1. Chemical Kinetics The Second Order Integrated Rate Equation

2. Second Order Rate Law
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3. This is the second order differential rate law.
Second Order (m+n=2)
For a general reaction
that obeys second order rate law.
a X + b Y  products
There are a number of ways for m+n=2. The most simple is
This is the second order differential rate law.
The rate of a second order reaction is proportional to the concentration of the reactant squared. Increasing the concentration of the reacting species will increase the rate of the reaction. (i.e. Doubling concentration increases the rate 4x.)
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4. This is the second order differential rate law.
Second Order (m+n=2)
For a general reaction
that obeys second order rate law.
a X + b Y  products
There are a number of ways for m+n=2. The most common is
This is the second order differential rate law.
When this equation is rearranged and both sides are integrated we get the following result.
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5. that obeys second order rate law.
For a general reaction
that obeys second order rate law.
a X + b Y  products
This is the second order differential rate law.
Separate the variables
Integrate
This is the second order integrated rate law.

6. This is the second order integrated rate law.
For a general reaction
that obeys second order rate law.
a X + b Y  products
This is the second order integrated rate law.
[X]0 is the concentration of X at time=0
[X]t is the concentration of X at time=t
A reaction is second order if the concentration of X decreases inversely with time.
A plot of is a straight line.
The slope of this resulting line is ak2
The unit of k2 is M-1s-1
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7. that obeys second order rate law.
For a general reaction
that obeys second order rate law.
a X + b Y  products
Integrated rate law
Differential rate law
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8. that obeys second order rate law.
For a general reaction
that obeys second order rate law.
a X + b Y  products
Integrated rate law
Differential rate law
A graph of “[X]-1 versus time” would yield a straight line.
Slope = ak2
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9. The more concentrated the reactant is, the shorter the half-life.
Second Order
For a general reaction
that obeys second order rate law.
a X + b Y  products
Integrated rate law
Half-life, 
When t= ,
Half-life of a second-order reaction depends on the inverse of the initial concentration of the reactant.
The more concentrated the reactant is, the shorter the half-life.
To test a set of data, could plot “ versus [X]o-1” to see if it yields a straight line with slope = (ak2)-1.

10. that obeys second order rate law.
Second Order (m+n=2)
For a general reaction
that obeys second order rate law.
a X + b Y  products
Overall second order;
First order with respect to X;
First order with respect to Y.
The integrated form is complicated.
In Experiment 3,
CH3COOC2H5 (aq) + OH- (aq)  CH3COO- (aq) + C2H5OH (aq)
Since the reaction is 1:1, we adjust the initial concentration of to be the same as that of CH3COOC2H5.
Rate = k [CH3COOC2H5] [OH-]
Then, the rate equation reduces to
Rate = k [CH3COOC2H5]2 or
Rate = k [OH-] 2
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11. Show that this reaction is 2nd order and not 1st order.
Second Order
Example: The decomposition of nitrogen dioxide
2 NO2 (g)  2 NO (g) + O2 (g)
Data collected in this kinetic study:
[NO2] (M)
Time (s)
1.00 x 10-2
8.00 x 10-3 50
6.47 x 10-3
100
4.76 x 10-3
200
3.74 x 10-3
300
Show that this reaction is 2nd order and not 1st order.
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12. Second Order Example: The decomposition of nitrogen dioxide
2 NO2 (g)  2 NO (g) + O2 (g)
Raw data:
What would you plot to test what order this reaction is?
Is it first-order?
Is it second-order?
[NO2] (M)
Time (s)
1.00 x 10-2
8.00 x 10-3 50
6.47 x 10-3
100
4.76 x 10-3
200
3.74 x 10-3
300
Graph “ln [NO2] versus time”
to see if it yields a straight line.
Graph “1 / [NO2] versus time”
to see if it yields a straight line.
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13. Second Order Example: The decomposition of nitrogen dioxide
2 NO2 (g)  2 NO (g) + O2 (g)
Original data
ln [NO2]
Time (s)
-4.605
-4.828 50
-5.041
100
-5.348
200
-5.589
300
To test for 1st order:
Take the natural log of [NO2] and plot it against time.
[NO2] (M)
Time (s)
1.00 x 10-2
8.00 x 10-3 50
6.47 x 10-3
100
4.76 x 10-3
200
3.74 x 10-3
300
To test for 2nd order:
Take the inverse of [NO2] and plot it against time.
1/[NO2] (M-1)
Time (s)
100.
125. 50
155.
100
210.
200
267.
300
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14. Second Order Example: The decomposition of nitrogen dioxide
2 NO2 (g)  2 NO (g) + O2 (g)
The “ln [NO2] versus Time” graph for this reaction does not yield a straight line, proving that it is not a first order reaction.
The “1 / [NO2] versus Time” graph for this reaction yields a straight line, proving that
it is a second order reaction.
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15. Concentration of HI (M)
Example:
We study the decomposition of hydrogen iodide at 508oC, by
Measuring the rate at which either H2 or I2 is formed, or the rate
at which HI is consumed.
2 HI(g)  H2(g) + I2(g)
Experimental Data:
Time (s)
Concentration of HI (M)
0.100 50
0.0716
100
0.0558
150
0.0457
200
0.0387
250
0.0336
300
0.0296
350
0.0265
Test the data to determine the order of the reaction.
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17. 2 HI(g)  H2(g) + I2(g)
Plot “[HI] vs. Time” to test for zeroth order
R² =
Plot “[HI] vs. Time” does not yield a straight line.
The decomposition of HI is not a zeroth order reaction.
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18. 2 HI(g)  H2(g) + I2(g)
Plot “ln [HI] vs. Time” to test for first order
R² = 0.965
Plot “ln [HI] vs. Time” does not yield a straight line.
The decomposition of HI is not a first order reaction.
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19. 2 HI(g)  H2(g) + I2(g)
Plot “1/[HI] vs. Time” to test for second order
Plot “1/[HI] vs. Time” yields a straight line.
The decomposition of HI is a second order reaction.
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20. Therefore, rate of disappearance of HI is
We study the decomposition of hydrogen iodide at 508oC, by
measuring the rate at which either H2 or I2 is formed, or the rate
at which HI is consumed.
2 HI(g)  H2(g) + I2(g)
Experimentally we find that this is a 2nd order reaction.
Question:
Calculate the rate at which HI disappears in the following
reaction at the moment when I2 is being formed at a rate of
1.8 x 10-6 M sec-1.
Answer:
Therefore, rate of disappearance of HI is
2(1.8 x 10-6) = 3.6 x 10-6 Msec-1
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21. Summary of the Rate Laws
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22. The Rate Law tells us the instantaneous rate (the slope of the curve) as a function of concentration.
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23. For a general reaction, a X + b Y  products Zeroth Order First Order
Second Order
Differential
Rate Law
Integrated Rate Law
Units of k
M s-1
s-1
M-1 s-1
Linear Plot
[X] vs. t
ln([X]) vs. t
1/[X] vs. t
Half-life
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24. The Integrated Rate Law tells us how the concentration behaves as a function of time.
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