chapter 13 chemical kinetics chemistry ii. kinetics is the study of the factors that affect the...
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Chapter 13Chemical Kinetics
Chemistry II
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• kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds
• Factors that influence the speed of a reaction: physical state of reactants, temperature, catalysts, concentration
The Rate of a Chemical Reaction
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The Rate of a Chemical ReactionDefining Rate
rate is how much a quantity changes in a given period of time, e.g.
time
distance Speed
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The Rate of a Chemical Reaction
• Chemical reaction - concentration change with time
• for reactants, a negative sign is used to show a decrease in concentration
• as time goes on, the rate of a reaction generally slows down and stops because the concentration of the reactants decreases
time
ionconcentrat Rate
time
[reactant]
time
[product] Rate
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at t = 0[A] = 8[B] = 8[C] = 0
at t = 0[X] = 8[Y] = 8[Z] = 0
at t = 16[A] = 4[B] = 4[C] = 4
at t = 16[X] = 7[Y] = 7[Z] = 1
25.0
061
04Rate
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125.0
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46Rate
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at t = 16[A] = 4[B] = 4[C] = 4
at t = 16[X] = 7[Y] = 7[Z] = 1
at t = 32[A] = 2[B] = 2[C] = 6
at t = 32[X] = 6[Y] = 6[Z] = 2
0625.0
061
12Rate
tt
ZZ
t
ZRate
12
12
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0625.0
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65Rate
tt
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125.0
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tt
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at t = 32[A] = 2[B] = 2[C] = 6
at t = 32[X] = 6[Y] = 6[Z] = 2
at t = 48[A] = 0[B] = 0[C] = 8
at t = 48[X] = 5[Y] = 5[Z] = 3
125.0
061
68Rate
tt
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CRate
12
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0625.0
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8
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08overall Rate
tt
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Coverall Rate
12
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80overall Rate
tt
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t
Aoverall Rate
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12
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9
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rate of change reactants = rate of change products
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The Rate of a Chemical ReactionStoichiometry
• e.g.
H2 (g) + I2 (g) 2 HI(g)
• for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different
• in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient
t
HI][
2
1
t
]I[
t
]H[ Rate 22
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The Rate of a Chemical ReactionStoichiometry
• e.g.
H2 (g) + I2 (g) 2 HI(g)
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The Rate of a Chemical ReactionAverage Rate
• the average rate is the change in measured concentrations in any particular time period linear approximation of a curve
• the larger the time interval, the more the average rate deviates from the instantaneous rate
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H2 I2
HI
Stoichiometry tells us that for every 1 mole/L of H2 used, 2 moles/L of HI are made.
Assuming a 1 L container, at 10 s, we used 0.181 moles of H2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles
At 60 s, we used 0.699 moles of H2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles
The average rate is the change in the concentration in a given time period.
In the first 10 s, the Δ[H2] is -0.181 M, so the rate is
s
M0181.0
s 10.000
M 181.0
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
0.000 1.000
10.000 0.819
20.000 0.670
30.000 0.549
40.000 0.449
50.000 0.368
60.000 0.301
70.000 0.247
80.000 0.202
90.000 0.165
100.000 0.135
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
0.000 1.000 0.000
10.000 0.819 0.362
20.000 0.670 0.660
30.000 0.549 0.902
40.000 0.449 1.102
50.000 0.368 1.264
60.000 0.301 1.398
70.000 0.247 1.506
80.000 0.202 1.596
90.000 0.165 1.670
100.000 0.135 1.730
Avg. Rate, M/s
Time (s) [H2], M [HI], M -[H2]/t
0.000 1.000 0.000
10.000 0.819 0.362 0.0181
20.000 0.670 0.660 0.0149
30.000 0.549 0.902 0.0121
40.000 0.449 1.102 0.0100
50.000 0.368 1.264 0.0081
60.000 0.301 1.398 0.0067
70.000 0.247 1.506 0.0054
80.000 0.202 1.596 0.0045
90.000 0.165 1.670 0.0037
100.000 0.135 1.730 0.0030
Avg. Rate, M/s Avg. Rate, M/s
Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t
0.000 1.000 0.000
10.000 0.819 0.362 0.0181 0.0181
20.000 0.670 0.660 0.0149 0.0149
30.000 0.549 0.902 0.0121 0.0121
40.000 0.449 1.102 0.0100 0.0100
50.000 0.368 1.264 0.0081 0.0081
60.000 0.301 1.398 0.0067 0.0067
70.000 0.247 1.506 0.0054 0.0054
80.000 0.202 1.596 0.0045 0.0045
90.000 0.165 1.670 0.0037 0.0037
100.000 0.135 1.730 0.0030 0.0030
