Transcript Rates and Rate Laws Reaction Rate The change in concentration of a reactant or product per unit of time [ A] at tim et2  [

Rates
and
Rate Laws
Reaction Rate
The change in concentration of a reactant or
product per unit of time
[ A] at tim et2  [ A] at tim et1
Rate
t2  t1
[ A]
Rate 
t
2NO2(g)  2NO(g) + O2(g)
Reaction Rates:
1. Can measure
disappearance of
reactants
2. Can measure
appearance of
products
3. Are proportional
stoichiometrically
2NO2(g)  2NO(g) + O2(g)
[NO2]
t
Reaction Rates:
4. Are equal to the
slope tangent to
that point
5. Change as the
reaction proceeds,
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]
(mol/L)
[Cl2]
(mol/L)
Rate
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]y
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
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]
 R = k[NO]2[Cl2]y
Writing a Rate Law
Part 1 – Determine the values for the exponents
in the rate law: R = k[NO]2[Cl2]y
Experiment
[NO]
(mol/L)
[Cl2]
(mol/L)
Rate
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 2 and 4, [NO] is constant
while [Cl2] doubles. The rate doubles, so
the reaction is first order with respect
to [Cl2]
 R = k[NO]2[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]
(mol/L)
[Cl2]
(mol/L)
Rate
Mol/L·s
1
0.250
0.250
1.43 x 10-6
2
mol
mol  
mol 

1.43 x10
 k  0.250
  0.250

Ls
L
L

 

6
2
 1.43 x106   mol   L3 
L
5
k 

9.15
x
10


3
3 
2
0.250
L

s
mol
mol
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: timevs. concentrationis linear
First Order: timevs. ln(concentration) is linear
1
is linear
Second Order: time vs .
concentrat ion
Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
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!!
(Click here to download my Rate Laws program
for theTi-83 and Ti-84)
Time vs. [H2O2]
Regression results:
y = ax + b
a = -2.64 x 10-4
b = 0.841
r2 = 0.8891
r = -0.9429
Time (s)
[H2O2]
0
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Time vs. ln[H2O2]
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
Regression results:
1200
-0.9943
y = ax + b
a = -8.35 x 10-4
b = -.005
r2 = 0.99978
r = -0.9999
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Time vs. 1/[H2O2]
Regression results:
y = ax + b
a = 0.00460
b = -0.847
r2 = 0.8723
r = 0.9340
Time (s)
1/[H2O2]
0
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
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 2 02 ]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.
 ln[H 2O2 ]  2.996
slope

t
3600s
4
slope  8.32 x10 s
1
Now remember:
ln[H 2O2 ]   kt  ln[H 2 02 ]0
 k = -slope
k = 8.32 x 10-4s-1
Time (s)
ln[H2O2]
0
0
120
-0.0943
300
-0.2485
600
-0.5276
1200
-0.9943
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Finding the Rate Constant, k
Method #2: Obtain k from the linear
regresssion analysis.
Regression results:
slope  8.32 x104 s 1
Now remember:
ln[H 2O2 ]   kt  ln[H 2 02 ]0
 k = -slope
k = 8.35 x 10-4s-1
y = ax + b
a = -8.35 x 10-4
b = -.005
r2 = 0.99978
r = -0.9999
Rate Laws Summary
Rate Law
Integrated
Rate Law
Plot the
produces a
straight line
Relationship of
rate constant
to slope of
straight line
Half-Life
Zero Order
First Order
Second Order
Rate = k
Rate = k[A]
Rate = k[A]2
[A] = -kt + [A]0
ln[A] = -kt + ln[A]0
1
1
 kt 
[ A]
[ A]0
[A] versus t
ln[A] versus t
1
versus t
[ A]
Slope = -k
Slope = -k
Slope = k
[ A]0
t1/ 2 
2k
0.693
t1/ 2 
k
1
t1/ 2 
k[ A]0