Take a sample reaction;
A + 3 B  2 C
rate =
–
D [A]
D t
=
1
3
–
D [B]
D t
=
1
2
D [C]
D t
• Average Rate average rate =
[A] f
– [A] init
D t

For most reactions, the rate depends on concentrations of the reactants. ∴ reaction rate changes with time.
In a
, the reaction rate is equal to the slope of the curve at any given time.
• See Figure 15.5 -- plot of [HI] vs. time for the reaction:
2 HI
(g)
--> H
2(g)
+ I
2(g)
Rate decreases with time (curve flattens out) because the concentration of the reactant decreases with time
•
rate: rate at time = 0 (easiest to measure)

slope = rise run instantaneous rate = slope of rate curve at any one point in time

• Monitor the change in concentration of one or more reactants or products with time as reaction proceeds reaction rate =
D (conc)
D (time)
–units: mole L -1 s -1 = mole
L sec
• products: conc increases with time
• reactants: conc decreases with time
The rates expressed for various reactants and products are related by the stoichiometry of the reaction.

• At a certain temperature, oxygen (O
2
) is produced by the following reaction at a rate of 0.30 mole/L sec. What are the rates of formation of the other products and the rate of disappearance of the reactant?
4 KNO
3(aq)
--> 2 K
2
O
(aq)
+ 2 N
2(g)
+ 5 O
2(g) rate (N
2
) =
0.30 mole O
2
L s x
2 mole N
2
5 mole O
2
=
0.12 mole N
2
L s
Likewise, rate (K
2
O) =
0.12 mole
L sec rate (KNO
3
) =
-0.30 mole O
2
L s x
4 mole KNO
3
5 mole O
2
=
-0.24 mole KNO
3
L s

For a general reaction: a A + b B --> products
• The “Rate Law” is the following type of relationship between rate and concentrations of reactants.
Rate = k[A] x [B] y
{All values of k, x, and y must be determined experimentally!}
• x and y are normally small integers (often 1, 2, or 0) x = “order of the reaction” with respect to A e.g. if x = 2, the rate depends on the square of [A] i.e. double [A], then rate increases by 4 times triple [A], then rate increases by 9 times, etc y = “order of the reaction” with respect to B
• Overall “order” of reaction = x + y

In general, the exponents x and y are
equal to or directly related to the coefficients in the balanced equation.
Rate = k[A] x [B] y where, k = “rate constant”
– (units depend on x and y values)
• Units: [A] and [B] are mole/L
Rate is mole/L s or mole L -1 s -1

Determination of a Rate Law from rate vs. concentration data in a kinetic study of the reaction,
2 NO
(g)
+ O
2(g)
--> 2 NO
The following rate data was obtained.
2(g)
Expt [NO] [O
2
] Rate (mole L -1 s -1 )
1
2
3
0.0010
0.0010
0.0030
0.0010
0.0040
0.0040
7.10
28.4
256 a) Determine the rate law for this reaction.
b) Determine the rate constant, with its proper units.

Expt [NO] [O
2
] Rate (mole L -1 s -1 )
1
2
3
0.0010
0.0010
0.0030
0.0010
0.0040
0.0040
7.10
28.4
256
• Part (a) Rate Law: rate = k[NO] x [O
2
] y
Compare Expts 1 and 2 with constant [NO] shows how rate changes with [O
2
] rate ∝ [O
2
] y or rate
2 rate
1
=
( )
2
]
1 y
28.4/7.10 = {0.0040/0.0010} y
4 = (4) y ∴ y = 1 (1st order in O
2
)

Expt [NO]
[O
2
] Rate (mole L -1 s -1 )
1
2
3
0.0010
0.0010
0.0030
0.0010
0.0040
0.0040
7.10
28.4
256
Compare Expts 2 and 3 with constant [O
2
] shows how rate changes with [NO] rate ∝ [NO] x or rate
3 rate
2
=
( )
2 x
256/28.4 = {0.0030/0.0010} x
9.0 = (3) x ∴ x = 2 (2nd order in NO)
• Overall Rate Law: rate = k[NO] 2 [O
2
]

1
2
3
• Part (b)
Expt [NO] [O
2
] Rate (mole L -1 s -1 )
0.0010
0.0010
0.0030
0.0010
0.0040
0.0040
7.10
28.4
256
Use data from any single expt to calc k
(all expts should give same value for k) e.g. from Expt 1: rate = k[NO] 2 [O
2
]
7.10 mole L -1 s -1 = k(0.0010 mole L -1 ) 2 (0.0010 mole L -1 )
∴ k = 7.1 x 10 9 L 2 mole -2 s -1
{Note: units of k must be determined algebraically}

• The Rate Law for a reaction shows how the reaction rate changes with concentration of reactants. However, what about changes in actual concentrations with time?
• Consider only for a first order reaction:
A --> products rate = k[A] = d[A]/dt ln {[A]
0
/[A] t
} = kt ln [A] t
= -kt + ln [A]
0 integrating this equation: and rearranging: y = mx +b !!
∴ plot of ln [A] vs t is a straight line with slope = -k and intercept = ln [A]
0