Ave. rate slows down as reaction proceeds
Rate of loss reactant = Rate gain product
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Concentration vs. Time for H2 + I2 --> 2HI
0.000
0.200
0.400
0.600
0.800
1.000
1.200
1.400
1.600
1.800
2.000
0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000
time, (s)
con
cen
trat
ion
, (M
)
[H2], M
[HI], M
average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI]
the average rate for the first 10 s is 0.0181 M/sthe average rate for the first 40 s is 0.0150 M/sthe average rate for the first 80 s is 0.0108 M/s
Ave. rate slows down as reaction proceeds
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The Rate of a Chemical Reaction Instantaneous Rate
• average rate becomes less accurate over longer time spans
• the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve
• determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function
for you calculus fans
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H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is:
s
M 0.0070 Rate
s 40
M 28.0 Rate
Using [HI], the instantaneous rate at 50 s is:
s
M 0.0070 Rate
s 40
M 56.0
2
1 Rate
rate reactants = rate products
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Generalized rate law:
aA + bB → cC + dD
t
D][1
t
C][1
t
B][1
t
A][1 Rate
dcba
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Ex 13.1 - For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the Δ[H+].
H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3
(aq) + 2 H2O(l)
Solve the equation for the Rate (in terms of the change in concentration of the Given quantity)
Solve the equation of the Rate (in terms of the change in the concentration for the quantity to find) for the unknown value
s
M3-104.40 Rate
s 10
M 000.1M 868.0
3
1
t
]I[
3
1 Rate
s
M3-s
M3- 108.80 104.402t
]H[
Rate2t
]H[
t
]H[
2
1 Rate
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Factors Affecting Reaction Rate- nature of reactants
• nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules;
gases tend to react faster than liquids which react faster than solids;
powdered solids are more reactive than “blocks” more surface area for contact with other reactants
certain types of chemicals are more reactive than others e.g., the activity series of metals
ions react faster than molecules no bonds need to be broken
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• increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the
speed of the reaction doubles
• there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later
Factors Affecting Reaction Rate- Temperature
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• catalysts are substances which affect the speed of a reaction without being consumed
• most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts
• homogeneous = present in same phase
• heterogeneous = present in different phase
• how catalysts work will be examined later
Factors Affecting Reaction Rate- Catalysts
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• generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the
gas higher pressure = higher concentration
• concentration of solutions depends on the solute to solution ratio (molarity)
Factors Affecting Reaction Rate- Reactant Concentration
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The Rate Law: Effect of Concentration On Reaction Rate
• Mathematical relationship between the rate of the reaction and the concentrations of the reactants
• for the reaction aA + bB products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant
mnk [B][A] Rate
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The Rate Law: Effect of Concentration On Reaction Rate
• sum of the exponents is called the order of the reaction • The rate law for the reaction:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
Rate = k[NO]2[O2]
The reaction is second order with respect to [NO], first order with respect to [O2],
and third order overall
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Reaction CH3CN CH3NC
CH3CHO CH4 + CO
2 N2O5 4 NO2 + O2
H2 + I2 2 HI
The Rate Law: Effect of Concentration On Reaction Rate
Rate Law Rate = k[CH3CN]
Rate = k[CH3CHO]
Rate = k[N2O5]2
Rate = k[H2][I2]
Sample Rate Laws
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The Rate Law: Effect of Concentration On Reaction Rate
Example:
A → Products
Rate = k[A]1 k = 0.015 / 0.10 = 0.15 s-1
If concentration of A doubles, the new rate, Rate2 = k[2A]1 = 2 k[A]1 = 2 x Rate
[A] (M) Initial Rate (M/s)
0.10 0.015
0.20 0.030
0.30 0.060
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The Rate Law: Effect of Concentration On Reaction Rate
Example:
Zero order: Second order:
Concentration of A doubles, the rate is constant Conc. A doubles, new rate
Rate1 = Rate2 = k Rate2 = k[2A]2 = 4 x k[A]2
Rate2 = 4 x Rate
[A] (M) Initial Rate (M/s)
0.10 0.015
0.20 0.015
0.30 0.015
[A] (M) Initial Rate (M/s)
0.10 0.015
0.20 0.060
0.30 0.240
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Reactant Concentration vs. TimeA Products
Rate = k[A]2
Rate = k[A]Rate = k
0: Concentration dec. linearly with time. Rate Is constant, reaction does not slow down as [A] dec.
