Entry Task: Dec 6 th Thursday Question : For the general rate law, Rate = k[A] [B] 2, what will...
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Transcript of Entry Task: Dec 6 th Thursday Question : For the general rate law, Rate = k[A] [B] 2, what will...
Entry Task: Dec 6th Thursday
Question :For the general rate law, Rate = k[A] [B]2, what
will happen to the rate of reaction if the concentration of A is tripled?
You have 5 minutes!
Agenda:
• Go through the answers to Rate Law ws #1• Walkthrough NOTES Ch. 14 sec 3 – The change
in concentration with time (integrated – graph)
• Rate Law ws #2
1. For 2 A + B C, we’ve determined the following experimental data:
a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 2
Rate = k[A]1[B]2 3
1.62 x10-5 = k[0.0100]1[0.0100]2 1.62 x10-5 = k= 16.2 M-1s-1
1.0 x10-6
2. For 2 A + B C, we’ve determined the following experimental data:
a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
2 1
Rate = k[A]2[B]1 3
2.80 x10-3 = k[0.026]2[0.015]1 2.80 x10-3 = k= 277 M-1s-1
1.01 x10-5
3. The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M.
3. Using these data, determine (a) the rate law for the reaction
Exp. 1 vs. Exp. 2, we doubled the concentration of H2, the rate doubled as well. This means [H2]1
3. Using these data, determine (a) the rate law for the reaction
Exp. 1 vs. Exp. 2, we doubled the concentration of H2, the rate doubled as well. This means [H2]1
Exp. 2 vs. Exp. 3, we doubled the concentration of NO and the rate quadrupled or 22 . This means that NO is 2nd order [NO]2
a) Rate = k[NO]2[H2]1
3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M. a) Rate = k[NO]2[H2]1
k=Rate
[NO]2[H2]1=
4.92 x 10-3 M/s[0.20]2[0.10]1
= 1.2 M-2/s-1
3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M.
a) Rate = k[NO]2[H2]1
b) k = 1.2 M1 s1
Rate = (1.2 M1 s1) (0.050 M)2(0.150)
Rate = 4.5 x 10-4 M/s
14.214. The following data were collected for the rate of disappearance of NO in the reaction:
2NO (g) + O2 (g) 2NO2 (g)
Experiment [NO] (M) [O2] (M) Initial Rate M/s
1 0.0126 0.0125 1.41 x 10-2
2 0.0252 0.0250 1.13 x 10-1
3 0.0252 0.0125 5.64 x 10-2
a) What is the rate law for the reaction?For [NO], if you doubled the concentration, the rate goes up by a factor of 4 so [NO]2
For [O2], if you doubled the concentration, the rate doubles so [O2]1
14.214. The following data were collected for the rate of disappearance of NO in the reaction: 2NO (g) + O2 (g) 2NO2 (g)
Experiment [NO] (M) [O2] (M) Initial Rate M/s
1 0.0126 0.0125 1.41 x 10-2
2 0.0252 0.0250 1.13 x 10-1
3 0.0252 0.0125 5.64 x 10-2
c) What is the average value of the rate constant calculated from the three data sets. SHOW YOUR WORK!!
k=Rate
[NO]2[H2]1=
5.64 x 10-2 M/s[0.0252]2[0.125]1
= 7.11x103 M-2/s-1
14.235. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C:
2NO (g) + Br2 (g) 2NOBr (g)
Experiment [NO] (M) [Br2] (M) Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
a) Determine the rate law?
For [NO], if you doubled the concentration, the rate goes up by a factor of 4 so [NO]2
For [Br2], if you doubled the concentration, the rate doubles so [Br2]
14.23Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g)
Experiment [NO] (M) [Br2] (M) Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets.
k=Rate
[NO]2[Br2]1=
150 M/s[0.25]2[0.20]1
= 1.2 x104 M-2/s-1
14.235. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g)
Experiment [NO] (M) [Br2] (M) Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2?
Rate = −11
[Br2]t
= 12
[NOBr]t
14.235. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g)
Experiment [NO] (M) [Br2] (M) Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
d) What is the rate of disappearance of Br2 when [NO] = 0.075M and [Br2] = 0.185 M?
= kRate [NO]2[Br2]1
[0.075]2[0.185]1= 2(1.2 x104 M-2/s-1)
= 6.1 M/s
ChemicalKinetics
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.
