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[object Object],[object Object],[object Object],[object Object]
Kinetics ,[object Object],[object Object]
Factors That Affect Reaction Rates ,[object Object],[object Object]
Factors That Affect Reaction Rates ,[object Object],[object Object],[object Object]
Factors That Affect Reaction Rates ,[object Object],[object Object],[object Object],[object Object]
Factors That Affect Reaction Rates ,[object Object],[object Object],[object Object]
Factors That Affect Reaction Rates ,[object Object],[object Object],[object Object]
 
[object Object]
Reaction Rates ,[object Object]
Reaction Rates  ,[object Object],C 4 H 9 Cl ( aq )  + H 2 O ( l )     C 4 H 9 OH ( aq )  + HCl ( aq )
Reaction Rates  ,[object Object],C 4 H 9 Cl ( aq )  + H 2 O ( l )     C 4 H 9 OH ( aq )  + HCl ( aq )   Average rate =  [C 4 H 9 Cl]  t
Reaction Rates  ,[object Object],[object Object],C 4 H 9 Cl ( aq )  + H 2 O ( l )     C 4 H 9 OH ( aq )  + HCl ( aq )
Reaction Rates  ,[object Object],[object Object],C 4 H 9 Cl ( aq )  + H 2 O ( l )     C 4 H 9 OH ( aq )  + HCl ( aq )
Reaction Rates  ,[object Object],[object Object],C 4 H 9 Cl ( aq )  + H 2 O ( l )     C 4 H 9 OH ( aq )  + HCl ( aq )
 
[object Object]
Reaction Rates and Stoichiometry  ,[object Object],[object Object],C 4 H 9 Cl( aq ) + H 2 O( l )     C 4 H 9 OH( aq ) + HCl( aq )  Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t
Reaction Rates and Stoichiometry ,[object Object],2 HI ( g )      H 2 ( g )  + I 2 ( g ) ,[object Object],Rate =  − 1 2  [HI]  t =  [I 2 ]  t
Reaction Rates and Stoichiometry ,[object Object],a A +  b B  c C +  d D Rate =  − 1 a  [A]  t =  − 1 b  [B]  t = 1 c  [C]  t 1 d  [D]  t =
For the reaction:  A   2 B , which of the following is a correct expression for the reaction rate? t   Rate [A] = t   Rate 2 [A] = 2.   t   Rate 2 [B] = 3.  4.  t   Rate 2 [B] = 1.
Correct Answer: In general for aA + bB    cC + dD t   Rate [A] = t   Rate 2 [A] = 2.  t   Rate 2 [B] = 3.  4.  t   Rate 2 [B] = 1. =    =    =   =
For the reaction:  2 A    3 B the rate of appearance of B,   [B]/  t , is 0.30  M /s.  What is the value of the rate of disappearance of A, –  [A]/  t ?   ,[object Object],[object Object],[object Object],[object Object],[object Object]
t t     Rate 3 [B] 2 [A] = = Correct Answer: t t     Rate 3 [B] 2 [A] = = /s 0.20 3 /s) 2(0.30 [A] M M t = = =   Rate ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object]
Concentration and Rate ,[object Object]
Concentration and Rate ,[object Object],NH 4 + ( aq )  + NO 2 − ( aq ) N 2 ( g )  + 2 H 2 O ( l )
Concentration and Rate ,[object Object],NH 4 + ( aq )  + NO 2 − ( aq ) N 2 ( g )  + 2 H 2 O ( l )
Concentration and Rate ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
For the reaction:  A + 2 B    B 2 A Doubling [A] doubles the rate of the reaction.  However, doubling [B] increases the rate 4  .  What is the correct rate law expression?   [A][B] Rate k = 3.  2. 4.  1.
For the reaction:  A + 2 B    B 2 A Doubling [A] doubles the rate of the reaction.  However, doubling [B] increases the rate 4  .  What is the correct rate law expression?   [A][B] Rate k = 3.  2. 4.  1. The rate order on A is one, because a doubling of [A] leads to a doubling of reaction rate. The rate order on B is two, because doubling [B] leads   to a 4    = (2) 2  increase in reaction rate. Rate =  k [A][B] 2
What is the rate law for the reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],A  +  2 B   products 2.0 x 10 – 4 0.20 0.20 1.0 x 10 – 4 0.20 0.10 1.0  x  10 -4 0.10 0.10 Initial Rate  M /s [B] [A]
What is the rate law for the reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],A  +  2 B   products 2.0 x 10 – 4 0.20 0.20 1.0 x 10 – 4 0.20 0.10 1.0  x  10 -4 0.10 0.10 Initial Rate  M /s [B] [A]
What is the rate law for the reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],2 A  +  B   products 8.0 x 10 -4 0.20 0.20 4.0 x 10 -4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate  M /s [B] [A]
What is the rate law for the reaction? ,[object Object],[object Object],[object Object],[object Object],[object Object],2 A  +  B   products 8.0 x 10 -4 0.20 0.20 4.0 x 10 -4 0.20 0.10 1.0 x 10 -4 0.10 0.10 Initial Rate  M /s [B] [A]
Rate Laws Rate =  k  [NH 4 + ] [NO 2 − ] ,[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Rate Laws Rate =  k  [NH 4 + ] [NO 2 − ]
 