1 and 2: Rate slows as reaction proceeds since [A] dec.
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The Integrated Rate Law
• can only be determined experimentally• graphically
rate = slope of curve [A] vs. time
if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope
if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope
if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope
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The Integrated Rate Law
• the half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value
• the half-life of the reaction depends on the order of the reaction
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Zero Order Reactions (n = 0)
Rate = -d[A] = k[A]0 = k dt
constant rate reactions
Solution: [A] = -kt + [A]0 (= integrated rate law) y = mx + b
graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0
[A] = [A0]/2, t ½ = [A0]/2k
Units: when Rate = M/sec, k = M/sec
[A]0
[A]
time
slope = - k
∫ -d[A]/dt = ∫ k∫ d[A] = -∫ k dt[A] = -kt + C, where C = [A]0
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First Order Reactions (n = 1)
Rate = -d[A] = k[A] dt
Solution: ln[A] = -kt + ln[A]0
graph of ln[A] vs. time gives straight line with
slope = -k and y-intercept = ln[A]0
used to determine the rate constant
[A] = [A0]/2, t½ = ln 2 [lna-lnb = ln(a/b)] k
the half-life of a first order reaction is constant
Units: when Rate = M/sec, k = sec-1
(dim. Analysis: M/s = k.M)
Rate slows as reaction proceeds since [A] dec.
∫ -d[A]/dt = ∫ k[A]∫ 1 d[A] = -∫ k dt [A]ln[A] = -kt + C, where C = ln[A]0
ln[A]0
ln[A]
time
slope = −k
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Rate Data for hydrolysis of C4H9Cl
Time (sec) [C4H9Cl], M
0.0 0.1000
50.0 0.0905
100.0 0.0820
150.0 0.0741
200.0 0.0671
300.0 0.0549
400.0 0.0448
500.0 0.0368
800.0 0.0200
10000.0 0.0000
Show reaction is first-order and find k
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C4H9Cl + H2O C4H9OH + 2 HCl Concentration vs. Time for the Hydrolysis of C4H9Cl
0
0.02
0.04
0.06
0.08
0.1
0.12
0 200 400 600 800 1000
time, (s)
conc
entr
atio
n, (
M)
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C4H9Cl + H2O C4H9OH + 2 HClRate vs. Time for Hydrolysis of C4H9Cl
0.0E+00
5.0E-05
1.0E-04
1.5E-04
2.0E-04
2.5E-04
0 100 200 300 400 500 600 700 800
time, (s)
Rat
e, (
M/s
)
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C4H9Cl + H2O C4H9OH + 2 HCl
LN([C4H9Cl]) vs. Time for Hydrolysis of C4H9Cl
y = -2.01E-03x - 2.30E+00
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
0 100 200 300 400 500 600 700 800
time, (s)
LN(c
once
ntra
tion)
slope = -2.01 x 10-3
k =2.01 x 10-3 s-1
s 345s 1001.2
693.0
693.0t
1-3
21
k
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Second Order Reactions
Rate = -d[A] = k[A] = k[A]2
dt
Solution: 1/[A] = kt + 1/[A]0
y = mx + b
graph 1/[A] vs. time gives straight line with
slope = k and y-intercept = 1/[A]0
used to determine the rate constant
t½ = 1 k[A0]
when Rate = M/sec, k = M-1∙sec-1
l/[A]0
1/[A]
time
slope = k
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Example Second-Order Reaction
Time (hrs.)PNO2,
(mmHg) ln(PNO2) 1/(PNO2)
0 100.0 4.605 0.01000
30 62.5 4.135 0.01600
60 45.5 3.817 0.02200
