kinetics “how fast a reaction occurs” factors 1.concentration 2.temperature 3.surface area...
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Kinetics
“How fast a reaction occurs”Factors1.Concentration2.Temperature3.Surface area4.Catalyst
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Measuring the Rate of a Reaction
Disappearance of a reactant
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
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Draw a graph of the following data (x=time)
Time(s) [C4H9Cl] (M)
0.0 0.100050.0 0.0905
100.0 0.0820150.0 0.0741200.0 0.0671300.0 0.0549400.0 0.0448500.0 0.0368800.0 0.0200
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Rate is the Slope y = m x + b[C4H9Cl]= (Rate)(t)
Rate = [C4H9Cl] t
Rate = d[C4H9Cl] dt
(Note how rate decreased with decreasing concentration)
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The problems on this page refer to the data table for C4H9Cl
a. Using the data in the table, calculate the rate of disappearance of C4H9Cl from 0 to 50 s.
b. Using the data in the table, calculate the rate of disappearance of C4H9Cl from 50 to 100 s.
Now graph the datac. Using the graph, calculate the rate at 100 s
(-1.64 X 10-4 M/s)d. Using the graph, estimate the rate at 0 seconds
and 300 s. (-2.0 X 10-4 M/s, -1.1 X 10-4 M/s)
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Rates and Stoichiometry
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Rate = -[C4H9Cl] = [C4H9OH]
t t
Rate of disappearance = Rate of appearance(only for 1:1 stoichiometry)
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2HI(g) H2(g) + I2(g)
Rate = -1 [HI] = [H2]= [I2]
2 t t t
aA + bB cC + dD
Rate = -1 [A] = -1 [B] = 1 [C] = 1 [D ]
a t b t c t d t
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Stoichiometry: Ex 1
How is the rate of the disappearance of ozone related to the appearance of oxygen according to:
2O3(g) 3O2(g)
Rate = -1 [O3] = [O2]
2 t 3 t-[O3] = 2 [O2] or RateO3 = -2/3 RateO2
t 3 t
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Stoichiometry: Ex 2
If the rate of appearance of oxygen at some instant is 6.0 X 10-5 M/s, calculate the rate of disappearance of ozone.
[O3] = -2 [O2]
t 3 t [O3] = -2 (6.0 X 10-5 M/s)
t 3 [O3] = -4.0 X 10-5 M/s
t
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Stoichiometry: Ex 3
The rate of decomposition of N2O5 at a particular instant is 4.2 X 10-7 M/s. What is the rate of appearance of NO2 and O2?
2N2O5(g) 4NO2(g) + O2(g)
Rate = -1 [N2O5] = 1 [NO2] = [O2]
2 t 4 t t-[N2O5] = 1 [NO2]
2 t 4 t
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[NO2] = – 4 [N2O5]
t 2 t[NO2] = – 2 [N2O5] = (-2)(4.2 X 10-7 M/s)
t t[NO2] = 8.4 X 10-7 M/s
t-[N2O5] = [O2]
2 t t-[O2] = (½)(4.2 X 10-7 M/s) = 2.1 X 10-7 M/s
t
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The Rate Law
aA + bB cC
Rate = k[A]m[B]n
k = Rate constant[A] and [B] = Initial concentrationsm and n = exponents
k, m, and n must be determined experimentally
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Rate Law: Ex 1
What is the rate law for the following reaction, given the following rate data:
NH4+(aq) + NO2
-(aq) N2(g) + 2H2O(l)
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Experiment Initial NH4+
Concentration (M)Initial NO2
-
Concentration (M)Initial Rate (M/s)
1 0.0100 0.200 5.4 X 10-7
2 0.0200 0.200 10.8 X 10-7
3 0.200 0.0202 10.8 X 10-7
4 0.200 0.0404 21.6 X 10-7
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Rate = k[NH4+]m[NO2
-]n
Divide Experiment 1 and 2Rate2 = k[NH4
+]m[NO2-]n
Rate1 = k[NH4+]m[NO2
-]n
10.8 X 10-7 = k[0.0200]m[0.200]n
5.4 X 10-7 = k[0.0100]m[0.200]n
2= [0.0200]m m = 11 =[0.0100]m
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Rate = k[NH4+]m[NO2
-]n
Divide Experiment 3 and 4Rate6 = k[NH4
+]m[NO2-]n
Rate5 = k[NH4+]m[NO2
-]n
21.6 X 10-7 = k[0.200]1[0.0404]n
10.8 X 10-7 = k[0.200]1[0.0202]n
2= [0.0404]n n = 11 =[0.0202]n
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Calculating kRate = k[NH4
+]1[NO2-]1
Consider Experiment 1 (can pick anyone)Rate1 = k[NH4
+]1[NO2-]1
5.4 X 10-7 = k[0.0100]1[0.200]1
k = 2.7 X 10-4 M-1s-1
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Rate Law: Ex 2
Calculate the rate law for the following general reaction: S + O2 SO2
Experiment [S] (M) [O2] (M) Initial Rate (M/s)
1 0.100 0.100 4.0 X 10-5
2 0.100 0.200 4.0 X 10-5
3 0.200 0.100 16.0 X 10-5
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ANS: Rate = (4.0 X 10-3 M-1s-1)[S]2
What is the rate of the reaction when [S] = 0.050 M and [O2] = 0.100 M?
