© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t y = b +mx y = b +mx

© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction

© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction

© University of South Carolina Board of Trustees Graphing a 1 st -Order Reaction straight line  this is a first-order reaction.

© University of South Carolina Board of Trustees [A] vs t Data  Rate Law Method of Initial Rates (Sec. 13.2)  Trial and Error with Common Laws (Sec. 13.3)

© University of South Carolina Board of Trustees ln [A] = ln [A] 0 - k t intercept = ln [A] 0

© University of South Carolina Board of Trustees ln [A] = ln [A] 0 - k t

© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life

© University of South Carolina Board of Trustees 1 st -Order Half-Life [A]  [A]/2 t 1/2

© University of South Carolina Board of Trustees 1 st -Order Half-Life [A]  [A]/2 t 1/2 5  s t 1/2

© University of South Carolina Board of Trustees 1 st -Order Half-Life [A]  [A]/2 t 1/2 5  s 3  s t 1/2

© University of South Carolina Board of Trustees Half-life [A]  [A]/2 t 1/2 5  s 3  s 1  s t 1/2

© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life always the same

© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life t 1/2 = ln 2 / k always the same

© University of South Carolina Board of Trustees Half-Life Problem I k = 6.2 x10 -5 s -1 for the reaction C 12 H 22 O 11 + H 2 O  C 6 H 12 O 6 + C 6 H 12 O 6 Calculate the half-life.

© University of South Carolina Board of Trustees 1 st -Order Rate Law Differential Form Rate = k [A]( t ) Integral Forms [A]( t ) = [A] 0 e - kt ln [A]( t ) = ln [A] 0 - k t Half-life t 1/2 = ln 2 / k always the same

© University of South Carolina Board of Trustees Common Rate Laws A  products a) First-order Rate = k [A] b) Second-order Rate = k [A] 2 c) Zero-order Rate = k [A] 0 = k

© University of South Carolina Board of Trustees 2 nd -Order Rate Law Differential Form Rate = k [A]( t ) 2

© University of South Carolina Board of Trustees 2 nd -Order Rate Law Differential Form Rate = k [A]( t ) 2 Integral Form

© University of South Carolina Board of Trustees Graphing a 2 nd -Order Reaction

© University of South Carolina Board of Trustees Graphing a 2 nd -Order Reaction not a straight line not a first-order reaction

© University of South Carolina Board of Trustees 2 nd -Order Rate Law Differential Form Rate = k [A]( t ) 2 Integral Form y = b + mx

© University of South Carolina Board of Trustees 1/ [A] = 1/ [A] 0 + k t straight line  this is a second- order reaction

© University of South Carolina Board of Trustees 1/ [A] = 1/ [A] 0 + k t 1/[A] 0 intercept = 1/[A] 0

© University of South Carolina Board of Trustees 1/ [A] = 1/ [A] 0 + k t yy xx 1/[A] 0

© University of South Carolina Board of Trustees 2 nd -Order Rate Law Differential Form Rate = k [A]( t ) 2 Half-life Integral Form

© University of South Carolina Board of Trustees 2 nd -Order Half-Life [A]  [A]/2 t 1/2

© University of South Carolina Board of Trustees 2 nd -Order Half-Life 1.0 s [A]  [A]/2 t 1/2 5  s

© University of South Carolina Board of Trustees 2 nd -Order Half-Life 4.7 s [A]  [A]/2 t 1/2 5  s 1   s

© University of South Carolina Board of Trustees 2 nd -Order Half-Life 4.7 s [A]  [A]/2 t 1/2 1.0 s 5  s 1   s reaction gets ‘slower’ as it proceeds

© University of South Carolina Board of Trustees 2 nd -Order Rate Law Differential Form Rate = k [A]( t ) 2 Half-life t 1/2 = 1 / k [A] varies - not so useful Integral Form

© University of South Carolina Board of Trustees Example: 2 nd -Order Reaction The reaction 2NOCl  2NO + Cl 2 obeys the rate law Rate = (0.020 M -1 s -1 ) [NOCl] 2 Calculate the concentration of NOCl after 30 minutes, if the initial concentration was M.

© University of South Carolina Board of Trustees Common Rate Laws A  products a) First-order Rate = k [A] b) Second-order Rate = k [A] 2 c) Zero-order Rate = k [A] 0 = k

© University of South Carolina Board of Trustees 0 th -Order Rate Law Differential Form Rate = k [A]( t ) 0 Rate = k

© University of South Carolina Board of Trustees 0 th -Order Rate Law Differential Form Rate = k [A]( t ) 0 Rate = k y = b + mx Integral Forms [A]( t ) = [A] 0 - kt

© University of South Carolina Board of Trustees Graphing a 0 th -Order Reaction straight line  this is a 0 th -order reaction.

© University of South Carolina Board of Trustees [A](t)=[A] 0 - kt yy xx

© University of South Carolina Board of Trustees 0 th -Order Rate Law Differential Form Rate = k [A]( t ) 0 Rate = k Half-life y = b + mx Integral Forms [A]( t ) = [A] 0 - kt

© University of South Carolina Board of Trustees 0 th -Order Rate Law Differential Form Rate = k [A]( t ) 0 Rate = k Half-life varies - not so useful Integral Forms [A]( t ) = [A] 0 - kt

© University of South Carolina Board of Trustees Summary: Differential Rate Laws Zero-Order First-Order Second-Order Rate vs Concentration

© University of South Carolina Board of Trustees Summary: Integral Rate Laws Zero-Order[A]( t ) = [A] 0 − kt First-Order[A]( t ) = [A] 0 e − kt Second-Order Concentration vs Time

© University of South Carolina Board of Trustees Summary: Graphical Tests Zero-Order[A]( t ) = [A] 0 − k t First-Orderln [A]( t ) = ln[A] 0 − k t Second-Order y = b + m x

© University of South Carolina Board of Trustees Summary: Half-Life Zero-Ordervaries First-Orderconstant: Second-Ordervaries

© University of South Carolina Board of Trustees Half-Life Problem II A sample of wood has 58% of the 14 C ( t 1/2 = 5730 yr) originally present. What is the age of the wood sample?

© University of South Carolina Board of Trustees Chapt. 13 Kinetics Sec. 4 Theory of the Rate Constant ( k )

© University of South Carolina Board of Trustees Bimolecular Rate Theory A + B  products Rate = k ( T ) [A][B] Experiment

© University of South Carolina Board of Trustees Bimolecular Rate Theory A + B  products Rate =frequency of collisions Rate = k ( T ) [A][B] Experiment

© University of South Carolina Board of Trustees Bimolecular Rate Theory A + B  products Rate =frequency of collisions( Z 0 [A][B]) Rate = pZ 0 e- [A][B] Theory Rate = k ( T ) [A][B] Experiment

© University of South Carolina Board of Trustees Bimolecular Rate Theory A + B  products Rate =frequency of collisions( Z 0 [A][B]) Rate = pZ 0 e- [A][B] Theory Rate = k ( T ) [A][B] Experiment correct conc. dependence rates much too large no temperature dependence

© University of South Carolina Board of Trustees k Changes with Temperature

© University of South Carolina Board of Trustees Activation Energy Diagram  G Thermodynamics Kinetics Reactants Products Transition State