90 35.7 3.576 0.02800
120 29.4 3.381 0.03400
150 25.0 3.219 0.04000
180 21.7 3.079 0.04600
210 19.2 2.957 0.05200
240 17.2 2.847 0.05800
Show that the reaction: NO2 → NO + O is incorrect according to the data below
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Rate Data Graphs ForNO2 → NO + O
Partial Pressure NO2, mmHg vs. Time
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
0 50 100 150 200 250
Time, (hr)
Pres
sure
, (m
mH
g)
Non-linear so not zero order
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Rate Data Graphs ForNO2 ® NO + O
ln(PNO2) vs. Time
2.4
2.6
2.8
3
3.2
3.4
3.6
3.8
4
4.2
4.4
4.6
4.8
0 50 100 150 200 250
Time (hr)
ln(p
ress
ure)
Non-linear so not first order
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Rate Data Graphs ForNO2 → NO + O
1/(PNO2) vs Time
1/PNO2 = 0.0002(time) + 0.01
0.00000
0.01000
0.02000
0.03000
0.04000
0.05000
0.06000
0.07000
0 50 100 150 200 250
Time, (hr)
Inve
rse
Pres
sure
, (m
mH
g-1
)
We can deduce actual reaction should be:2NO2 → 2NO + O2
k = 2 x 10-4 M-1 s-1
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Tro, Chemistry: A Molecular Approach 47
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Ex. 13.4 – The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M
the new concentration is less than the original, as expected
[SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1
[SO2Cl2]
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
[SO2Cl2][SO2Cl2]0, t, k
0ln[A]tln[A] :processorder 1st afor k
M 0.0175 ]Cl[SO
4.043.790.251]Clln[SO0.0225lns 865s 102.90]Clln[SO
]Clln[SOt]Clln[SO
(-4.04)22
22
1-4-22
02222
e
k
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49
Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Write a general rate law including all reactants
Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not
mnk [CO]][NO Rate 2
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Determine by what factor the concentrations and rates change in these two experiments.
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
2M 10.0
M 20.0
][NO
][NO
1expt 2
2expt 2 4 0021.0
0082.0
Rate
Rate
sM
sM
1expt
2expt
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Determine to what power the concentration factor must be raised to equal the rate factor.
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
2M 10.0
M 20.0
][NO
][NO
1expt 2
2expt 2 4 0021.0
0082.0
Rate
Rate
sM
sM
1expt
2expt
242
Rate
Rate
][NO
][NO
1expt
2expt
1expt 2
2expt 2
n
n
n
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Repeat for the other reactants
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
2M 10.0
M 20.0
[CO]
[CO]
2expt
3expt 1 0082.0
0083.0
Rate
Rate
sM
sM
2expt
3expt
012
Rate
Rate
[CO]
[CO]
2expt
3expt
2expt
3expt
m
m
m
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Substitute the exponents into the general rate law to get the rate law for the reaction
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
mnk [CO]][NO Rate 2n = 2, m = 0
22
022
][NO Rate
[CO]][NO Rate
k
k
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)
given the data below.
Substitute the concentrations and rate for any experiment into the rate law and solve for k
Expt.
