CHEM 102, Fall 2015, LA TECH Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours: M.W &F, 8:00-9:00 & 11:00-12:00 and Tu,Th 8: :00 am. am. or by appointment Test Dates Chemistry 102 Fall 2015 Sept. 29, 2015 (Test 1): Chapter 13 Oct. 22, 2015 (Test 2): Chapter 14 &15 Nove. 17, 2015 (Test 3): Chapter 16 &17 Nvember 19, 2015 (Make-up test) comprehensive: Chapters 13-17

CHEM 102, Fall 2015, LA TECH Chapter 13. Chemical Kinetics 13.1 Catching Lizards The Rate of a Chemical Reaction The Rate Law: The Effect of Concentration on Reaction Rate The Integrated Rate Law: The Dependence of Concentration on Time The Effect of Temperature on Reaction Rate Reaction Mechanisms Catalysis 593

CHEM 102, Fall 2015, LA TECH Chemical Kinetics Definitions and Concepts a) The Rate of a Chemical Reaction a) The Rate of a Chemical Reaction b) rate law b) rate law b) rate constant b) rate constant c) order c) order d) differential rate law d) differential rate law c) integral rate law c) integral rate law d) Half-life law d) Half-life law

CHEM 102, Fall 2015, LA TECH a) Every chemical reaction has a Rate Law b) The rate law is an expression that relates the rate of a chemical reaction to a constant the rate of a chemical reaction to a constant (rate constant-k) and concentration of reactants raised to a power. c) The power of a concentration is called the order with respect to a particular reactant. Rate Law

CHEM 102, Fall 2015, LA TECH Rate Law E.g. aA + bB -----> cC rate  [A] l [B] m rate  [A] l [B] m rate = -1/a d[A]/dt = k [A] l [B] m ; k = rate constant rate = -1/a d[A]/dt = k [A] l [B] m ; k = rate constant [A] = concentration of A [A] = concentration of A [B] = concentration of B [B] = concentration of B l = order with respect to A l = order with respect to A m = order with respect to B m = order with respect to B l & m have nothing to do with stoichiometric coefficients l & m have nothing to do with stoichiometric coefficients

CHEM 102, Fall 2015, LA TECH Differential Rate Law E.g. E.g. 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) rate= - ½ d[N 2 O 5 ]/dt  [N 2 O 5 ] 1 rate= - ½ d[N 2 O 5 ]/dt  [N 2 O 5 ] 1 rate = - ½ d[N 2 O 5 ]/dt = k [N 2 O 5 ] 1 rate = - ½ d[N 2 O 5 ]/dt = k [N 2 O 5 ] 1 k = rate constant k = rate constant [N 2 O 5 ] = concentration of N 2 O 5 [N 2 O 5 ] = concentration of N 2 O 5 1 = order with respect to N 2 O 5 1 = order with respect to N 2 O 5 Rate and the order are obtained by experiments Rate and the order are obtained by experiments

CHEM 102, Fall 2015, LA TECH Order The power of the concentrations is the order with The power of the concentrations is the order with respect to the reactant. E.g. a A + b B -----> c C E.g. a A + b B -----> c C If the rate law: rate = k [A] 1 [B] 2 If the rate law: rate = k [A] 1 [B] 2 The order of the reaction with respect to A is one (1). The order of the reaction with respect to B is two (2). Overall order of a chemical reaction is equal to the sum of all orders (3).

CHEM 102, Fall 2015, LA TECH Graphical method Order R ate Law Integrated Rate LawGraph X vs. time Slope 0rate = k [A] t = -kt + [A] 0 [A] t -k-k 1 rate = k[A]ln[A] t = -kt + ln[A] 0 ln[A] t -k-k 2 rate=k[A] 2 = kt + k 1 [A] 0 1 [A] t 1 [A] t

CHEM 102, Fall 2015, LA TECH Chemical Kinetics Definitions and Concepts a) The Rate of a Chemical Reaction a) The Rate of a Chemical Reaction b) rate law b) rate law b) rate constant b) rate constant c) order c) order d) differential rate law d) differential rate law c) integral rate law c) integral rate law d) Half-life law d) Half-life law

