Chapter15: Chemical Kinetics

1
Chapter15 Chemical Kinetics
2
Rates of Reactions (Kinetics)

• in the previous chapter we discussed reaction
  spontaneity.
• we also have discussed equilibrium composition
• (i.e. should it change, and if so what will it
  look like when it is done)
• thus far we have ignored the element of time.
• e.g. CO(g) NO(g) -----gt CO2(g) 1/2N2(g)
• ?G -343.8kJ
• Keq 1060
• both of these values would lead you to believe
  this to be a very favorable reaction

3
Kinetics Deals With Three Points

• 1. At what rate does a chemical system undergo
  change under a given set of conditions?
• 2. How will changing conditions affect the rate?
• 3. Given the above, what can we say about the
  details of the chemical change?

4
Rate of Reaction

• change that occurs in a given period of time
• Cl2 2I- -----gt 2Cl- I2
• e.g. measure appearance of I2
• ?I2 I2t later - I2t start
  and ?t tlater - tstart
• ? ???????????I2
• rate -----------------
• ? ????????????t
• could also have measured disappearance of Cl2
• ???????????Cl2
• rate -----------------
• ? ????????????t

will make rate a positive quantity
5
Rate of Reaction

• for the reaction CH3Cl(aq)
  I-(aq) -----gt CH3I(aq) Cl-(aq)
• I- t(min)
• 0.50 0
• 0.45 180
• 0.41 360
• 0.35 720
• 0.27 1440
• what is the rate from start of the reaction to
  180 min?
• ???????????I- -(0.45 - 0.50)M
• rate ------------ --------------------
  ---- 2.77 x 10-4 mole/L.min
• ? ????????????t 180 - 0 min
• from 360 to 1440 minutes?
• -(0.27 - 0.41)M
• rate ------------------------ 1.29 x
  10-4 mole/L.min
• ?? 1440 - 360 min

6
Rate of Reaction

• notice that the rate has slowed down later in the
  reaction. Why?
• rates depend on the concentrations of one or more
  of the reactants (since they are consumed, their
  concentrations decrease and the rates decrease).
• because the rate is continually decreasing, the
  rate measured for any ?t is only the average
  rate.
• most often report initial rates.
• NOTES
• 1. can use any time interval be sure to use
  common units.
• 2. do not have to measure concentration directly
  can measure any quantity which changes as the
  concentration changes (e.g. Pressure, absorbance,
  etc.)

7
Reaction Rate and Concentration

• one goal of kinetics is to establish the exact
  relationship between concentration and the rate
  of reaction.
• this relationship can be established ONLY BY
  EXPERIMENTATION!
• the equation that describes this relationship is
  called the Rate Law.
• e.g. 2H2O2 -----gt 2H2O O2
• rate is found experimentally to be proportional
  to H2O2
• therefore, the rate law is rate kH2O2
• e.g. 2NOCl -----gt 2NO Cl2
• rate is found experimentally to be proportional
  to NOCl2
• therefore, the rate law is rate kNOCl2
• NOT RELATED TO OVERALL STOICHIOMETRY!

8
Rate Order

• the exponents define a characteristic of a
  reaction, called the rate order
• rate kH2O2 describes a first order reaction
• rate kNOCl2 describes a second order
  reaction
• NO2 CO -----gt NO CO2
• follows the rate law rate kNO2CO
• is first order in NO2 and CO, and second order
  overall
• 2NO O2 -----gt 2NO2 rate kNO2O2
• H2 Br2 -----gt 2HBr rate kH2Br21/2
• Remember Rate Law can NOT be found by
  inspection, only by experimentation.

9
Determining Rate Orders

• in many cases, can find the rate order by
  measuring the initial rates of reaction at
  different concentrations
• e.g. O3(g) NO(g) -----gt O2(g) NO2(g)
• O3 NO rate
• 1. 2.1x10-6 2.1x10-6 1.6
• 2. 4.2x10-6 2.1x10-6 3.2
• 3. 6.3x10-6 2.1x10-6 4.8
• 4. 6.3x10-6 4.2x10-6 9.6
• 5. 6.3x10-6 6.3x10-6 14.4
• compare experiments where one component changes
  and all others are constant.
• in experiments 1 2, double O3 and rate
  doubles
• in experiments 1 3, triple O3 and rate
  triples
• therefore, rate is proportional to O3
• in experiments 3 4, double NO and rate
  doubles
• in experiments 3 5, triple NO and rate
  triples
• therefore, rate is proportional to NO
• RATE LAW rate kO3NO

