1. Entry Task: Nov 5th Wednesday
Sign off on Expression and Law ws
Grab Integrated Law notes by door
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2. Agenda: Go through Expression and Law ws
Walkthrough NOTES Ch. 14 sec 3 – The change in concentration with time (integrated – graph)
Integrated Rate Law ws
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3. MAYHAN

4. A decreases twice as fast
1. Given the reaction: 2A (g) --->B (g) + C (g)
a. Express the rate of reaction in terms of the change in concentration of each of the reactants and products.
Δ[B] Δt 1
Δ[C] Δt
- 1 2
Δ[A] Δt 1 = =
b. When [C] is increasing at 2.0 mol.L–1.s–1 , how fast is [A] decreasing? SHOW WORK!
Δ[A] Δt 1
Δ[2.0] Δt
- 1 2 = 2 1
Δ[2.0] Δt
Δ[A] Δt
4 mol.L-1s-1 = =
A decreases twice as fast
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5. Δ[F] Δt 1 2 Δ[G] Δt Δ[H] Δt Δ[D] Δt -1 3 Δ[E] Δt -1 1 1 - 1 2 = = = =
2. Given the reaction: 2D (g) + 3E (g) + F (g) ----> 2G (g) + H (g)
a. Express the rate of reaction in terms of the change in concentration of each of the reactants and products.
Δ[F] Δt 1 2
Δ[G] Δt
Δ[H] Δt
Δ[D] Δt -1 3
Δ[E] Δt -1 1 1
- 1 2 = = = =
b. When [E] is decreasing at 0.10 mol.L–1.s–1, how fast is [G] increasing? SHOW WORK!
- 1 3
Δ[0.10] Δt 1 2
Δ[G] Δt = 2 3
Δ[0.10] Δt
Δ[G] Δt
0.067 mol.L-1s-1 = =
2/3 as fast
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6. Double the concentration of H3PO4 or I-
3. For the reaction 3 ClO – (aq) → ClO3 – (aq) + 2 Cl – (aq) doubling the concentration of ClO – quadruples the initial rate of formation of ClO3-. What is the rate law for the reaction?
Rate = k [ClO –]2
4. The reaction C6H5N2Cl (aq) + H2O (l) → C6H5OH (aq) + N2 (g) + HCl (aq) is first order in C6H5N2Cl and zero order in H2O. What is the rate law for this reaction?
Rate = k [C6H5N2Cl]
5. For the reaction H3PO4 (aq) + 3I – (aq) + 2 H+ (aq) → H3PO3 (aq) + I3- (aq) + H2O (l) the rate law under certain conditions is given by Rate = k[H3PO4][I -][H+]2. What method(s) could be used if you want to double the reaction rate?
Double the concentration of H3PO4 or I-
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7. NO is a second order Cl2 is 1st order Rate = k [NO]2[Cl2]
6. What is the overall order of reaction for each of the following?
a) Rate = k[NO2]2 _______ b) Rate = k _______
c) Rate = k[Br2] _______ d) Rate = k[NO]2[O2] ______
2nd
Zero
1st
3rd
7. For the reaction 2 NO (g) + Cl2 (g) → 2 NOCl (g) If the concentration of NO is tripled, the rate of the reaction increases by a factor of nine. If the concentration of Cl2 is cut in half, the rate of the reaction is decreased to half the original rate. Find the order of reaction for each reactant and write the rate law for the reaction.
NO is a second order Cl2 is 1st order
Rate = k [NO]2[Cl2]
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8. Rate would increase by a factor of 9
8. In the decomposition of ammonia on a platinum surface at 856°C , 2 NH3 (g) → N2 (g) + 3 H2 (g) , changing the concentration of NH3 has no effect on the rate. Write the rate law for the reaction.
Rate = k
9. A reaction has the experimental rate law of Rate = k[A]2.
a) What happens to the rate if the concentration of A is tripled?
b) What happens to the rate if the concentration if A is reduced to one third the initial concentration?
Rate = k [3]2
Rate would increase by a factor of 9
Rate = k [1/3 ]2
Rate would decrease by a factor of 1/9
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9. A would be 1st order and B is zero order
10. For the reaction 2 A + B → C + D, if the concentration of A is doubled, the reaction rate doubles. If the concentration of B is halved, there is no change in the reaction rate. Determine the order of reaction with respect to each reactant and the overall order of reaction. Write the rate law for the reaction.
A would be 1st order and B is zero order
Rate = k [A]
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10. MAYHAN

11. I can…
Graph the relationship of time with amount of reactant concentrations and integrate this with rates of reactions.
Determine the graphical relationship between time and the rate order.
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12. Equation Sheet Under thermochemistry and kinetics
1st order
2nd order
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13. Two Types of Rate Laws
Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law
Integrated- Data table contains TIME AND CONCENTRATION DATA. Uses graphical methods to determine the order of the given reactant. K=slope of best fit line found through linear regressions
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14. Integrated Rate Law
Can be used when we want to know how long a reaction has to proceed to reach a predetermined concentration of some reagent
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15. Graphing Integrated Rate Law
Time is always on x axis
Plot concentration on y axis of 1st graph
Plot ln [A] on the y axis of the second graph
Plot 1/[A] on the y axis of third graph
You are in search of a linear graph
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16. Zero order Reactions-
Use A  B as an example. What happens when we double [A], what happens to the rate of reaction that is zero order? So does the concentration affect rate? Y/N____ What would the rate law be for a zero order?
The rate of reaction does not change No
Rate = k
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17. Sketch a graph with rate on Y and concentration on X axis- Label axis!!
As concentration increases, the rate of reaction remains the same.
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18. Sketch a graph with concentration on Y and time on X axis- Label axis!!