[object Object]
[object Object],[object Object]
 
[object Object],[object Object]
[object Object],[object Object],[object Object]
What are the units of the rate constant for a reaction with a rate law of: ,[object Object],[object Object],[object Object],[object Object],[object Object],Rate =  k [A][B]
What are the units of the rate constant for a reaction with a rate law of: ,[object Object],[object Object],[object Object],[object Object],[object Object],Rate =  k [A][B]
[object Object],[object Object],[object Object],[object Object]
Answers:   (a)  rate =  k [NO] 2 [H 2 ];  (b)  k  =1.2  M –2 s –1 ;  (c)  rate = 4.5    10 –4   M /s (a)  Determine the rate law for this reaction.  (b)  Calculate the rate constant.  (c)  Calculate the rate when  [NO] = 0.050  M  and [H 2 ] = 0.150  M .  BELLWORK The following data were measured for the reaction of nitric oxide with hydrogen:
Rate laws (a.k.a.  Differential rate laws ) are used to calculate reaction rates when given reactant concentrations. Integrated rate laws  are used to calculate the amount of reactants left after a certain amount of time has passed.
Integrated Rate Laws ,[object Object],Where [A] 0  is the initial concentration of A. [A] t  is the concentration of A at some time,  t , during the course of the reaction. ln [A] t [A] 0 =  − k t
Integrated Rate Laws ,[object Object],ln [A] t   −  ln [A] 0  =  −   k t ln [A] t  =  −   k t + ln [A] 0 … which is in the form y   =  mx  +  b ln [A] t [A] 0 =  − k t
First-Order Processes ,[object Object],ln [A] t  =  - k t + ln [A] 0
First-Order Processes ,[object Object],CH 3 NC CH 3 CN
First-Order Processes ,[object Object],CH 3 NC CH 3 CN
First-Order Processes ,[object Object],[object Object],[object Object],[object Object]
Second-Order Processes ,[object Object],also in the form y   =  mx +  b 1 [A] t =  − kt  + 1 [A] 0
Second-Order Processes ,[object Object],1 [A] t =  − kt  + 1 [A] 0
Second-Order Processes The decomposition of NO 2  at 300 ° C is described by the equation and yields data comparable to this: NO 2   ( g ) NO  ( g )  + 1/2 O 2   ( g ) Time ( s ) [NO 2 ],  M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380
Second-Order Processes ,[object Object],[object Object],Time ( s ) [NO 2 ],  M ln [NO 2 ] 0.0 0.01000 − 4.610 50.0 0.00787 − 4.845 100.0 0.00649 − 5.038 200.0 0.00481 − 5.337 300.0 0.00380 − 5.573
Second-Order Processes ,[object Object],[object Object],Time ( s ) [NO 2 ],  M 1/[NO 2 ] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263
The plot given below indicates a component whose reaction order is __. ,[object Object],[object Object],[object Object],[object Object]
The plot given below indicates a component whose reaction order is __. ,[object Object],[object Object],[object Object],[object Object],Second-order reactions will exhibit a linear plot of 1/[reactant] vs. time.
a)  b)
[object Object],a)  b)
[object Object],[object Object],a)  b)
Half-Life ,[object Object],[object Object],[object Object]
Half-Life ,[object Object],ln 0.5 =  − kt 1/2 − 0.693 =  − kt 1/2 NOTE:  For a first-order process, the half-life does not depend on [A] 0 . 0.5 [A] 0 [A] 0 ln =  − kt 1/2 =  t 1/2 0.693 k
Half-Life ,[object Object],1 0.5 [A] 0 =  kt 1/2  +  1 [A] 0 2 [A] 0 =  kt 1/2  +  1 [A] 0 2  −  1 [A] 0 =  kt 1/2 1 [A] 0 = =  t 1/2 1 k [A] 0
For a first order reaction,  k  = 0.0693 s  1  and the initial concentration of the reactant is 0.10  M .  Calculate the half-life of the reactant. ,[object Object],[object Object],[object Object],[object Object],[object Object]
For a first order reaction,  k  = 0.0693 s  1  and the initial concentration of the reactant is 0.10  M .  Calculate the half-life of the reactant. ,[object Object],[object Object],[object Object],[object Object],[object Object],Because the reaction is first order, half-life depends only on the rate constant according to the equation: t 1/2  = 0.693/ k  = 0.0693/0.0693 s  1 t 1/2  = 10. s
=  t 1/2 1 k [A] 0
[object Object],=  t 1/2 1 k [A] 0
If  t 1/2  = 500 s for a first-order reaction, how long will it take for the concentration of a reactant to decrease to 1/8 of its original value?  ,[object Object],[object Object],[object Object],[object Object],[object Object]
If  t 1/2  = 500 s for a first-order reaction, how long will it take for the concentration of a reactant to decrease to 1/8 of its original value?  ,[object Object],[object Object],[object Object],[object Object],[object Object]
SUMMARY Order of reaction Zero First Second Rate law Rate = k Rate = k[A] Rate = k[A] 2 Integrated rate law [A] t  = [A] 0  - kt ln[A] t - ln[A] 0  = -kt 1/[A] t  - 1/[A] 0  = kt Units of k M/s 1/s 1/(M •s) Linear plot [A] vs. t ln[A]  vs.  t 1/[A]  vs.  t slope -k -k k Half life t 1/2 = 1/2[A] 0 k t 1/2  = 0.693/k t 1/2 = 1/(k[A] 0 )
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Bellwork a)What is the order of the reaction? b)What is the rate law? c) What effect will doubling the initial concentration of A have on the reaction rate?
Temperature and Rate ,[object Object],[object Object]
The Collision Model ,[object Object],[object Object]
The Collision Model ,[object Object]
 