Number
Initial [NO2], (M)
Initial
[CO], (M)
Initial Rate
(M/s)
1. 0.10 0.10 0.0021
2. 0.20 0.10 0.0082
3. 0.20 0.20 0.0083
4. 0.40 0.10 0.033
1-1-
2s
M
2s
M
22
sM 21.0M 01.0
0.0021M 10.0 0.0021
1expt for ][NO Rate
k
k
k
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The Effect of Temperature on Reaction Rate
• Rate constant k is temperature dependent
• Arrhenius investigated this relationship and showed that:
RT
Ea
eAk
R is the gas constant in energy units, 8.314 J/(mol∙K)
where T is the temperature in kelvin
A is a constant called the frequency factor
Ea is the activation energy, the extra energy needed to start the molecules reacting
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As x (temperature) increases e-1/x will increase up to a maximum value of 1, k increases
As Ea increases k will decrease (follows e-x graph)
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The Effect of Temperature on Reaction RateActivation Energy and the Activated Complex
• Ea is an energy barrier to the reaction
• amount of energy needed to convert reactants into the activated complex aka transition state
• the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy
because partial bonds
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The Effect of Temperature on Reaction RateThe Exponential Factor
• e-Ea/RT is a number between 0 and 1• it represents the fraction of reactant molecules with sufficient energy
to make it over the energy barrier• that extra energy comes from converting the KE of motion to PE in the
molecule when the molecules collide
• e-Ea/RT decreases as Ea increases
Reaction rate inc.:-Increasing T increases the Ave. KE of the molecules-Increases no. of molecules with sufficient energy to overcome the energy barrier
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The Effect of Temperature on Reaction RateArrhenius Plots
• the Arrhenius Equation can be algebraically solved:
AR
Ek a ln
T
1)ln(
y = mx + bwhere y = ln(k) and x = (1/T)
a graph of ln(k) vs. (1/T) is a straight line
slope of the line = -Ea/R so Ea = -mR
ey-intercept = A, (unit is the same as k …why?)
k = Ae-Ea/RT
lnk = ln(Ae-Ea/RT)lnk = lnA + lne-Ea/RT
lnk = lnA – Ea/RT
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Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:
Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-1
600 3.37 x 103 1300 7.83 x 107
700 4.83 x 104 1400 1.45 x 108
800 3.58 x 105 1500 2.46 x 108
900 1.70 x 106 1600 3.93 x 108
1000 5.90 x 106 1700 5.93 x 108
1100 1.63 x 107 1800 8.55 x 108
1200 3.81 x 107 1900 1.19 x 109
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Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:
use a spreadsheet to graph ln(k) vs. (1/T)
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Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:
Ea = m∙(-R)
solve for Ea
molkJ
molJ4
KmolJ4
1.93
1031.9314.8K 1012.1
a
a
E
E
A = ey-intercept
solve for A11-11
118.26
sM 1036.4
1036.4
A
eA
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The Effect of Temperature on Reaction RateThe Collision Model
• for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other.
• once molecules collide they may react together or they may not, depending on two factors –
1. whether the collision has enough energy to "break the bonds holding reactant molecules together";
2. whether the reacting molecules collide in the proper orientation for new bonds to form.
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The Effect of Temperature on Reaction RateThe Collision Model
Effective Collisions
• collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions
• the higher the A value (frequency of effective collisions), the higher k value and the faster the reaction rate
• when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state
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The Effect of Temperature on Reaction RateThe Collision Model
Orientation Effect
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The Effect of Temperature on Reaction RateThe Collision Model
• A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier
• there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)
RT
E
RT
E aa
pzeeAk
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The Effect of Temperature on Reaction RateThe Collision Model
Orientation Factor
• proper orientation is when the atoms are aligned so that old bonds can break and the new bonds can form
• the more complex the reactants, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p
slightly less than 1
• for most reactions, the orientation factor is less than 1 For many, p << 1
e.g. H(g) + I(g) → HI(g)
H2(g) + I2(g) → 2HI(g)
HCl(g) + HCl(g) → H2(g) + Cl2(g) Smallest p
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Reaction Mechanisms
• we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules
• but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible
• most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules
• describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism
• knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism
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Reaction Mechanisms
• Overall reaction:
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) • Mechanism:
1) H2(g) + ICl(g) HCl(g) + HI(g)
2) HI(g) + ICl(g) HCl(g) + I2(g)
• the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps
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1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g)
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
Reaction MechanismsIntermediates
• notice that the HI is a product in Step 1, but then a reactant in Step 2
• since HI is made but then consumed, HI does not show up in the overall reaction
• materials that are products in an early step, but then a reactant in a later step are called intermediates
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Reaction MechanismsMolecularity
• the number of reactant particles in an elementary step is called its molecularity
• a unimolecular step involves 1 reactant particle
• a bimolecular step involves 2 reactant particles
though they may be the same kind of particle
• a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps
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Reaction MechanismsRate Laws for Elementary Steps
• each step in the mechanism is like its own little reaction – with its own activation energy and own rate law
• the rate law for an overall reaction must be determined experimentally
• but the rate law of an elementary step can be deduced from the equation of the step
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl] 2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]
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Reaction MechanismsRate Laws for Elementary Steps
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Reaction MechanismsRate Determining Step
• in most mechanisms, one step occurs slower than the other steps
• the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction
• we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy
• the rate law of the rate determining step determines the rate law of the overall reaction
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Another Reaction MechanismNO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 slow
NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] fast
NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2
The first step in this mechanism is the rate determining step.