• Half Life (t
1/2 to disappear
) = time required for one-half of a given reactant for first order reaction: t
1/2 ln 2 k
0.693
k for second order reaction: t
1/2
=
1 k[A]
0 for zero order reaction: t
1/2
=
[A]
0
2k
More often, we’re interested in how much of a reactant remains after given time e.g. half-life of iodine-131 is 8 days, therefore:
# half-lives 0
# days 0
%I-131 left 100
1
8
50
2
16
25
3
24
4
32
5
40
12.5
6.25 3.13

1. In the reaction, 2A + B ⇌ 3C, if [A] decreases at a rate of
0.090 mol/L s, then [C] increases at a rate of _____ mole/L s.
2. The half-life of a first order reaction is 45 min. After a period of 3 hours, the percent of the original reactant present is ________%.
The rate constant (k) for this same reaction is ______ s -1 .

1. In the reaction, 2A + B ⇌ 3C, if [A] decreases at a rate of
0.090 mol/L s, then [C] increases at a rate of _____ mole/L s.
Answer: 0.14
2. The half-life of a first order reaction is 45 min. After a period of 3 hours, the percent of the original reactant present is ________%.
The rate constant (k) for this same reaction is ______ s -1 .
Answer: 6%, 2.6 x 10 -4

• Rate of a reaction depends on the frequency of
collisions between the reactant molecules.
• To be “effective” collisions must occur with proper orientation and sufficient energy .
Activation Energy (E a
) = minimum energy required for reaction to occur

Activation energy: minimum energy required for reaction to occur

Reaction coordinate diagrams illustrate the energy changes that occur as reactants combine to form products.
• E a
= energy difference between reactants and the
“activated complex” or “ transition state ”

Relationship between temperature and rate constant: k = Ae –Ea/RT (A = “frequency factor”)
More useful in logarithmic form: ln k = ln A -
E a
RT y = mx +b !!
Plot of ln k vs 1/T is a straight line with slope = -E a
/R
( R = 8.314 x 10 -3 kJ/mole K )

ln k = ln A - E a
/RT
If rate constant is measured at two temps: ln k
2 k
1
=
-E a
R
1
T
2
1
T
1
∴ given two k’s and T’s, calculate E a e.g.
Given: k
1 k
2
= 3.20 L mol -1 s -1 at T
1
= 23.0 L mol -1 s -1 at T
2
= 350 ºC (623 K)
= 400 ºC (673 K)
• Find E a
-- plug in k’s, T’s, and R = 8.314 x 10 -3 kJ/mole K
E a
= 1.4 x 10 2 kJ/mole {watch the units!!!}

A “reaction mechanism” is a
sequence of steps, called
converted into products.
* by which the reactants are
Rate-Determining Step : slow step in the reaction mechanism determines the overall rate law
Relationship between mechanism and kinetics:
• For a proposed mechanism, it is possible to predict the rate law that is consistent with the mechanism
• The predicted rate law can then be compared with the
rate law. If the predicted and experimental rate laws are the same, then the proposed mechanism is possible.
* Never more than one or two molecules or ions reacting in a single step!

For an elementary process, the exponents in the rate expression are equal to the coefficients in the balanced equation for that individual process.
e.g. for the overall reaction,
NO
2
+ CO --> NO + CO
2
The following mechanism is proposed.
slow: NO
2 fast: NO
3
+ NO
2
--> NO +
+ CO --> NO
2
NO
+ CO
2
3 net: NO
2
+ CO --> NO + CO
2
(an “intermediate”)
The slow step, predicts: rate = k[NO
2
] 2
The actual, experimental rate law: rate = k[NO
2
] 2
∴ this is a possible mechanism!
The slow step is the
.

What about an alternate, one-step mechanism?
Slow: NO
2
+ CO --> NO + CO
2
• This would predict: rate = k[NO
2
][CO] which is not observed experimentally.
∴ this is not a possible mechanism!
What about an alternate, two-step mechanism?
Fast: NO
2
Slow: NO
3
+ NO
2
--> NO +
+ CO --> NO
2
NO
+ CO
3
2
(rds)
Net: NO
2
+ CO --> NO + CO
2
In this case, the predicted rate law is: rate = k[NO
2
] 2 [CO]
∴ this is not a possible mechanism!

A study of the kinetics of the gas-phase reaction of nickel carbonyl, Ni(CO)
4
, with phosphorus trifluoride, PF
3
, gave the following initial rate vs concentration data.
Ni(CO)
4
+ PF
3
--> Ni(CO)
3
PF
3
+ CO
Expt [Ni(CO)
4
] [PF
3
]
(1) 0.10
0.25
(2)
(3)
(4)
0.20
0.50
0.50
0.50
0.50
0.25
Rate (mole/L s)
9.60 x 10 -4
1.92 x 10 -3
4.80 x 10 -3
4.80 x 10 -3 a) Determine the rate law for this reaction. b) Determine the value for the rate constant (k) for the above reaction.

