The area of chemistry that concerns reaction rates and reaction mechanisms .

Reaction Rate
The change in concentration of a reactant or product per unit of time
Rate

[ ]
2

[ ]
1 t
 t
2 1
Rate


[ ]
 t

2NO
2
(g)  2NO(g) + O
2
(g)
Reaction Rates:
1. Can measure disappearance of reactants
2. Can measure appearance of products
3. Are proportional stoichiometrically

2NO
2
(g)  2NO(g) + O
2
(g)
Reaction Rates:
4. Are equal to the slope tangent to that point

[NO
2
]
 t
5. Change as the reaction proceeds, if the rate is dependent upon concentration

[ NO
2
]
 t
 constant

Rate Laws
Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
The differential rate law is usually just called “the rate law.”
Integrated rate laws express (reveal) the relationship between concentration of reactants and time

Writing a (differential) Rate Law
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:
2 NO(g) + Cl
2
(g)  2 NOCl(g)
Experiment
3
4
1
2
[NO]
(mol/L)
0.250
0.500
0.250
0.500
[Cl
2
]
(mol/L)
0.250
0.250
0.500
0.500
Rate
Mol/L·s
1.43 x 10 -6
5.72 x 10 -6
2.86 x 10 -6
11.4 x 10 -6

Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law: R = k[NO] x [Cl
2
] y
Experiment
3
4
1
2
[NO]
(mol/L)
0.250
0.500
0.250
0.500
[Cl
2
]
(mol/L)
0.250
0.250
0.500
0.500
Rate
Mol/L·s
1.43 x 10 -6
5.72 x 10 -6
2.86 x 10 -6
1.14 x 10 -5
In experiment 1 and 2, [Cl while [NO] doubles.
2
] is constant
The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO] 2 [Cl
2
] y

Writing a Rate Law
Part 1 – Determine the values for the exponents in the rate law: R = k[NO] 2 [Cl
2
] y
Experiment
3
4
1
2
[NO]
(mol/L)
0.250
0.500
0.250
0.500
[Cl
2
]
(mol/L)
0.250
0.250
0.500
0.500
Rate
Mol/L·s
1.43 x 10 -6
5.72 x 10 -6
2.86 x 10 -6
1.14 x 10 -5
In experiment 2 and 4, [NO] is constant while [Cl
2
] doubles. The rate doubles, so the reaction is first order with respect to [Cl
2
]  R = k[NO] 2 [Cl
2
]

Writing a Rate Law
Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:
R = k[NO] 2 [Cl
2
]
Experiment
1
[NO]
(mol/L)
0.250
[Cl
2
]
(mol/L)
0.250
Rate
Mol/L·s
1.43 x 10 -6

6 mol
 


0.250
mol
L
 
0.250
mol
L
 k


 0.250
3

6



 mol





L
3 mol
3


 
5
L
2 mol
2  s

Writing a Rate Law
Part 3 – Determine the overall order for the reaction.
R = k[NO] 2 [Cl
2
]
2 + 1 = 3
 The reaction is 3 rd order
Overall order is the sum of the exponents, or orders, of the reactants

Determining Order with
Concentration vs. Time data
(the Integrated Rate Law)
Zero Order:
First Order:
.ln( )
Second Order:
1
.

Time (s)
0
120
300
600
1200
1800
2400
3000
3600
Solving an Integrated Rate Law
[H
2
O
2
] (mol/L)
1.00
0.91
0.78
0.59
0.37
0.22
0.13
0.082
0.050
Problem : Find the integrated rate law and the value for the rate constant, k
A graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. [H
2
O
2
]
Regression results: y = ax + b a = -2.64 x 10 -4 b = 0.841
r 2 = 0.8891
r = -0.9429
Time (s) [H
2
O
2
]
0 1.00
120
300
0.91
0.78
600
1200
1800
2400
3000
3600
0.59
0.37
0.22
0.13
0.082
0.050

Time vs. ln[H
2
O
2
]
Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005
r 2 = 0.99978
r = -0.9999
Time (s)
0
120
300
600
1200
1800
2400
3000
3600 ln[H
2
O
2
]
0
-0.0943
-0.2485
-0.5276
-0.9943
-1.514
-2.04
-2.501
-2.996

Time vs. 1/[H
2
O
2
]
Regression results: y = ax + b a = 0.00460
b = -0.847
r 2 = 0.8723
r = 0.9340
Time (s) 1/[H
2
O
2
]
0 1.00
120
300
1.0989
1.2821
600
1200
1800
2400
3000
3600
1.6949
2.7027
4.5455
7.6923
12.195
20.000

And the winner is… Time vs. ln[H
2
O
2
]
1. As a result, the reaction is 1 st order
2. The (differential) rate law is:
R k H O ]
2 2
3. The integrated rate law is: ln[ H O
2 2
] kt ln[ H O ]
2 2 0
4. But…what is the rate constant, k ?