The first step is slower than the second step because its activation energy is larger.
The rate law of the first step is the same as the rate law of the overall reaction.
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Reaction MechanismsValidating a Mechanism
in order to validate (not prove) a mechanism, two conditions must be met:
1. the elementary steps must sum to the overall reaction
2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law
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Reaction Mechanisms Mechanisms with a Fast Initial Step
• when a mechanism contains a slow initial step, the rate law will not contain intermediates
• when a mechanism contains a fast initial step, the rate limiting step, and hence the rate law may contain intermediates
• We can express[intermediate] in terms of [reactant]• If first step is fast, intermediate products build up (limited by slower step down
the line) as they build up they react to re-form reactants reaches equilibrium, the forward and reverse reaction rates are equal – so
the concentrations of reactants and products of the step are related• substituting into the rate law of the RDS will produce a rate law in terms of just
reactants
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An Example
2 NO(g) ⇌ N2O2(g) Fast
H2(g) + N2O2(g) H2O(g) + N2O(g) Slow (rate limiting)
H2(g) + N2O(g) H2O(g) + N2(g) Fast
k1
k-1
2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g)
Experimentally observed Rateobs = k [H2][NO]2
Proposed mechanism:
2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g)
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Reaction Mechanisms Mechanisms with a Fast Initial Step
• Is the mechanism valid?
1. steps must sum to the over all reaction
2. rate law predicted by mechanism must be consistent with exp. observation
Since 2nd step is rate limiting, Rate = k2[H2][N2O2]
BUT! Contains intermediate [N2O2] , not consistent with observation
Since 1st step is in equilibrium we can express [intermediate] in terms of reactants
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An Example
2 NO(g) ⇌ N2O2(g) Fast
H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]
H2(g) + N2O(g) H2O(g) + N2(g) Fast
k1
k-1
2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
for Step 1 Rateforward = Ratereverse
2
1
122
2212
1
[NO]]O[N
]O[N [NO]
k
k
kk
22
1
12
2
1
122
2222
][NO][HRate
[NO]][HRate
]O][N[HRate
k
kk
k
kk
k
Let k2k1/k-1 = k, we have obs rate law
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Ex 13.9 Show that the proposed mechanism for the reaction 2 O3(g) 3 O2(g) matches the observed rate law
Rate = k[O3]2[O2]-1
O3(g) O⇌ 2(g) + O(g) Fast
O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O]
k1
k-1
for Step 1 Rateforward = Ratereverse
123
1
1
2131
]][O[O[O]
][O][O ][O
k
k
kk
1-2
23
1
12
1-23
1
132
32
][O][ORate
]][O[O][ORate
][O][ORate
k
kk
k
kk
k
(Slow rate has intermediate)
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Catalysts
• catalysts are substances that affect the rate of a reaction without being consumed
• catalysts work by providing an alternative mechanism for the reaction with a lower activation energy
• catalysts are consumed in an early mechanism step, then made in a later step
mechanism without catalyst
O3(g) + O(g) 2 O2(g) V. Slow
mechanism with catalyst
Cl(g) + O3(g) O⇌ 2(g) + ClO(g) Fast
ClO(g) + O(g) O2(g) + Cl(g) Slow
O3(g) + O(g) 2 O2(g)
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CatalystsDemo
http://academics.rmu.edu/~short/chem1215/1215-demos/oscillating-methanol-7mins.mov
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
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Ozone Depletion over the Antarctic
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Catalysts
polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals
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CatalystsHomorgeneous and Heterogeneous Catalysts
• homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g)
• heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system
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Catalysts Enzymes
• because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate
• protein molecules that catalyze biological reactions are called enzymes
• enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction
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Enzymatic Hydrolysis of Sucrose