Chapter 13 - Chemical Kinetics II Integrated Rate Laws Reaction Rates and Temperature
Reaction Order - Graphical Picture A ->Products
Integrated Rate Laws
Zero Order Reactions Rate = k[a] 0 = k (constant rate reactions) [A] = -kt +[A]0 (integrated rate law) graph of [A] vs t is straight line (slope = -k and y intercept = [A]0) t½ = [A]0/2k [A] 0 [A] slope = - k time
First Order Reactions Rate = k[a] ln[a] = -kt +ln[a]0 (integrated rate law) graph of ln[a] vs t is straight line (slope = -k and y intercept = ln[a]0) t½ = ln2/k = (0.693)/k (constant half-life) ln[a] 0 [A] 0 [A] ln[a] slope = - k time
Second Order Reactions Rate = k[a] 2 1/[A] = -kt + 1/[A]0 (integrated rate law) graph of 1/[A] vs t is straight line (slope = +k and y intercept = 1/[A]0) t½ = 1/(k[A]0) 1/[A] slope = +k 1/[A] 'me
Half-life, t½ the length of time it takes for the concentration of the reactants to fall to one-half its initial value depends on the order of the reaction
Zero Order Reactions Determining Rate Graphically First Order Reactions Second Order Reactions
Data for 2NO2 (g) 2NO (g) + O2 (g) ln[no2] vs time 1/[NO2] vs time
Problems Using Integrated Rate Laws
1) N2O decomposes at 1050 K through 1st order kinetics: 2 N2O > 2 N2 + O2 Rate = k[n2o], k = 3.4 sec -1 For an initial concentration of [N2O] of 0.20 M, what is the [N2O] after 100 msec. kt = ln[a]0 - ln[a] kt = ln [A]0 [A] (3.4 sec -1 )(100 x 10-3 sec)= ln [A] 0 [A] (3.4 sec e -1 )(100 x 10-3 sec) [A]0 = [A] (.34) [A]0 e = [A] 1.40 = 0.20 M x x = 0.14 M
2) N2O5 decomposes at 25 ºC through 1st order kinetics: 2 N2O5 > 4 NO2 + O2 Given the following concentration vs time data, determine the rate constant for the reaction. tmin 0 200 400 600 800 1000 [N2O5] 0.015 0.0096 0.0062 0.004 0.0025 0.0016 0.01500 0.01125 [N2O5] [N2O5] vs time ln[n2o5] -4.20-4.65-5.08-5.52-5.99-6.43 ln[n2o5] vs time 0 0.00750-1.75 0.00375 0 0 250 500 750 1000 [A] 0-3.50 slope = -k -5.25 -slope = -2.23 60,000 sec -( ) -7.00 0 250 500 750 1000 slope = 3.7 x 10-5 sec -1
3) Cyclopropane is converted to propene (a rearrangement) at 500ºC with first order kinetics with a rate constant of 6.7 x 10-4 sec -1. C3H6 > C3H6 How long does it take for the concentration of cyclopropane to drop to 1/2 of its original concentration? to 1/4? t ½ = 0.69/(6.7 x 10-4 sec -1 ) = 1.0 x 10 3 sec t 1/4 = 2 x t ½ = 2.0 x 10 3 sec 4) N2O5 decomposes at 25 ºC through 1st order kinetics: 2 N2O5 > 4 NO2 + O2 k = 5.2 x 10-3 sec -1 at 65º How long will it take for a sample which is 20 mm N2O5 to decrease to 2 mm? kt = ln [A]0 t = ln [A]0 [A] k [A] 1 1 = ln 5.2 x 10 20mM -3 2 mm = (192)ln(10) = 442 sec
5) HI decomposes at 700 K through 2nd order kinetics: 2 HI > H2 + I2 k = 1.8 x 10-3 M -1 sec -1 How long will it take for 1/2 of a 0.10 M sample to decompose? for 3/4?. t ½ = 1/(1.8 x 10-3 M -1 sec -1) (0.010 M) t ½ = 5.6 x 10 4 sec t = 1/(1.8 x 10-3 M -1 sec -1 )(0.005 M) 1/2 1/4 t 1/2 1/4 = 11.2 x 10 4 sec Total time for 3/4 to decompose = 1.7 x 10 5 sec
Reaction Rates and Temperature
Temperature: Changing the temperature changes the rate constant in the rate law. These factors are incorporated into rate constant k by the Arrhenius equation: R = universal gas constant (in J/mol K) T = temperature (in kelvin) A is a collisional frequency factor. (Includes frequency of collisions and an orientation factor) E a = Activation energy, The minimum energy of molecular collisions required to break bonds in reactants, leading to formation of products.
Collision Theory In order for a chemical reaction to occur, reactant molecules must collide with sufficient energy and in the proper orientation. Effective collisions involve both of these factors. When two molecules have an effective collision, a temporary, high-energy (unstable) chemical species, the reactive intermediate*, is formed. *also referred to as an activated complex or transition state
Arrhenius equation: Activation Energy Frequency Factor Exponential Factor
Reaction Energy Profile High-energy transition state ( activated complex ) Activation energy (E a )
Reaction Energy Profile Partially broken and partially formed bonds The amount of energy necessary to form the activated complex
The Reaction of CH3Br and OH-
Reaction Energy Diagram for the Reaction of CH3Br and OH-
The Arrhenius equation can be rearranged: by
Graphical Determination of E a (Arrhenius Plot) Slope = E a /R Intercept = ln A According to the rearranged Arrhenius equation, the activation energy for a reaction can be determined by plotting k vs 1/T which yields a line whose slope is -E a /R and whose intercept is lna.
Mathematical Determination of E a Consider determination of k two different T s: By subtraction,
The Exponential Factor k = (A)(e - Ea RT) A number between 0 and 1, representing the fraction of reactant molecules with sufficient energy to make it over the energy barrier
The Effect of E a on the Fraction of Collisions with Sufficient Energy to Allow Reaction Temp E a (J/mol) -E a /RT -E a /RT e 523 K 10,000-2.30 0.100 523 K 523 K 20,000 50,000-4.60 0.001-11.5 0.00001 k = (A)(e - Ea RT) The higher the energy barrier, the fewer molecules that have sufficient energy to overcome the energy barrier.
The Effect of E a on the Fraction of Collisions with Sufficient Energy to Allow Reaction
The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction E a (J/mol) Temp -E a /RT -E a /RT e 10,000 298 K -4.04 0.0176 10,000 523 K -2.30 0.100 10,000 773 K -1.56 0.211 k = (A)(e - Ea RT) The extra kinetic energy of the reactants is converted into potential energy when molecules collide.
The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction Increased temperature increases kinetic energy of molecules and molecular collisions.
Frequency Factor k = (A)(e - Ea RT) A number representing the number of reactant molecules which can approach each other with the correct orientation and sufficient energy to make it over the energy barrier
Frequency Factor A=(p)(z) p Orientation Factor for reactions of single atoms, p=1 for most molecules p<<1 z Collision Frequency Factor for a collision to overcome the energy barrier, the reactants mus have sufficient kinetic energy so that the collision leads to the formation of an activated complex
c
Effective Collisions-Kinetic Energy Effect