ChemicalKinetics
Equation SheetUnder thermochemistry and kinetics
1st order
2nd order
Arrhenius Equation
ChemicalKinetics
Chapter 14 section 3 Notes
ChemicalKinetics
Two Types of Rate Laws
1. Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law
2. 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
ChemicalKinetics
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
ChemicalKinetics
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
ChemicalKinetics
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
ChemicalKinetics
Sketch a graph with rate on Y and concentration on X axis- Label axis!!
As concentration increases, the rate of reaction remains the same.
ChemicalKinetics
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.
ChemicalKinetics
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)
ChemicalKinetics
First Order Reactions
AB in a reaction.
① Write the rate expression for reactant A. (sec. 1stuff)
Rate = - ∆[A]∆t
ChemicalKinetics
14.3- The Change of Concentration with Time
② Write the rate law for reactant A. (sec 2 stuff)
Rate = k[A]1
ChemicalKinetics
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
ChemicalKinetics
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.
ChemicalKinetics
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
ChemicalKinetics
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 ConcentrationIn [A]
Slope is negative (-k)
ChemicalKinetics
14.3- The Change of Concentration with Time
Take equations and and smash ① ②them together.
Rate = - ∆[A]∆t
= k[A]
ChemicalKinetics
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.
ChemicalKinetics
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
ChemicalKinetics
First-Order Processes
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.
ln [A]t = -kt + ln [A]0
Relate this equation to the slope.
ChemicalKinetics
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 107 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
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
ln[insecticide]t -1 yr = 1.45 + (14.51)
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
t = 1 year
ln [insecticide]t-1yr = [X]
-(1.45 yr-1)
SET IT UP
(1 year)ln [insectacide]t-1yr = + ln [5.0 x 10-7g/cm3]
Get rid of ln by ex on both sides
ln[insecticide]t - 1 yr = 15.96
[insecticide]t = 1 yr = e15.96 = 1.2 107 g/cm3
ChemicalKinetics
The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 107 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
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
k = 1.45 yr-1
ln [A]0 = [5.0 x 10-7g/cm3]
t = X
ln [3.0 10-7]t = -(1.45 yr-1)
SET IT UPXln [3.0 10-7]t = + ln [5.0 x 10-7g/cm3]
Get X by itself- move to left side
1.45 yr-1
=Xln [3.0 10-7]t - ln [5.0 x 10-7g/cm3]1.45 yr
=0.35 years15.02 - -14.51
ChemicalKinetics
Practice ExerciseThe 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?
Continued
Sample Exercise 14.5 Using the Integrated First-Order Rate Law
ln[torr]t = 0.9656 + (4.91)
k = 6.8 x 10-4-s-1
ln [A]0 = [135 torr]
t = 1420 s
ln [X torr]t = [X]
-(6.8 x 10-4)
SET IT UP
(1420 s)ln [X torr]t = + ln [135 torr]
Get rid of ln by ex on both sides
ln[torr]t = 3.94
[torr]t = e3.94= 51.6 torr
ChemicalKinetics
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
ChemicalKinetics
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.
ChemicalKinetics
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
ChemicalKinetics
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 Concentration1/[A]
And the slope is positive (k)
ChemicalKinetics
Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get
1[A]t
= kt +1
[A]0also in the form
y = mx + b
Provide the second order equation.
ChemicalKinetics
Second-Order Processes
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.
1[A]t
= kt +1
[A]0
ChemicalKinetics
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
ChemicalKinetics
Second-Order ProcessesThe 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 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
ChemicalKinetics
Second-Order Processes• Graphing ln [NO2] vs. t
yields:
Time (s) [NO2], M ln [NO2]
0.0 0.01000 −4.610
50.0 0.00787 −4.845
100.0 0.00649 −5.038
200.0 0.00481 −5.337
300.0 0.00380 −5.573
• The plot is not a straight line, so the process is not first-order in [A].
ChemicalKinetics
Second-Order Processes• Graphing ln
1/[NO2] vs. t, however, gives this plot.
Time (s) [NO2], M 1/[NO2]
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
• Because this is a straight line, the process is second-order in [A].
ChemicalKinetics
kt[A][A] 0 ln[A]
0[A]
1kt
[A]
1
0ln[A]kt
s
M
s
1sM
1
t vs.[A] t vs.ln[A] t vs.[A]
1
ChemicalKinetics
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 ln[A] 0ln[A]kt
ChemicalKinetics
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 kt[A][A] 0
ChemicalKinetics
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
0[A]
1kt
[A]
1
ChemicalKinetics