Ch 12 Rates and Rate Laws

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Ch 12 RatesandRate Laws
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Reaction Rate
The change in concentration of a reactant or
product per unit of time
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2NO2(g) ? 2NO(g) O2(g)
Reaction Rates
1. Can measure disappearance of reactants
2. Can measure appearance of products
3. Are proportional stoichiometrically
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2NO2(g) ? 2NO(g) O2(g)
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
?t
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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
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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/Ls
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
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Writing a Rate Law
Part 1 Determine the values for the exponents
in the rate law
R kNOxCl2y
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
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 kNO2Cl2y
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Writing a Rate Law
Part 1 Determine the values for the exponents
in the rate law
R kNO2Cl2y
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
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 kNO2Cl2
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Writing a Rate Law
Part 2 Determine the value for k, the rate
constant, by using any set of experimental data
R kNO2Cl2
Experiment NO (mol/L) Cl2 (mol/L) Rate Mol/Ls
1 0.250 0.250 1.43 x 10-6
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Writing a Rate Law
Part 3 Determine the overall order for the
reaction.
R kNO2Cl2
3
2

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? The reaction is 3rd order
Overall order is the sum of the exponents, or
orders, of the reactants
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Determining Order withConcentration vs. Time data
(the Integrated Rate Law)
Zero Order
First Order
Second Order
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Rate Ex. Warm Up

• Rate kM
• Rate kMN
• Rate kN
• Rate kM2N2
• Rate kMN2

• What is the order of the reaction for each?

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Rate Ex. Warm Up Matching

• Rate kM
• Rate kMN
• Rate kN
• Rate kM2N2
• Rate kMN2

1. Doubling the concentration of M has no effect
2. Doubling the concentration of M and N increases
   reaction rate by 2
3. Doubling the concentration of M only quadruples
   the reaction rate

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Rate Ex. Warm Up Matching

• Rate kM
• Rate kMN
• Rate kN2
• Rate kM2N2
• Rate kMN2

1. Doubling the concentration of M has no effect
2. Doubling the concentration of M and N increases
   reaction rate by 2
3. Doubling the concentration of M only quadruples
   the reaction rate

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Example (p. 536)
NH4(aq) NO2-(aq) ? N2(g) 2H2O(l)
Experiment Initial NH4 Initial NO2- Initial Rate (mol/L.s)
1 0.100 M 0.0050 M 1.35 x 10-7
2 0.100 M 0.010 M 2.70 x 10-7
3 0.200 M 0.010 M 5.40 x 10-7
What is the rate law, overall order of reaction
value for k?
Rate k NH4 NO2-
Order 11 2nd order
k 2.7 x 10-4 L/mol.s
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Solving an Integrated Rate Law
Rate as a function of time
Problem Find the integrated rate law and the
value for the rate constant, k
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
A graphing calculator with linear regression
analysis greatly simplifies this process!!
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Time vs. H2O2
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
Regression results
y ax b a -2.64 x 10-4 b 0.841 r2
0.8891 r -0.9429
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Time vs. lnH2O2
Time (s) lnH2O2
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
Regression results
y ax b a -8.35 x 10-4 b -.005 r2
0.99978 r -0.9999
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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
2400 7.6923
3000 12.195
3600 20.000
Regression results
y ax b a 0.00460 b -0.847 r2 0.8723 r
0.9340
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And the winner is Time vs. lnH2O2
1. As a result, the reaction is 1st order
2. The (differential) rate law is
3. The integrated rate law is
4. Butwhat is the rate constant, k ?
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Finding the Rate Constant, k
Method 1 Calculate the slope from the Time vs.
lnH2O2 table.
Time (s) lnH2O2
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
Now remember
? k -slope
k 8.32 x 10-4s-1
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Finding the Rate Constant, k
Method 2 Obtain k from the linear regresssion
analysis.
Regression results
y ax b a -8.35 x 10-4 b -.005 r2
0.99978 r -0.9999
Now remember
? k -slope
k 8.35 x 10-4s-1
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Rate Laws Summary
Zero Order First Order Second Order
Rate Law Rate k Rate kA Rate kA2
Integrated Rate Law A -kt A0 lnA -kt lnA0
Plot the produces a straight line A versus t lnA versus t
Relationship of rate constant to slope of straight line Slope -k Slope -k Slope k
Half-Life
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Homework

• Graphical Methods for determining reaction order
  and rate constant

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Answers