[object Object]
Activation Energy ,[object Object],[object Object]
Reaction Coordinate Diagrams ,[object Object]
Reaction Coordinate Diagrams ,[object Object],[object Object],[object Object],[object Object]
Reaction Coordinate Diagrams ,[object Object]
Maxwell – Boltzmann Distributions ,[object Object],[object Object]
Maxwell – Boltzmann Distributions ,[object Object],[object Object]
Maxwell – Boltzmann Distributions ,[object Object],[object Object]
Maxwell – Boltzmann Distributions ,[object Object],[object Object],f  =  e − E a / RT
 
[object Object]
Arrhenius Equation ,[object Object],[object Object],[object Object]
Arrhenius Equation ,[object Object],[object Object],y   =  mx  +  b Therefore, if  k  is determined experimentally at several temperatures,  E a   can be calculated from the slope of a plot of ln  k  vs. 1/ T . 1 RT
The plot below of ln  k  vs. 1/ T  is useful because it allows determination of:   ,[object Object],[object Object],[object Object]
The plot below of ln  k  vs. 1/ T  is useful because it allows determination of:   ,[object Object],[object Object],[object Object],The slope of the line is equal to    R / E a ,  where  E a   is the activation energy.
Reaction Mechanisms ,[object Object]
Reaction Mechanisms ,[object Object],[object Object]
Reaction Mechanisms ,[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
The following mechanism has been proposed for: 2 NO + Cl 2    2 NOCl NO + Cl 2     NOCl 2   NOCl 2  + NO    2 NOCl Identify the reaction intermediate. ,[object Object],[object Object],[object Object],[object Object]
Correct Answer: NOCl 2  is not observed in the overall chemical reaction; therefore it is a short-lived reaction intermediate. ,[object Object],[object Object],[object Object],[object Object]
 
