lect w2 152 - rate laws_alg
TRANSCRIPT
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General Chemistry IIGeneral Chemistry IICHEM 152CHEM 152
Week 2
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Week 2 Reading Assignment
Chapter 13 – Sections 13.2 (rate), 13.3 (rate law), 13.4 (integrated
rate law)
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A more refined descriptionThe average reaction rate gives us information about the speed of the
reaction over a certain period of time. But what if we are interested in a more
detailed description.How do we
calculate the instantaneous reaction rate
at any particular
time?
0
0.001
0.002
0.003
0.0040.005
0.006
0.007
0.008
0.009
0.01
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Time, min
Conc
entra
tion,
mol
/L
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START
END
Different Rates During a Reaction
Overall averagerate
InitialRate
InstantaneousRate
(slope of the lineAt a point)
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Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)
time
.
From your lab experience, how could we follow
the course of this reaction? Measure rates? What’s changing? What could we measure?
Measuring RatesConsider this reaction:
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Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)
time
393 nmlight
Detector
[Br2] Absorption3
93 n
m
Br2 (aq)
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Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)
Average rate = [Br2]t
= -[Br2]final – [Br2]initial
tfinal - tinitial
slope oftangent
slope oftangent slope of
tangent
Instantaneous rate = rate for specific tiny instance in time
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Br2(aq) + HCOOH(aq) HBr(aq) + CO2(g)
RATE
rate [Br2]
rate = k [Br2] + 0.0
k = rate[Br2]
= rate constant
= 3.50 x 10-3 s-1
Y=mX+b
Constant = Slope = k
How do the RATES
change with [Br2]?
Rate = k [Br2]1 this is called a 1st order reaction
Rate LawThe rate law for a
reaction tells us how rate varies with
concentration of the reactants.
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SOLUTION:
For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2](b)CH3CHO(g) CH4(g) + CO(g);
rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) +
2H2O(l); rate = k[H2O2][I-]
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
Your TurnYour Turn
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SOLUTION:
For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2](b)CH3CHO(g) CH4(g) + CO(g);
rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) +
2H2O(l); rate = k[H2O2][I-]
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
Your TurnYour Turn
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Consider the reaction:
2A + 2B + C → 2D + E
If the rate law for this reaction is Rate = k[A][B]2
What is the effect on the rate if the concentration of A is doubled?
1. No change
2. x23. x44. x85. x16
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What is the effect on the rate if the concentration of A is doubled?
Rate = k[A][B]2
1. No change2. x23. x44. x85. x16
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What is the effect on the rate if the concentrations of A, B & C are all
doubled?
Rate = k[A][B]2
1. No change2. x23. x44. x85. x16
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Measured Reaction Rates
• The rate of reaction changes with time, as the concentrations of reactants and products change.
• The rate constant, but not the rate, is independent of time.
• The instantaneous rate is the rate at any given time and is equal to the slope of a line tangent to the plot of concentration vs. time.
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Determining Reaction Orders
• The rate law for a given reaction must be determined experimentally.
• The order must be found for each species involved.
• The two major methods are1. Initial Rate Method
2. Integrated Rate Law Method
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Rate LawsIn general, forIn general, for
a A + b B a A + b B x X x X with a catalyst C with a catalyst C
Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp
The The exponentsexponents m, n, and p are the m, n, and p are the reaction order with respect each reaction order with respect each
reactantreactant
OverallOverall reaction order= m + n + p reaction order= m + n + p
These numbers can be 0, 1, 2 or These numbers can be 0, 1, 2 or fractions and fractions and must be determined by must be determined by
experimentexperiment!!
They are They are NOTNOT related to the related to the stoichiometric stoichiometric coefficientscoefficients a,b,x a,b,x
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Experiment
Initial Reactant Concentrations (mol/L) Initial Rate
(mol/L*s)
1
2
3
4
5
O2 NO
1.10x10-2 1.30x10-2 3.21x10-3
1.10x10-2 3.90x10-2 28.8x10-3
2.20x10-2
1.10x10-2
3.30x10-2
1.30x10-2
2.60x10-2
1.30x10-2
6.40x10-3
12.8x10-3
9.60x10-3
Your TurnYour Turn
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Determining Reaction Orders Using Initial Rates
Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate.