ANS: 1.0 X 10-5 M/s
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Rate Law: Ex 3
A particular reaction varies with [H+] as follows. Calculate the Rate Law:
[H+] (M) Initial Rate (M/s)0.0500 6.4 X 10-7
0.100 3.2 X 10-7
0.200 1.6 X 10-7
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Rate = 3.2 X 10-8 M2s-1[H+]-1
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Rate Law: Ex 4
Calculate the rate if the [H+] = 0.400 M
Rate = 8 X 10-8 M/s
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Calculate the rate law for the following reaction:
BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) ----> 3 Br2(l) +
H2O(l)
Trial [BrO3-] [Br-] [H+] Initial
rate (mol/Ls)
1 0.10 0.10 0.10 8.0 x 10-
4
2 0.20 0.10 0.10 1.6 x 10-3
3 0.10 0.20 0.10 1.6 x 10-3
4 0.10 0.10 0.20 3.2 x 10-
3
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Reaction Order
• m and n are called reaction orders• m + n + …. = overall reaction order• Units of k
– k must always have units that allow rate to have a unit of M/s
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Reaction Order: Ex 1
What is the overall reaction order for the following reaction. What unit will the rate constant (k) have?
CHCl3 + Cl2 CCl4 + HCl
Rate=k[CHCl3][Cl2]1/2
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Zeroth Order
• Reaction order can be zero• Concentration does not affect the rate• 2NH3 (g) N2(g) + 3H2(g)
Rate = k[NH3]0
Rate = k
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Recognizing Zeroth Order
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Concentration and Time: First Order
• First Order Reaction – reaction whose rate varies with the concentration of a single reactant to the first power
• Often the decay(chemical or nuclear) or decomposition of one substance
ln[A]t = -kt + ln[A]0
y = mx + b
First order eqns - linear ln[Conc] vs time graph
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Example of a First Order Reaction
• First order in CH3NC
• ln[CH3NC]t = -kt + ln[CH3NC]0
• Can use this equation to calculate the concentration at any time.
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This graph is not linear This graph is[A]t = [A]0e-kt ln[A]t = -kt + ln[A]0
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First Order: Ex 1
The first order rate constant for the decomposition of an insecticide is 1.45 yr-1. If the starting concentration of the compound in a lake is 5.0 X 10-7 g/cm3, what will be the concentration the following year? In two years?
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ln[A]t = -kt + ln[A]0
ln[A]t = -(1.45 yr-1 )(1 yr) + ln(5.0 X 10-7 g/cm3)
ln[A]t = -15.96
[A]t = e-15.96
[A]t = 1.2 X 10-7 g/cm3
or[A]t = [A]0e-kt
[A]t = (5.0 X 10-7 g/cm3)e-(1.45 yr-1 )(1yr)
[A]t = 1.2 X 10-7 g/cm3
([A]2yr = 2.8 X 10-8 g/cm3)
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First Order: Ex 2
How long will it take the concentration to drop to 3.0 X 10-7 g/cm3?
ln[A]t = -kt + ln[A]0
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First Order: Ex 3
The decomposition of dimethyl ether is a first order process with k=6.8 X10-4s-1. If the initial pressure is 135 torr, what is the partial pressure after 1420s?
(CH3)2O(g) CH4(g) + H2(g) + CO(g)
(ANS: 51 torr)
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First Order Half-life
• Half-life – time required for the concentration of a reactant to drop to one half the initial value– Medicine in the body– Nuclear decay
• [A]t = ½[A]0
t½ = 0.693
k
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Half-Life: Ex 1From the figure below, calculate the half-life and k
for the reaction of C4H9Cl with water.
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Sodium-24 (used in some medical tests) has a half-life of 14.8 hours. What is the rate constant for its decay?
(ANS: 0.0468 hr-1)
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Half-Life• Half-life - The time during which one-half of a
radioactive sample decays – Ranges from fraction of a second to billions of years.– You can’t hurry half-life.
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Carbon-14 dating
• 14C in traces of once-living organisms– E.g., a 5730 years after death, only half of the 14C remains.