Multistep Mechanisms ,[object Object],[object Object]
Slow Initial Step ,[object Object],[object Object],[object Object],[object Object],NO 2   ( g )  + CO  ( g )     NO  ( g )  + CO 2   ( g )
Slow Initial Step ,[object Object],[object Object],[object Object],[object Object],[object Object]
Fast Initial Step ,[object Object],[object Object],[object Object],2 NO  ( g )  + Br 2   ( g )     2 NOBr  ( g )
Fast Initial Step ,[object Object],Step 2:  NOBr 2  + NO    2 NOBr  (slow) Step 1:  NO + Br 2 NOBr 2   (fast)
[object Object],[object Object],[object Object],[object Object],Step 2:  NOBr 2  + NO    2 NOBr  (slow) Step 1:  NO + Br 2 NOBr 2   (fast)
[object Object],[object Object],[object Object],[object Object],Step 2:  NOBr 2  + NO    2 NOBr  (slow) Step 1:  NO + Br 2 NOBr 2   (fast)
[object Object],[object Object]
For 2 NO + Cl 2    2 NOCl, the following mechanism has been proposed. ,[object Object],[object Object],NO + Cl 2     NOCl 2   NOCl 2  + NO    2 NOCl The rate law is Rate =  k [NO] 2 [Cl 2 ], which implies which step is the rate-limiting step?
Correct Answer: If the first step were rate limiting, the observed rate law would be: Rate  = k [NO][Cl 2 ] The square dependence on [NO] indicates the second step must be slow relative to the first step. ,[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
Catalysts ,[object Object],[object Object]
Catalysts ,[object Object]
Homogenous catalysts are in the same phase as the reactants. Heterogeneous catalysts are in a different phase than the reactants.  Pt, Pd, & Rh metals are used to catalyze gas reactions in a catalytic converter.
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Correct Answer: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Enzymes ,[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Chapter 14 Lecture- Chemical Kinetics

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  • 21. For the reaction: A  2 B , which of the following is a correct expression for the reaction rate? t   Rate [A] = t   Rate 2 [A] = 2. t   Rate 2 [B] = 3. 4. t   Rate 2 [B] = 1.
  • 22. Correct Answer: In general for aA + bB  cC + dD t   Rate [A] = t   Rate 2 [A] = 2. t   Rate 2 [B] = 3. 4. t   Rate 2 [B] = 1. =  =  = =
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  • 30. For the reaction: A + 2 B  B 2 A Doubling [A] doubles the rate of the reaction. However, doubling [B] increases the rate 4  . What is the correct rate law expression? [A][B] Rate k = 3. 2. 4. 1.
  • 31. For the reaction: A + 2 B  B 2 A Doubling [A] doubles the rate of the reaction. However, doubling [B] increases the rate 4  . What is the correct rate law expression? [A][B] Rate k = 3. 2. 4. 1. The rate order on A is one, because a doubling of [A] leads to a doubling of reaction rate. The rate order on B is two, because doubling [B] leads to a 4  = (2) 2 increase in reaction rate. Rate = k [A][B] 2
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  • 47. Answers:   (a) rate = k [NO] 2 [H 2 ]; (b) k =1.2 M –2 s –1 ; (c) rate = 4.5  10 –4 M /s (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H 2 ] = 0.150 M . BELLWORK The following data were measured for the reaction of nitric oxide with hydrogen:
  • 48. Rate laws (a.k.a. Differential rate laws ) are used to calculate reaction rates when given reactant concentrations. Integrated rate laws are used to calculate the amount of reactants left after a certain amount of time has passed.
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  • 57. Second-Order Processes The decomposition of NO 2 at 300 ° C is described by the equation and yields data comparable to this: NO 2 ( g ) NO ( g ) + 1/2 O 2 ( g ) Time ( s ) [NO 2 ], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380
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  • 70. = t 1/2 1 k [A] 0
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  • 74. SUMMARY Order of reaction Zero First Second Rate law Rate = k Rate = k[A] Rate = k[A] 2 Integrated rate law [A] t = [A] 0 - kt ln[A] t - ln[A] 0 = -kt 1/[A] t - 1/[A] 0 = kt Units of k M/s 1/s 1/(M •s) Linear plot [A] vs. t ln[A] vs. t 1/[A] vs. t slope -k -k k Half life t 1/2 = 1/2[A] 0 k t 1/2 = 0.693/k t 1/2 = 1/(k[A] 0 )
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  • 76. Bellwork a)What is the order of the reaction? b)What is the rate law? c) What effect will doubling the initial concentration of A have on the reaction rate?
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  • 118. Homogenous catalysts are in the same phase as the reactants. Heterogeneous catalysts are in a different phase than the reactants. Pt, Pd, & Rh metals are used to catalyze gas reactions in a catalytic converter.
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Notas do Editor

  1. Need Figure 14.4 ?
  2. Need Figure 14.4 ?
  3. NEED FIGURE 14.7?
  4. NEED FIGURE 14.7?
  5. NEED FIGURE 14.7?