A study of the kinetics of the gas-phase reaction of nickel carbonyl, Ni(CO)
4
, with phosphorus trifluoride, PF
3
, gave the following initial rate vs concentration data.
Ni(CO)
4
+ PF
3
--> Ni(CO)
3
PF
3
+ CO
Expt [Ni(CO)
4
] [PF
3
]
(1) 0.10
0.25
(2)
(3)
(4)
0.20
0.50
0.50
0.50
0.50
0.25
Rate (mole/L s)
9.60 x 10 -4
1.92 x 10 -3
4.80 x 10 -3
4.80 x 10 -3 a) Determine the rate law for this reaction. b) Determine the value for the rate constant (k) for the above reaction.
Answer: rate = k[Ni(CO)
4
], 9.6 x 10 -3 s -1

continued…
Ni(CO)
4
+ PF
3
--> Ni(CO)
3
PF
3
+ CO
Expt [Ni(CO)
4
] [PF
3
]
(1) 0.10
0.25
(2)
(3)
(4)
0.20
0.50
0.50
0.50
0.50
0.25
Rate (mole/L s)
9.60 x 10 -4
1.92 x 10 -3
4.80 x 10 -3
4.80 x 10 -3 c) Given the experimental rate law, as determined above, is it possible for this reaction to occur by a simple one-step mechanism? Why or why not?
d) If the answer to c) is No, then propose a mechanism for this process that is consistent with the rate law.

continued…
Ni(CO)
4
+ PF
3
--> Ni(CO)
3
PF
3
+ CO
Expt [Ni(CO)
4
] [PF
3
]
(1) 0.10
0.25
(2)
(3)
(4)
0.20
0.50
0.50
0.50
0.50
0.25
Rate (mole/L s)
9.60 x 10 -4
1.92 x 10 -3
4.80 x 10 -3
4.80 x 10 -3 c) Given the experimental rate law, as determined above, is it possible for this reaction to occur by a simple one-step mechanism? Why or why not?
d) If the answer to c) is No, then propose a mechanism for this process that is consistent with the rate law.
Answer: c) No, because both reactants would be involved, but the rate law shows they’re not. d) e.g. CO dissociation mechanism.

• Nature of the Reactants:
– Molecular structure, bond polarity, physical state, etc.
• Ability of reactants to come in contact
–
– hetero geneous reaction: reactants in different phases, e.g. solid/gas or solid/liquid homo geneous reaction: all reactants in same phase, e.g. reaction in solution
• Concentrations of the reactants
• Temperature of the system
• Presence of a
– (substance that increases the reaction rate without being consumed itself)

A
is a substance that increases the rate of a reaction without itself being consumed.
– Two types: homo geneous or hetero geneous
Mode of action:
• Catalyst alters the reaction mechanism by providing a lower energy (smaller E a
) pathway for the reaction.
• Catalyst often activate (weaken) one or more bonds in the reactants and/or help to properly orient the reacting molecules (especially on surface of
geneous catalysts), thus promoting their reaction.

At O °C, a study of the kinetics of the solution-phase reaction of bromosilane, H
3
SiBr, with fluoride ion gave the following initial rate vs concentration data.
H
3
SiBr + F – --> H
3
SiF + Br –
Expt [H
3
SiBr] [F – ] Rate (mole/L s)
(1) 0.15
0.15
1.75 x 10 -3
(2)
(3)
(4)
0.90
0.90
0.75
0.15
1.20
0.75
1.05 x 10 -2
8.40 x 10 -2
4.38 x 10 -2 a) Determine the rate law for the above reaction. Clearly show how you arrive at your answer.
b) Determine the value for the rate constant (k) for this reaction. SHOW ALL WORK, including proper units.

4. continued…
H
3
SiBr + F – --> H
3
SiF + Br –
Expt [H
3
SiBr] [F – ]
(1) 0.15
0.15
(2)
(3)
(4)
0.90
0.90
0.75
0.15
1.20
0.75
Rate (mole/L s)
1.75 x 10 -3
1.05 x 10 -2
8.40 x 10 -2
4.38 x 10 -2 c) Propose a simple 2-step mechanism for this reaction that is consistent with
experimental rate law in part (a).
Clearly indicate the rate-determining step in your mechanism. d) When experiment number (4) above was repeated at 25 ºC, using the same concentrations, the reaction rate was 0.125 mole/L s. Calculate the energy of activation (E a reaction.
) for this

a) rate = k[H
3
SiBr] x [F – ] y
Use expts 1&2 to determine x: 90/15 = 6, and
(1.05 x 10 –2 )/(1.75 x 10 –3 ) = 6, so x = 1
Use expts 2&3 to determine y: 1.20/0.15 = 8, and
(8.40 x 10 –2 )/(1.05 x 10 –2 ) = 8, so y = 1
Rate = k[H
3
SiBr][F – ] b) (1.75 x 10 –3 M/s) = k(0.15 M)(0.15 M)
Numerically, k = 7.8 x 10 –2
Units? (mol 2 /L 2 )k = (mol/L s)
So k = (L/mol s) c) One possible answer is:
Slow: H
3
SiBr + F – --> H
3
SiBrF –
Fast: H
3
SiBrF – --> H
3
SiF + Br – d) E a
= 28 kJ/mol