O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n
Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2
n
k [O2]1m[NO]1
n=
rate2
rate1 =
[O2]2m
[O2]1m
=
6.40x10-3mol/L*s
3.21x10-3mol/L*s
[O2]2
[O2]1
m
=1.10x10-2mol/L
2.20x10-2mol/L m; 2 = 2m m = 1
Do a similar calculation for the other reactant(s).
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Experiment
Initial Reactant Concentrations (mol/L) Initial Rate
(mol/L*s)
1
2
3
4
5
O2 NO
1.10x10-2 1.30x10-2 3.21x10-3
1.10x10-2 3.90x10-2 28.8x10-3
2.20x10-2
1.10x10-2
3.30x10-2
1.30x10-2
2.60x10-2
1.30x10-2
6.40x10-3
12.8x10-3
9.60x10-3
Your TurnYour Turn
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Expected
Solution:
Another ExampleConsider this reaction inside a car engine:
NO2(g) + CO(g) NO(g) + CO2(g)
How can we use this data to determine the rate law and the rate orders with respect to
each reactant?
Exp.
Initial Rate (mol/L*s)
Initial [NO2] Initial [CO]
123
0.00500.0800.0050
0.10
0.100.400.10
0.100.20
rate = k [NO2]m[CO]n
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0.080
0.0050
rate 2
rate 1
[NO2] 2
[NO2] 1
m=
k [NO2]m2[CO]n
2
k [NO2]m1 [CO]n
1
=
0.40
0.10=
m
16 = 4m and m = 2
The reaction is
2nd order in NO2.
First, choose two experiments in which [CO]
remains constant and the [NO2] varies.
Exp.
Initial Rate (mol/L*s)
Initial [NO2] Initial [CO]
123
0.00500.0800.0050
0.10
0.100.400.10
0.100.20
One Variable at a Time
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Now, choose two experiments in which
[NO2] remains constant and the [CO]
varies.
Exp.
Initial Rate (mol/L*s)
Initial [NO2] Initial [CO]
123
0.00500.0800.0050
0.10
0.100.400.10
0.100.20
One Variable at a Time
k [NO2]m3[CO]n
3
k [NO2]m1 [CO]n
1
[CO] 3
[CO] 1
n=
rate 3
rate 1=
0.0050
0.0050=
0.20
0.10
n; 1 = 2n and n = 0
The reaction is zero order in CO.
rate = k [NO2]2[CO]0 = k [NO2]2
Overall order
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Your TurnExp.
Initial Rate (mol/L*s)
Initial [NO2] Initial [CO]
123
0.00500.0800.0050
0.10
0.100.400.10
0.100.20
rate = k [NO2]2
What is the value of the constant, k, in this reaction?
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What are the UNITS of k in this reaction?
1. s-1
2. L/(mol s)3. mol/(L s)4. L2/(mol2
s)
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Overall Reaction Order
Units of k (t in seconds)
1 1/s (or s-1)
0 mol/(L*s) (or mol L-1 s-1) M/s
2 L/(mol*s) (or L mol -1 s-1) 1/Ms
3 L2 /(mol2 *s) (or L2 mol-2 s-1) 1/M2s
Units: Rate Constant k
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Concentration-Time Concentration-Time RelationsRelations
Consider a Consider a FIRST ORDER REACTION:FIRST ORDER REACTION: AA BB
The rate law isThe rate law is Rate= k[A] Rate= k[A]
RATE= RATE= -d[A]/dt-d[A]/dt= = k[A]k[A]
How can we predict the concentration of reactants at any moment in time?
How do you find How do you find [A][A] = = f(t)f(t)??
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First-Order Rate Law
A Bk
k[A]dt
d[A]
[A]
[A]
t
00
kdt[A]d[A]
or kdt[A]d[A]
kt0
0
e[A] [A] or kt[A][A]
ln
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Integrating Integrating - (d[A]/d t)- (d[A]/d t) = = k [A],k [A], we we getget
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] = - k tln[A]o
naturallogarithm [A] at time = 0
[A] / [A][A] / [A]00 =fraction remaining after =fraction remaining after time t has elapsed.time t has elapsed.