Finding the Rate Constant, k
Method #1: Calculate the slope from the
Time vs. ln[H
2
O
2
] table.
Time (s) ln[H
2
O
2
] slope slope


 ln[ H O
 t
2 2
8.32 10
]


2.996
s
3600

4

1 s
0
120
300
600
0
-0.0943
-0.2485
-0.5276
Now remember: ln[ H O
2 2
]
  ln[ H O ]
2 2 0
 k = -slope
1200
1800
2400
3000
3600
-0.9943
-1.514
-2.04
-2.501
-2.996
k = 8.32 x 10 -4 s -1

Finding the Rate Constant, k
Method #2: Obtain k from the linear regresssion analysis.
Regression results:

4

1
8.35 10 s ln[
Now remember:
H O
2 2
]
  ln[ H O
 k = -slope
]
2 2 0 y = ax + b a = -8.35 x 10 -4 b = -.005
r 2 = 0.99978
r = -0.9999
k = 8.35 x 10 -4 s -1

Rate Law
Integrated
Rate Law
Plot that produces a straight line
Relationship of rate constant to slope of straight line
Half-Life
Rate Laws Summary
Zero Order
Rate = k
[A] = -kt + [A]
0
First Order
Rate = k[A] ln[A] = -kt + ln[A]
0
Second Order
Rate = k[A] 2
1 1
A
0
[A] versus t ln[A] versus t
1 versus t
Slope = -k t
1/ 2

A
0
2 k
Slope = -k t
1/ 2

0.693
k
Slope = k t
1/ 2

1 k A
0

The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
 The sum of the elementary steps must give the overall balanced equation for the reaction
 The mechanism must agree with the experimentally determined rate law

In a multi-step reaction, the slowest step is the rate-determining step.
It therefore determines the rate of the reaction.
The experimental rate law must agree with the rate-determining step

Identifying the Rate-Determining Step
For the reaction:
2H
2
(g) + 2NO(g)  N
2
(g) + 2H
2
O(g)
The experimental rate law is:
R = k[NO] 2 [H
2
]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1 H
2
(g) + 2NO(g)  N
2
O(g) + H
2
O(g)
Step #2 N
2
O(g) + H
2
(g)  N
2
(g) + H
2
O(g)
Step #1 agrees with the experimental rate law

Identifying Intermediates
For the reaction:
2H
2
(g) + 2NO(g)  N
2
(g) + 2H
2
O(g)
Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)
Step #1 H
2
(g) + 2NO(g)  N
2
O(g) + H
2
O(g)
Step #2 N
2
O(g) + H
2
(g)  N
2
(g) + H
2
O(g)
2H
2
(g) + 2NO(g)  N
2
(g) + 2H
2
O(g)
 N
2
O(g) is an intermediate

Key Idea : Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?

1.
Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy).
2.
Colliding particles must be correctly oriented to one another in order to produce a reaction.

Factors Affecting Rate
Increasing temperature always increases the rate of a reaction.
 Particles collide more frequently
 Particles collide more energetically
Increasing surface area increases the rate of a reaction
Increasing Concentration USUALLY increases the rate of a reaction
Presence of Catalysts , which lower the activation energy by providing alternate pathways

Endothermic Reactions

Exothermic Reactions

k

Ae

E a
/ RT
 k
= rate constant at temperature T

A
= frequency factor

E a
= activation energy

R
= Gas constant, 8.314 J/K·mol

The Arrhenius Equation, Rearranged
 
E
R a
1
 

 Simplifies solving for
E a

-E a
/ R is the slope when (
1/T
) is plotted against ln(k)
 ln(A) is the y-intercept
 Linear regression analysis of a table of
(1/T) vs. ln(k) can quickly yield a slope

E a
= -R(slope)

•Catalyst : A substance that speeds up a reaction without being consumed
•Enzyme : A large molecule (usually a protein) that catalyzes biological reactions.
•Homogeneous catalyst : Present in the same phase as the reacting molecules.
•Heterogeneous catalyst : Present in a different phase than the reacting molecules
.

Lowering of Activation Energy by a Catalyst

Catalysts Increase the Number of
Effective Collisions

Heterogeneous Catalysis
Step #1:
Adsorption and activation of the reactants.
Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface

Heterogeneous Catalysis
Step #2:
Migration of the adsorbed reactants on the surface.
Carbon monoxide and nitrogen monoxide arranged prior to reacting

Heterogeneous Catalysis
Step #3:
Reaction of the adsorbed substances.
Carbon dioxide and nitrogen form from previous molecules

Heterogeneous Catalysis
Step #4:
Escape, or desorption, of the products .
Carbon dioxide and nitrogen gases escape
(desorb) from the platinum surface