# Kinetics

Document Sample

```					        Chemical Kinetics
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

[ A] at timet2  [ A] at timet1
Rate 
t2  t1
[ A]
Rate 
t
Reaction Rates:
2NO2(g)  2NO(g) + O2(g)

1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products

3. Are proportional
stoichiometrically
Reaction Rates:
2NO2(g)  2NO(g) + O2(g)
4. Are equal to the
slope tangent to
that point

5. Change as the
[NO2]
reaction proceeds,
t                     if the rate is
dependent upon
concentration
[ NO2 ]
 constant
t
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) + Cl2(g)  2 NOCl(g)
Experiment    [NO]      [Cl2]      Rate
(mol/L)   (mol/L)    Mol/L·s
1            0.250     0.250     1.43 x 10-6
2            0.500     0.250     5.72 x 10-6
3            0.250     0.500     2.86 x 10-6
4            0.500     0.500     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[Cl2]1
2    y

Experiment      [NO]        [Cl2]           Rate
(mol/L)     (mol/L)         Mol/L·s
1               0.250       0.250        1.43 x 10-6
2               0.500       0.250        5.72 x 10-6
3               0.250       0.500        2.86 x 10-6
4               0.500       0.500        1.14 x 10-5

In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with
respect to [NO]
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with
respect to [Cl2]
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[Cl2]
Experiment          [NO]         [Cl2]            Rate
(mol/L)      (mol/L)          Mol/L·s
1                  0.250        0.250          1.43 x 10-6

2
mol
6            mol         mol 
1.43 x10      k  0.250       0.250     
Ls             L           L 

 1.43 x106   mol   L3                    L2
k                             9.15 x105
   0.2503   L  s   mol 3               mol 2  s
Writing a Rate Law
Part 3 – Determine the overall order for the
reaction.
R = k[NO]2[Cl2]

2 + 1 = 3
 The reaction is 3rd 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:    time vs. concentrationis linear

First Order:   timevs.ln(concentration) is linear
1
Second Order:   time vs.               is linear
concentration
Solving an Integrated Rate Law
Time (s)   [H2O2] (mol/L)
Problem: Find the
0          1.00         integrated rate law
120         0.91
and the value for the
300         0.78
rate constant, k
600         0.59
1200         0.37
A graphing calculator
1800         0.22
with linear regression
2400         0.13
analysis greatly
3000         0.082
simplifies this process!!
3600         0.050

theTi-83 and Ti-84)
Time vs. [H2O2]
Time (s)   [H2O2]
0       1.00
120      0.91
300      0.78
600      0.59
1200      0.37

Regression results:       1800      0.22
2400      0.13
y = ax + b               3000      0.082
a = -2.64 x 10-4         3600      0.050
b = 0.841
r2 = 0.8891
r = -0.9429
Time vs. ln[H2O2]
Time (s)   ln[H2O2]

0        0
120      -0.0943
300      -0.2485
600      -0.5276
1200      -0.9943
Regression results:     1800      -1.514
y = ax + b             2400       -2.04

a = -8.35 x 10-4       3000      -2.501

b = -.005              3600      -2.996

r2 = 0.99978
r = -0.9999
Time vs. 1/[H2O2]
Time (s)   1/[H2O2]
0      1.00
120    1.0989
300    1.2821
600    1.6949
1200   2.7027
1800   4.5455
Regression results:       2400   7.6923
y = ax + b              3000   12.195
a = 0.00460             3600   20.000
b = -0.847
r2 = 0.8723
r = 0.9340
And the winner is… Time vs. ln[H2O2]

1. As a result, the reaction is 1st order
2. The (differential) rate law is:

R  k[ H 2O2 ]
3. The integrated rate law is:
ln[ H 2O2 ]   kt  ln[ H 2O2 ]0
4. But…what is the rate constant, k ?
Finding the Rate Constant, k
Method #1: Calculate the slope from the
Time vs. ln[H2O2] table.
Time (s)   ln[H2O2]
 ln[ H 2O2 ] 2.996            0          0
slope               
t        3600 s          120      -0.0943

4 1              300      -0.2485
slope   8.32 x10 s                   600      -0.5276
1200      -0.9943
Now remember:                        1800      -1.514

ln[ H 2O2 ]   kt  ln[ H 2O2 ]0    2400       -2.04
3000      -2.501
 k = -slope                    3600      -2.996

k = 8.32 x 10-4s-1
Finding the Rate Constant, k
Method #2: Obtain k from the linear
regresssion analysis.
Regression results:
4 1
slope  a   8.35 x10 s            y = ax + b
a = -8.35 x 10-4
Now remember:                     b = -.005
ln[ H 2O2 ]   kt  ln[ H 2O2 ]0   r2 = 0.99978
r = -0.9999
 k = -slope

k = 8.35 x 10-4s-1
Rate Laws Summary
Zero Order           First Order         Second Order

Rate Law             Rate = k           Rate = k[A]          Rate = k[A]2
Integrated
1           1
Rate Law
[A] = -kt + [A]0   ln[A] = -kt + ln[A]0         kt 
[ A]        [ A]0
1
Plot the            [A] versus t        ln[A] versus t            versus t
produces a                                                   [ A]
straight line
Relationship of
rate constant
Slope = -k           Slope = -k            Slope = k
to slope of
straight line

Half-Life                  [ A]0               0.693                  1
t1/ 2              t1/ 2               t1/ 2 
2k                   k                  k[ A]0
Reaction Mechanism

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
Rate-Determining Step
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:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the
rate-determining (slowest) step?

Step #1   H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2    N2O(g) + H2(g)  N2(g) + H2O(g)

Step #1 agrees with the experimental rate law
Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)

Which species in the reaction mechanism are
intermediates (do not show up in the final,
balanced equation?)

Step #1    H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2    N2O(g) + H2(g)  N2(g) + H2O(g)
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)

 N2O(g) is an intermediate
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
Collision Model
Collisions must have sufficient energy to
1. produce the reaction (must equal or
exceed the activation energy).

Colliding particles must be correctly
2.
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
The Arrhenius Equation

 Ea / RT
k  Ae
 k = rate constant at temperature T
 A = frequency factor
 Ea = activation energy
 R = Gas constant, 8.31451 J/K·mol
The Arrhenius Equation, Rearranged
Ea  1 
ln(k )      ln( A)
R T 
 Simplifies solving for Ea
 -Ea / 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
 Ea = -R(slope)
Catalysis
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

Carbon monoxide and
nitrogen monoxide
activation of the
reactants.
Heterogeneous Catalysis

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

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

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

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 23 posted: 6/29/2011 language: English pages: 36
How are you planning on using Docstoc?