• ~ 50,000 years.• 15% margin of error• Mummies, the Dead Sea Scrolls, Shroud of Turin
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Half-Life
Isotope Half-life
Uranium-238 4.51x109 years
Lead-210 20.4 years
Polonium-214 1.6x10-4 seconds
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Half-life: Example 1
Carbon-14 has a half-life of 5730 years and is used to date artifacts. How much of a 26 g sample will exist after 3 half-lives? How long is that?
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Half-life: Example 2
Tritium undergoes beta decay and has a half life of 12.33 years. How much of a 3.0 g sample of tritium remains after 24.66 years?
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Half-life: Example 3
Cesium-137 has a half-life of 30 years. If you start with a 200 gram sample, and you now have 25 grams left, how much time has passed?
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Rate Law: Ex 1
Uranium-238 has a half-life of 4.5 X 109 yr. If 1.000 mg of a 1.257 mg sample of uranium-238 remains, how old is the sample?
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Rate Law: Ex 2
A wooden object is found to have a carbon-14 activity of 11.6 disintegrations per second. Fresh wood has 15.2 disintegrations per second. If the half-life of 14C is 5715 yr, how old is the object?
ANS: 2230 yr
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After 2.00 yr, 0.953 g of a 1.000 g sample of strontium-90 remains.
a. Calculate k (0.0241 y-1)b.Calculate how much remains after 5.00 years.
(0.886 g)
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Ex 4
A sample for medical imaging contains 18F (1/2 life = 110 minutes). What percentage of the original sample remains after 300 minutes?
ANS: 15.1%
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Concentration and Time: Second Order
Second Order Reaction – reaction whose rate varies with the concentration of a single reactant to the second power
1 = kt + 1
[A]t [A]0
y = mx + b
2nd order eqns give a linear 1/[Conc] vs time graph
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Second Order Half-life
t½ = 1
k[A]0
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Second Order: Ex 1
The following data is for the decomposition of nitrogen dioxide: NO2(g) NO(g) + ½ O2(g)
Is the reaction first or second order?
Time (s) [NO2] (M)
0.0 0.0100050.0 0.00787
100.0 0.00649200.0 0.00481300.0 0.00380
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Plot both ln[NO2] vs time and 1/[NO2] vs time
Which is linear?
Time (s) [NO2] (M) ln[NO2] 1/[NO2]
0.0 0.01000 -4.610 10050.0 0.00787 -4.945 127
100.0 0.00649 -5.038 154200.0 0.00481 -5.337 208300.0 0.00380 -5.573 263
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Pick a convenient point 1 = kt + 1
[A]t [A]0
Calculate the slope.
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Second Order: Ex 2
What is the half life for the above reaction?
t½ = 1
k[A]0
t½ ~ 1
(0.5 M-1s-1)(0.0100 M)t½ ~ 200 s
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Ex 3
Time (s) [CH3NCl](M)
0 0.01652000 0.0115000 0.00598000 0.00314
12000 0.0013715000 0.00074
Is the following reaction first or second order? Draw two graphs, and calculate k.
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Temperature and Rate
• Rate of most chemical reactions increases with temperature– Bread rising– Plants growing– Light sticks
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• Rate = k[A]m[B]n…• k increases with temperature• Orders (m, n, etc..) do not change
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Collision Model
For an effective collision to occur, molecules must collide:
1.Often (increases with temperature)2.Proper orientation3.Enough Activation Energy (increases with
temperature)
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Proper vs. Improper Orientation
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Insufficient vs. Sufficient Activation Energy
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Activation Energy
• Energy needed to form activated complex or transition state
• Energy needed to start a reaction
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a.Rank the following reactions from slowest to fastestb.Rank the reverse reactions from slowest to fastest
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Reaction Mechanisms
Elementary Processes – only one step involved– Unimolecular – Bimolecular– Termolecular(rare)
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Multistep Reactions
NO2(g) + CO(g) NO(g) + CO2(g)
NO2(g) + NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g) NO2(g) + CO2(g)
NO2(g) + CO(g) NO(g) + CO2(g)
NO3(g) is an intermediate
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Reaction Mechanisms: Ex 1
Ozone decomposes in the following steps. Describe the molecularity of each step and write the overall equation:
O3(g) O2(g) + O(g)
O3(g) + O(g) 2O2(g)
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O3(g) O2(g) + O(g) unimolecular
O3(g) + O(g) 2O2(g) bimolecular
2O3(g) 3O2(g)
O(g) is an intermediate
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Reaction Mechanisms: Ex 2
The following steps:Mo(CO)6 Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3 Mo(CO)5P(CH3)3
Are proposed for the reaction:Mo(CO)6 + P(CH3)3 Mo(CO)5P(CH3)3 + CO
Is the proposed mechanism consistent with the reaction?