CHEM 102, Fall 2015, LA TECH Rate Law Differential Rate Law Integral Rate rate  k [A] 0  [A]/  t = k ; ([A] 0 =1)[A] f -[A] 0 = -kt  d [A]/dt = k ; ([A] 0 =1 [A] f = -kt + [A] 0  d [A]/dt = k ; ([A] 0 =1 [A] f = -kt + [A] 0 [A] f - [A] 0 = -kt [A] f - [A] 0 = -kt rate  k [A] 1   [A]/  t = k [A] ln [A] t /[A] 0 = - kt rate  k [A] 1   [A]/  t = k [A] ln [A] t /[A] 0 = - kt d  [A]/dt = - k [A] d  [A]/dt = - k [A] rate  k [A] 2  [A]/  t = k [A] 2 1/ [A] f - 1/[A] 0 = kt  d  [A]/dt = - k [A] 2  d  [A]/dt = - k [A] 2 1/ [A] f = kt + 1/[A] 0 1/ [A] f = kt + 1/[A] 0 Differential and Integral Rate Law

CHEM 102, Fall 2015, LA TECH Integral Law [A] f -[A] 0 = -kt ln [A] t /[A] 0 = -kt 1/[A] f = kt + 1/[A] 0 Differential, Integral and Half-life forms First order -d [A]/ d t = k [A] 0 -d [A]/ d t = k/[A] 1 -d [A]/ d t = k/[A] 1 -d [A]/ d t = k[A] 2 Second order t ½ Law t ½ = [A] o / 2k t ½ = / k t ½ = 1 / k [A] o t ½ = 1 / k [A] o Zero order Differential Law

CHEM 102, Fall 2015, LA TECH Integral from Differential Half life from Intergral

CHEM 102, Fall 2015, LA TECH

CHEM 102, Fall 2015, LA TECH

CHEM 102, Fall 2015, LA TECH 1) The reaction A ---> B + C is known to follow the rate law: rate = k [A] 1 What are the differential, integral and half-life (t ½ ) form of this rate law?

CHEM 102, Fall 2015, LA TECH This plot of ln[cis-platin] vs. time produces a straight line, suggesting that the reaction is first-order. Comparing graphs

CHEM 102, Fall 2015, LA TECH Time / min[N 2 O 5 ] / moldm -3 ln [ N 2 O 5 ] Using graphical method, show that 2 N 2 O 5 ---> 4 NO 2 + O 2, is a first order reaction.

CHEM 102, Fall 2015, LA TECH Method of initial rates The order for each reactant is found by: Changing the initial concentration of that reactant. Holding all other initial concentrations and conditions constant. Measuring the initial rates of reaction The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant. Finding rate laws by Initial rates

CHEM 102, Fall 2015, LA TECH Decomposition Reaction

CHEM 102, Fall 2015, LA TECH Graphical Ways to get Order

CHEM 102, Fall 2015, LA TECH Initial rate

CHEM 102, Fall 2015, LA TECH How do get order of reactants E.g. a A + b B -----> c C Hold [B] constant and change (double) [A] a A + b B -----> c C a A + b B -----> c C If the rate law: rate = k [A] x [B] y If the rate law: rate = k [A] x [B] y rate = k [A] 1 k 1 rate = k [A] 1 k 1 First order: 1 x rate = k [2A] 1 k 1 = k 2 1 [A] 1 k 1 rate 1 = k [A ] 1 k 1 rate 1 = 1 rate 2 = k 2 1 [A ] 1 k 1 rate 2 = 2 1 (doubles) Second order: 2 x rate = k [2A] 1 k 1 = k 2 2 [A] 2 k 1 Second order: 2 x rate = k [2A] 1 k 1 = k 2 2 [A] 2 k 1 rate 1 = k [A ] 2 k 1 rate 1 = 1 rate 2 = k 2 2 [A ] 2 k 1 rate 2 = 2 2 (quadruples)