Integrated Rate laws
We look for straight lines. This provides a “clean” visual about the relationship of concentration and time.
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19. Sketch a graph with concentration on Y and time on X axis- Label axis!!
So when we plot our data table and get a negative straight line it is ____________order!
Slope is negative (-k)
Zero
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20. First Order Reactions
AB in a reaction. ① Write the rate expression for reactant A. (sec. 1 stuff)
Rate = -
∆[A] ∆t
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21. 14.3- The Change of Concentration with Time
② Write the rate law for reactant A. (sec 2 stuff)
Rate =
k[A]1
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22. 14.3- The Change of Concentration with Time
Describe a First Order reaction.
Double the concentration the reaction doubles.
Low amount of reactant = low rate of reaction
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23. Sketch a graph with rate on Y and concentration on X axis- Label axis!!
As we double our concentration , the rate doubles. It’s a direct relationship.
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24. Sketch a graph with concentration on Y and time on X axis- Label axis!!
Integrated Rate laws
We look for straight lines. This provides a “clean” visual about the relationship of concentration and time. This does not provide a straight line
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25. Natural log x Concentration
Sketch a graph with concentration on Y and time on X axis- Label axis!!
Integrated Rate laws
We can manipulate the data to provide a straight line plot.
Change how we plot concentration.
Natural log x Concentration
In [A]
Slope is negative (-k)
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26. 14.3- The Change of Concentration with Time
Take equations ① and ② and smash them together.
Rate = -
∆[A] ∆t
= k[A]
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27. 14.3- The Change of Concentration with Time
How do you use this equation to solve for concentration?
Using calculus to integrate the rate law for a first-order process gives us ln
[A]t
[A]0
= −kt
Where
[A]0 is the initial concentration of A.
[A]t is the concentration of A at some time, t, during the course of the reaction.
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28. Integrated Rate Laws Manipulating this equation produces… ln [A]t [A]0
= −kt
ln [A]t − ln [A]0 = − kt
ln [A]t = − kt + ln [A]0
…which is in the form
y = mx + b
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29. First-Order Processes
Relate this equation to the slope.
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.
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30. Sample Exercise 14.5 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of
5.0  107 g/cm3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of
the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to
decrease to 3.0  10–7 g/cm3? PLUG & CHUG
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
SET IT UP
ln [insecticide]t-1yr = [X]
t = 1 year
ln [insectacide]t-1yr =
-(1.45 yr-1)
(1 year)
+ ln [5.0 x 10-7g/cm3]
ln[insecticide]t -1 yr =  (14.51)
ln[insecticide]t - 1 yr = 15.96
Get rid of ln by ex on both sides
[insecticide]t = 1 yr = e15.96 = 1.2  107 g/cm3
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31. Sample Exercise 14.5 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of
5.0  107 g/cm3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of
the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to
decrease to 3.0  10–7 g/cm3? PLUG & CHUG
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
SET IT UP
ln [ ]t =
ln [ ]t =
-(1.45 yr-1) X
+ ln [5.0 x 10-7g/cm3]
t = X
Get X by itself- move to left side
-15.02
ln [ ]t
- ln [5.0 x 10-7g/cm3] =X
=0.35 years
-1.45 yr
-1.45 yr-1
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32. Sample Exercise 14.5 Using the Integrated First-Order Rate Law
Continued
Practice Exercise
The decomposition of dimethyl ether, (CH3)2O, at 510 ºC is a first-order process with a rate constant of 6.8  10–4 s–1:
(CH3)2O(g)  CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is its pressure after 1420 s?
k = 6.8 x 10-4-s-1
ln [A]0 = [135 torr]
SET IT UP
ln [X torr]t = [X]
t = s
ln [X torr]t =
-(6.8 x 10-4)
(1420 s)
+ ln [135 torr]
ln[torr]t =  (4.91)
ln[torr]t = 3.94
Get rid of ln by ex on both sides
[torr]t = e3.94= 51.4 torr
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33. 14.3- The Change of Concentration with Time
Describe a second-order reaction.
When you double the reactant the rate increases by a power of 2, to quadruple the rate
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34. Sketch a graph with rate on Y and concentration on X axis- Label axis!!
The relationship is more pronounced.
Double your concentration and the rate goes up by the power of 2.
Hence- second order.
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35. Sketch a graph with concentration on Y and time on X axis- Label axis!!
Integrated Rate laws
We look for straight lines. This provides a “clean” visual about the relationship of concentration and time. This does not provide a straight line
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36. Sketch a graph with concentration on Y and time on X axis- Label axis!!
Integrated Rate laws
We can manipulate the data to provide a straight line plot.
Change how we plot concentration.
1 divided by Concentration
1/[A]
And the slope is positive (k)
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37. Second-Order Processes
Provide the second order equation.
Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1
[A]t
= kt +
[A]0
also in the form
y = mx + b
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38. Second-Order Processes1
[A]t
= kt +
[A]0
So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.
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39. 14.3- The Change of Concentration with Time
What does second order reactions depend on?
A second order reaction is one whose rate depends on the initial reactant concentration
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40. Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields data comparable to this:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0
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41. Second-Order Processes
Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
−4.610
50.0
−4.845
100.0
−5.038
200.0
−5.337
300.0
−5.573
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42. Second-Order Processes
Graphing ln 1/[NO2] vs. t, however, gives this plot.
Because this is a straight line, the process is second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263
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44. Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph.
What order is this reaction and what formula would I use to calculate various times/concentrations?
First order and
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45. Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph.
What order is this reaction and what formula would I use to calculate various times/concentrations?
Zero order and
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46. Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph.
What order is this reaction and what formula would I use to calculate various times/concentrations?
Second order and
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