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Rate Determining Step
• Consider two toll plazas
Plaza A limits the Plaza B limits theflow of traffic flow of traffic
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St 1: NO2 + NO2 NO3 + NO (slow)
St 2: NO3 + CO NO2 + CO2 (fast)
NO2 + CO NO + CO2
Since Step 1 is much slower, it controls the rate (rate determining step)
Rate = k[NO2]2
(agrees with experiment)
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Mechanisms with an Initial Fast Step
Proposed mechanismSt 1: NO + Br2 NOBr2 (fast)
St 2: NOBr2 + NO 2NOBr (slow)
Rate = k[NOBr2][NO]
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• NOBr2 is intermediate
• Quickly breaks up into NO and Br2
Rate = k[NOBr2][NO]
Rate = k[NO][Br2][NO]
Rate = k[NO]2[Br2]
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Reaction Mechanisms: Ex 3
Show that the following mechanism is consistent with the rate law: Rate = k[NO]2[Br2]
St 1: NO + NO N2O2 (fast)
St 2: N2O2 + Br2 2NOBr (slow)
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Catalyst
• Catalyst – substance that increases the speed of a reaction without being used up in the process
• Can increase the rate of reaction by thousands of times
• Lowers the activation energy• Ex:2KClO3(s) 2KCl(s) + 3O2(g)
• KClO3 is stable
• Adding MnO2 allows the reaction to go fast
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• Red line shows pressure during a decomposition without a catalyst (HCOOH CO2 + H2)
• Blue line shows with a catalyst
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Ex: 2H2O2(aq) 2H2O(l) + O2(g)
• Br-(aq) acts as a catalyst• Brown Br2 formed in between
• Br- regenerated at the end
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Catalytic Convertor
• heterogeneous (different phase) catalyst• Use Platinum and Rhodium ($$)• Speed up:
CO + O2 CO2
NO, NO2 N2
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Enzymes
• Biochemical catalysts• Large protein molecules • Catalase – in liver to decompose H2O2
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• Active Site – “lock and key” model• Turnover numbers of 1000 to 10 million per
second
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2. B 2A16. 0.033 mol 8.3 X 10-4
0.055 mol 5.5 X 10-4
0.070 mol 3.8 X 10-4
0.080 mol 2.5 X 10-4
18.8.3 X 10-5 M/s 7.0 X 105 M/s6.9 X 10-5 3.5 X 105 M/s5.2 X 10-5
3.4 X 10-5
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20. a) -[H2O]/2t = [H2]/2t = [O2]/t
b) -[SO2]/2t = [O2]/t = [SO3]/2t
c) -[NO]/2t = -[H2]/2t = [N2]/t = [H2O]/2t
22. RateCO2 = RateH2O = 0.050 M/s
24.a) rate = k[A][C]2 b) rate double c) no change d) 9X e) 27 times
26. a) rate = k[H2][NO]2 b) 1.1 M/s
c) 6.0 M/s28.a,b) rate = k[C2H5Br][OH-], k = 3.6X10-5 M-1s-1
c) ¼ of old rate ( ½ X ½ )
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30. a) Rate = k[ClO2]2[OH-]
b) k = 2.3 X 102 M-2s-1
c) Rate = 0.12 M/s
32. a) Rate = k[NO]2[O2]
b and c) k = 7.11 X 103 M-2s-1
d) Rate = 0.400 M/se) Rate = 0.200 M/s
38.a) 2.56 s b) 0.0125 M40. a) 0.0032 mol N2O5 b) 2.2 min c) 1.69 min
42. K = 2.08 X 10-4 s-1 t = 3.33 X 103s
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44. a) First order b) k = 0.0101 s-1
c) half-life = 68.7 s46. a) First order b) 3.68 X 10-3 min-1
52. a) Ea = 154 kJ b) 18 kJ
54. a) 70 kJ b) 45 kJ c) 45 kJ(b) and (c) are faster than (a)
56. Ea = 1.15 kJ/mol
58. Ea = 130 kJ/mol, A = 1.0 X 1010.
84. 1.75 X 10-7 (for S)2(1.75 X 10-7) = 3.5 X 10-7 for H+ and Cl-
87. time = 48 s 95. Ea = 150 kJ/mol
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• GOGGLES• Do one trial for each of the three ratios• We will compile class data to get an average time• Color is more of a light orange than a yellow• Record the color every 30 seconds (rather than
just guessing at the change time)
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The initial rate of the reaction has been measured at the reactant concentrations shown (in mol/L):
BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) ----> 3 Br2(l) +
H2O(l)
Experiment [BrO3
-] [Br-] [H+] Initial rate (M/s) 1 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.10 0.20 0.10 1.6 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3
According to these results what would be the initial rate (in M/s) if all three concentrations are:
[BrO3-]=[Br-]=[H+]=0.20 M?