10
Determining Rate Laws All Purpose Method

• e.g. 2X Z -----gt P
• X Z rate
• I. 0.1 0.1 4.6x10-4
• II. 0.2 0.1 9.1x10-4
• III. 0.3 0.1 1.3x10-3
• IV. 0.1 0.2 1.8x10-3
• rateII kXIIYZIIQ XIIY
  XII Y 0.2 Y 2Y 9.1 x 10-4
• rateI kX IYZIQ X IY
  XI 0.1 4.6 x 10-4
• 2Y 1.978 Y 1
• rateIV kXIVYZIVQ ZIVQ
  ZIV Q 0.2 Q 2Q 1.8 x 10-3
• rateI kX IYZIQ Z IQ
  ZI 0.1 4.6 x 10-4
• 2Q 3.91 Q 2
• rate kXZ2

rate kXYZQ
( )
( )
( )
( )
11
Dealing With Exponents

• what happens if you cant exactly double or
  triple concentrations?
• XZ Y
• logXZ logY
• Z logX logY
• Z logY/ logX

12
Unusual Situations

• what happens if varying a concentration results
  in no change in rate?
• reaction is zero order in that component
• sometimes we cant vary the concentrations of all
  of the components of a reaction
• e.g. CH3I(aq) H2O -----gt CH3OH(aq)
  HI(aq)
• follows the rate law rate kCH3IH2O
• but since the reaction is carried out in water,
  cant vary H2O
• express rate law as rate kCH3I
• called a pseudo-first order rate law

13
Rate Laws Reaction Mechanisms

• What does the following mean?
• 2O3(g) -----gt 3O2(g)
• that in order for 3O2 molecules to be produced,
  2O3 must be present does NOT imply that the
  ozones necessarily react together.
• This process actually occurs in two steps
• O3(g) -----gt O2(g) O(g)
• O3(g) O(g) -----gt 2O2(g)
• A reaction mechanism is a description on a
  molecular level of all the changes that reactants
  undergo during a reaction.
• Usually a number of simple steps called
  elementary reactions.
• Since O is not included among the final products,
  called intermediate.

14
Elementary Reactions

• IMPORTANT!
• The overall reaction simply gives stoichiometry.
• Elementary reactions occur as written!
• The rate law can not be predicted from
  stoichiometry.
• Reason? Rate law is derived from elementary
  reactions.
• Can write rate laws for elementary reactions
• O3(g) -----gt O2(g) O(g) rate kO3
• O3(g) O(g) -----gt 2O2(g) rate kO3O
• 2CH3 -----gt C2H6 rate kCH32

15
Rate-Limiting Step

• When one elementary reaction proceeds at a slower
  rate than the others, it is known as the
  Rate-Limiting Step.
• The rate of this step determines the overall rate
  of the complex reaction, no matter at which point
  in the reaction it occurs.
• Obtaining the rate law from the mechanism is
  simplified if you realize that the reaction rate
  is not influenced by steps that occur after the
  rate limiting step.
• H2O2 I- -----gt H2O IO- SLOW
• IO- H -----gt HOI FAST
• HOI H I- -----gt H2O I2 FAST
• I- I2 -----gt I3- FAST
• H2O2 3I- 2H -----gt 2H2O
  I3- overall stoichiometry
• rate limiting step is first one rate k
  H2O2 I-

Can be verified experimentally
16
Pre-Equilibria and Rate Laws

• Given the reaction
• 2NO ltgt NO-ON FAST
• NO-ON NO -----gt N2O NO2 SLOW
• rate limiting step is second elementary reaction
• rate kNO-ONNO
• can we measure NO-ON? NO!
• But, NO-ON
• Keq
  or NO-ON KeqNO2
• NO2
• so, rate kKeqNO2NO kNO3
• both constants

17
Reactant Concentration and Time

• a different form of the rate law can be used to
  calculate the course of a reaction with time.
• e.g. first order reaction follows rate
  kA
• if we assume that the reaction starts at t0 with
  A Co, then the relationship between between
  A and time is
• C -kt
• log ------- -----------
• Co 2.30
• C A at time t
• called integrated rate equation (derived via
  integral calculus)
• note valid only for first order reactions

Know This!
18
Integrated Rate Equation Examples

• Cl2O7(g) -----gt Cl2(g) 7/2O2(g)
  follows first-order kinetics
• a) After 55 sec the pressure of Cl2O7 falls from
  0.062 to 0.044atm. What is k?
• b) What would the pressure of Cl2O7 be after
  100sec?
• C - 6.2 x 10-3
  sec-1(100 sec)
• log ---------- ---------------------
  ------------ C 0.033 atm
• 0.062
  2.30
• c) What time would be required for the pressure
  to fall to one-tenth its original value?
• 0.0062 - 6.2 x 10-3
  sec-1 t
• log ---------- ---------------------
  ------------ t 370 sec
• 0.062
  2.30