[A] / [A][A] / [A]00 =fraction remaining after =fraction remaining after time t has elapsed.time t has elapsed.
Called the Called the integratedintegrated first-order rate first-order rate lawlaw
Integrated Rate Laws
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Do you have to memorize all of the kinetics equations?
1. Yes2. No
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• Plot the experimental ln[A] vs. time• If the graph is linear, the reaction is
first-order.• The slope of this line = -k
y = mx + b
First-Order Rate Law
ln [A] = -kt + ln [A]0
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Examining a reaction
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC CH3CN
This reaction is suspected to be first order…
Design an experiment to determine the order of the reaction.
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First-Order Processes
• When ln P is plotted as a function of time, a straight line results.– The process is first-order.– k is the negative slope: 5.1 10-5 s-1.
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First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln[A]0
[A]0/2
k=t½
ln2k
=0.693
k=
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
t½ln2k
=0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
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A plot of [N2O5] vs. time for three half-lives
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Second-Order Rate Law2A Bk
2eff
2eff [A]k
dtd[A]
or [A]kR
[A]
[A]
t
0eff2eff2
0
dtk[A]d[A]
or dtk[A]d[A]
0effeff
0 [A]1
tk[A]1
or tk[A]
1[A]1
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• Plot the experimental 1/[A] vs. time• If the graph is linear, the reaction is
second-order.• The rate constant is k = slope
y = mx + b
0][
1
][
1
Ak t
A
Second-Order Rate Law
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Second-Order Half Life
0[A]1
kt[A]
12
1
21
0021 [A]
1kt
[A]1
21
0k[A]1
t2
1
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Zero-Order Reactions
A product rate = -D[A]Dt
rate = k [A]0 = k
k = rate[A]0
= M/sD[A]Dt
= k-
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ =[A]0
2k
[A] = [A]0 - kt
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Consider the following data for the reaction:
N2O5 → 2NO2 + ½ O2
Graphical Analysis
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Summary
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
[A]t = -kt + [A]0
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At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction,
with the very high rate constant of 87 s-
1, to two molecules of ethylene (C2H4).
If the initial C4H8 concentration is 2.00M, what is the concentration after
0.010 s?
_____________ M
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At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).
(b) What fraction of C4H8 has decomposed in this time?
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SOLUTION:
ln2.00
[C4H8] = -(87s-1)(0.010s)
[C4H8] = 0.84 mol/L
ln[C4H8]
0
[C4H8]t= -kt
(a)
(b) [C4H8]0 - [C4H8]t
[C4H8]0
=
2.00M - 0.84M 2.00M = 0.58
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PLAN:
Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via the following first-order reaction: CH2
H2C CH2(g)
H3C CH CH2 (g)
The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?
One-quarter of the initial value means two half-lives have passed.
Use t1/2 = ln2/k to find the half-life
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SOLUTION:
t1/2 = 0.693/9.2s-1 = 0.075s(a)
2 t1/2 = 2(0.075s) = 0.15 s(b)
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0th Order 1st Order 2nd Order
Plot for straight line
Slope, y-intercept
Half-life
Rate law
rate = k rate = k [A] rate = k [A]2
Units for k mol/(L*s) 1/s L/(mol*s)
Int. rate law (straight-line form)
[A]t =
-k t + [A]0
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 1/[A]0
[A]t vs. t ln[A]t vs. t 1/[A]t vs t
-k, [A]0 -k, ln[A]0k, 1/[A]0
[A]0/(2k) (ln 2)/k 1/(k [A]0)
Overview
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Summary Activity:N2O5(g) NO3(g) + NO2(g)
Consider the following graphs and reaction data to predict the concentration of N2O5 at 275 sec.
Time (s)
[N2O5]
0 1.000
25 0.822
50 0.677
75 0.557
100 0.458
125 0.377
150 0.310
175 0.255
200 0.210
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UA GenChem
What is the concentration of N2O5 after 275 sec?
______________ mol/L