s = 20 minutes ln2 0.693 = k 5.7 x 10-4 s-1 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 How do you know decomposition is first order? units of k (s-1) The half-life of a 1st-order reaction is independent of the initial concentration of the reactant. 14.3

A product # of half-lives [A] = [A]0/n First-order reaction 1 2 2 4 3 8 4 16 n= 2 for each half-life elapsed 14.3

14.3

A product rate = [A]2 M/s D[A] 1 1 - M2 = k [A]2 = + kt Dt [A] [A]0 t½ = D[A] rate = - Dt 1 k[A]0 Second-Order Reactions rate = k [A]2 = 1/M•s k = [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 14.3

1 1 = + kt [A] [A]0 Distinguishing 1st and 2nd order reactions NO2(g) NO(g) + ½ O2(g) at 300oC • Is the rate law 1st or 2nd order? • If (a) plot gives a straight line, then 1st order and rate = k[A] • If (b) plot gives a straight line, then 2nd order and rate = k[A]2 ln[A] = ln[A]0 - kt

A product rate [A]0 D[A] - = k Dt [A]0 t½ = D[A] 2k rate = - Dt Zero-Order Reactions rate = k [A]0 = k = M/s k = [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 14.3

Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2 14.3

Temperature and reaction rates • Rates of most chemical reactions increase as the temperature rises • Dough rises faster at r.t. than when refrigerated • Plants grow more rapidly in warm weather than in cold

Temperature and reaction rates • This effect of temperature on reaction rate can be seen directly by observing a chemiluminescent reaction

Temperature affects rate of chemiluminescence reaction in CyalumeTM light sticks • Left: light stick in hot water • Right: light stick in cold water • At the higher temperature, the reaction is initially faster and produces a brighter light • Although the light stick glows more brightly initially, its luminescence also dies out more rapidly

Temperature Dependence of the Rate Constant 1st order reaction, ln[CH3NC]t = -kt + ln[CH3NC]0 [A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt 14.4

Temperature Dependence of the Rate Constant 1st order reaction, ln[CH3NC]t = -kt + ln[CH3NC]0 14.4

For a successful put, we need: • Enough energy to get over the hill • Directionality • For a successful reaction, we need: • Enough energy to get over the activation barrier • Collisions between reacting molecules • Reacting molecules with the right orientation

A + B C + D Endothermic Reaction Exothermic Reaction The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction. 14.4

Cl + NOCl NO + Cl2

lnk = - + lnA Ea 1 T R Temperature Dependence of the Rate Constant k = A •exp( -Ea/RT ) (Arrhenius equation) Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor 14.4

lnk = - + lnA Ea 1 T R 14.4

2NO (g) + O2 (g) 2NO2 (g) Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. N2O2 is detected during the reaction! 14.5

Chemical Kinetics