19
Half-Lives

• there are many situations where we want to know
  the time required for half of the starting
  quantity to be consumed.
• 0.5Co -kt
  0.693
• log ------- -----------
  or t1/2 -----------------
• Co 2.30
  k

20
Half-Lives

• there are many situations where we want to know
  the time required for half of the starting
  quantity to be consumed.
• 0.5Co -kt
  0.693
• log ------- -----------
  or t1/2 -----------------
• Co 2.30
  k
• e.g. radioactive decay is a first-order process.
  14C has a half life of 5760 yrs. A geiger counter
  measures 14.7 cpm vs. 15.3 cpm for living
  organisms. Is this papyrus genuine?
• 0.693 0.693
• k ------- -----------
  1.20 x 10-4 yr-1
• t1/2 5760yr
• 14.7cpm - 1.2 x 10-4 yr-1
  t
• log ---------- -------------------
  -------- t 333 yr
• 15.3cpm
  2.30

Egypt in 21B.C.
21
Graphical Representations

• integrated first-order equation
• C -kt
• log ------- -----------
• Co 2.30
• -kt
• log C - log Co -----------
• 2.30
• -kt
• log C ----------- log Co
• 2.30
• y mx b
  a plot of logC vs t is a straight
  line!
• integrated second order equation
• 1 1
  1
• ------- kt -------
  a plot of ------ vs t is a straight
  line!
• C Co
  C

22
Graphical Determination of Rate Order

• e.g. decomposition of HI
• t(hr) HI logHI 1/HI
• 0 1.00 0.00 1.00
• 2 0.50 -0.30 2.00
• 4 0.33 -0.48 3.00
• 6 0.25 -0.60 4.00

Not First-Order
Second-Order Reaction!
Not Zero-Order
time (hrs)
1/HI
HI
logHI
time (hrs)
time (hrs)
23
Reaction Rate Limitations

• The following reaction is favored
  thermodynamically. Why doesnt it occur
  instantaneously?
• NO(g) O3(g) -----gt NO2(g) O2(g)
• 1. Reactions only take place when 2 molecules
  collide.
• Takes time for molecules to find one another.
• Rate is therefore limited by rate at which
  collisions occur.
• 2. A reaction will only take place (in this case)
  if the molecules collide in a specific way.
• 
  O
• N---O O O No Reaction
• 
  O
• O---N O O Get Reaction

24
Reaction Rate Limitations (cont.)

• 3. Most important reason why most collisions
  dont result in a chemical change is natural
  electrostatic repulsion.
• To overcome this, the molecules have to approach
  each other with relatively high kinetic energies.
• Plus, even more energy is needed so that the
  electronic changes that lead to the formation of
  products takes place.
• When a collision takes place with enough energy
  to allow formation of products, the system is in
  a transition state.
• The difference in energy between the reactants
  and those molecules in the transition state is
  called the activation energy, Ea.

transition state
Ea
reactants
E
products
25
Reaction Rate and Temperature

• Why would heating the reactants increase the
  rate?
• Kinetic Theory of Gases.

T1
Ea
At T2 a larger fraction of molecules have
sufficient energy
Number of molecules
T2
Kinetic Energy
At T1 a small fraction of molecules have
sufficient energy to react
26
Arrhenius Equation

• Relationship between Ea and k first proposed by
  Svante Arrhenius (1889)
• k ce
• -Ea
• or, log k log C
  (y mx b)
• 2.30 RT
• plot of logk vs T is a straight line with slope
  -Ea/ (2.30 (8.31 J/mole K))
• Two-Point Equation
• Ea
  1 1
• logk2 - logk1
  -
• 2.30 (8.31)
  T1 T2
• good for determining Ea from just two
  temperatures
• knowing Ea can determine k at alternate
  temperatures.

Describes fraction of molecules with minimum
energy required for the reaction
-Ea RT
Frequency factor (depends on collisions with
correct geometry)
( )
27
Catalysts

• Speed up the rate of reaction without being
  consumed.
• Do change during the reaction, but are
  regenerated by the end.
• e.g. 2H2O2 -----gt O2 2H2O
• in the absence of catalyst proceeds very slowly
• add a little I- and reaction occurs readily.
• H2O2 I- -----gt IO- H2O
• H2O2 IO- -----gt O2 H2O I-
• 2H2O2 -----gt O2 2H2O
• a catalyst speeds up the reaction by lowering Ea
• does not change energies of reactants and
  products
• does not affect the position of an equilibrium
• will speed up rate at which equilibrium is
  reached

No catalyst
With catalyst
R
P