CHEM 102, Fall 2015, LA TECH How do you find order? A + B -----> C A + B -----> C rate = k [A] l [B] m ; rate = k [A] l [B] m ; Hold concentration of other reactants constant If [A] doubled, rate doubled 1st order, [2A] 1 = 2 1 x [A] 1, 2 1 = 2 1st order, [2A] 1 = 2 1 x [A] 1, 2 1 = 2 b) If [A] doubled, rate quadrupled 2nd order, [2A] 2 = 2 2 x [A] 2, 2 2 = 4 2nd order, [2A] 2 = 2 2 x [A] 2, 2 2 = 4 c) If [A] doubled, rate increased 8 times c) If [A] doubled, rate increased 8 times 3rd order, [2A] 3 = 2 3 x [A] 3, 2 3 = 8 3rd order, [2A] 3 = 2 3 x [A] 3, 2 3 = 8

CHEM 102, Fall 2015, LA TECH Rate data

CHEM 102, Fall 2015, LA TECH 3. For the reaction: A ---> D, Find the order of [A] for each case. It was found in separate experiments that a) The rate doubled when [A] doubled b) The rate tripled when [A] tripled c) The rate quadrupled when [A] doubled d) The rate increased 8 times when [A] doubled

CHEM 102, Fall 2015, LA TECH Units of the Rate Constant (k) 1 first order: k ─── s -1 first order: k = ─── = s -1 s L second order k ─── second order k = ─── mol s mol s L 2 L 2 third order k ─── third order k = ─── mol 2 s mol 2 s

CHEM 102, Fall 2015, LA TECH 4. For the chemical reaction: A + B ----> C Using the following initial data to deduce: a) Order of each reactant b) Rate constant [A],mol/L [B],mol/L rate,mol/Ls _____________________________

CHEM 102, Fall 2015, LA TECH Overall order

CHEM 102, Fall 2015, LA TECH Rate Constant E.g. a A + b B -----> c C E.g. a A + b B -----> c C rate  [A] l [B] m rate  [A] l [B] m rate = k [A] l [B] m ; rate = k [A] l [B] m ; k = rate constant k = rate constant proportionality constant of the rate law proportionality constant of the rate law Larger the k faster the reaction Larger the k faster the reaction It is related inversely to t ½ It is related inversely to t ½

CHEM 102, Fall 2015, LA TECH Determining K, Rate Constant

CHEM 102, Fall 2015, LA TECH First Order Reactions and t ½ A ----> B

CHEM 102, Fall 2015, LA TECH Radio Activity and Nuclear Kinetics Nuclear reactions? FusionFission What kinetics fission follow?

CHEM 102, Fall 2015, LA TECH Half-life t ½ RadioisotopeHalf-life Polonium seconds Bismuth seconds Sodium-2415 hours Iodine days Cobalt years Carbon years Radium years Uranium billion years

CHEM 102, Fall 2015, LA TECH Nuclear Reactions : First order kinetics

CHEM 102, Fall 2015, LA TECH t 1/2 equation t 1/2 equation  =k t 1/ t 1/2 = ---- t 1/2 = ---- k

CHEM 102, Fall 2015, LA TECH The half-life and the rate constant are related. t 1/2 = t 1/2 = Half-life can be used to calculate the first order rate constant. For our N 2 O 5 example, the reaction took 1900 seconds to react half way so: k = = = 3.65 x s k t 1/ s Half-life - t 1/2

CHEM 102, Fall 2015, LA TECH 5. The rate constant for the first-order conversion of A to B is 2.22 hr -1. How much time will be required for the concentration of A to reach 75% of its original value?

CHEM 102, Fall 2015, LA TECH 6) The half-life of a radioactive (follows first order rate law) isotope is 10 days. How many days would be required for the isotope to degrade to one eighth of its original radioactivity?

CHEM 102, Fall 2015, LA TECH 7) The rate constant for the first order decomposition of SO 2 Cl 2 (SO 2 Cl 2  SO 2 +Cl 2 ) at very high temperature is 1.37 × min -1. If the initial concentration is M, predict the concentration after five hours (300 min).