2. First-Order Rate = k[A] Integrated: ln[A] = –kt + ln[A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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3. Plot of ln[N2O5] vs Time
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4. k = rate constant First-Order
Time required for a reactant to reach half its original concentration
Half–Life:
k = rate constant
Half–life does not depend on the concentration of reactants.
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5. A first order reaction is 35% complete at the end of 55 minutes
A first order reaction is 35% complete at the end of 55 minutes. What is the value of k?
ln(0.65) = –k(55) + ln(1)
k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1.
Note: Use the red box animation to assist in explaining how to solve the problem.
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6. Second-Order Rate = k[A]2 Integrated:
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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7. Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time

8. Second-Order Half–Life: [A]o = initial concentration of A
k = rate constant
[A]o = initial concentration of A
Half–life gets longer as the reaction progresses and the concentration of reactants decrease.
Each successive half–life is double the preceding one.
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9. EXERCISE!
For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively.
Write the rate law for this reaction.
b) Calculate k.
c) Calculate [A] at t = 525 minutes.
a) rate = k[A]2
We know this is second order because the second half–life is double the preceding one.
b) k = 8.0 x 10-3 M–1min–1
25 min = 1 / k(5.0 M)
c) [A] = 0.23 M
(1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M)
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10. Zero-Order Rate = k[A]0 = k Integrated: [A] = –kt + [A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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11. Plot of [A] vs Time
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12. Zero-Order Half–Life: k = rate constant
[A]o = initial concentration of A
Half–life gets shorter as the reaction progresses and the concentration of reactants decrease.
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13. CONCEPT CHECK!
How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs?
For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent).
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14. Rate Laws To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
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15. Summary of the Rate Laws
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16. Zero order First order Second order EXERCISE!
Consider the reaction aA  Products. [A]0 = 5.0 M and k = 1.0 × 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is:
Zero order
First order
Second order
a) 4.7 M
[A] = –(1.0×10–2)(30.0) + 5.0
b) 3.7 M
ln[A] = –(1.0×10–2)(30.0) + ln(5.0)
c) 2.0 M
(1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0)
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