brdy 6ed ch14 chemicalkinetics
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Chap 14 BrdyTRANSCRIPT
Chapter 14: Chemical Kinetics
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Speeds at Which Reactions Occur Kinetics: Study of factors that govern How rapidly reactions occur and How reactants change into products
Rate of Reaction: Speed with which reaction occurs How quickly reactants disappear and products
form
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Important Questions in Kinetics
Practical value!! Chemical /pharmaceutical manufacturers Is it practical to make drug? Made on manageable time scale? Can we adjust conditions to improve rate and
yield? Mechanism of Reaction Series of individual steps leading to overall
observed reaction How do reactants change into products? Detailed sequence of events
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors that Affect Reaction Rates
1. Chemical nature of reactants What elements, compounds, salts are
involved? What bonds must be formed, broken? What are fundamental differences in
chemical reactivity?
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors that Affect Reaction Rates 2. Ability of reactants to come in contact If two or more reactants must meet in order to react Gas or solution phase facilitates this
Reactants mix and collide with each other easily Homogeneous reaction
All reactants in same phase Occurs rapidly
Heterogeneous reaction Reactants in different phases Reactants meet only at interface between phases Surface area determines reaction rate ↑ area, ↑ rate ↓ area, ↓ rate
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors that Affect Reaction Rates
3. Concentrations of reactants Rates of both homogeneous and
heterogeneous reactions affected by [X] Collision rate between A and B ↑ if we ↑ [A] or
↑ [B]. ∴ Often (but not always) Reaction rate ↑ as [X]↑
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors that Affect Reaction Rates
4. Temperature Rates are often very sensitive to T Cooking sugar
Raising T usually makes reaction faster for two reasons:a. Faster molecules collide more often and
collisions have more Eb. Most reactions, even exothermic reactions,
require E to get going Crude Rule of Thumb: Rate doubles if ↑ T by 10 °C (10 K)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Factors that Affect Reaction Rates
5. Presence of Catalysts Catalysts Substances that ↑ rates of chemical and
biochemical reactions without being used up Rate-accelerating agents Speed up rate dramatically
Rate enhancements of 106 not uncommon
Chemicals that participate in mechanism but are regenerated at the end
Ex. Enzymes
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Measuring Rate of Reaction Rate = ratio with time unit in denominator
Ex. Rate of pay =
Rate of Chemical Reaction ↓ in [X] of particular species per unit time.
Always with respect to (WRT) given reactant or product
[reactants] ↓ w/ time [products] ↑ w/ time
reaction rate=Δ [ reactant ]
Δ time
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10 dollarshr
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rate of Reaction with Respect to Given Species X
Concentration in M Time in s Units on rate: Ex. [product] ↑ by 0.50 mol/L per second ⇒ rate =
0.50 M/s [reactant] ↓ by 0.20 mol/L per second ⇒ rate =
0.20 M/s10
Rate WRT X =[ X ]t2
−[ X ]t1
t2−t1
¿Δ[ X ]
Δt
mol/Ls
=molL⋅s
=Ms
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rate of Reaction Always + Whether something is ↑ or ↓ in [X].
Reactants Need – sign to make rate + Reactant consumed So ∆[X] = –
Products Produced as reaction goes along So ∆[X] = + Thus Rate = +
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Rate=−Δ [ reactant ]
Δt
Rate=Δ [ product ]
Δt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rates and Coefficients Relative rates at which reactants are
consumed and products are formed Related by coefficients in balanced chemical
equation. Know rate with respect to one product or reactant Can use equation to determine rates WRT all other
products and reactants. C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
Rate of Reaction
12
=−Δ [C3 H 8 ]
Δt=−
15
Δ [O2 ]
Δt=
13
Δ[ CO2 ]
Δt=
14
Δ[ H 2O ]
Δt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rates and Coefficients
O2 reacts 5 times as fast as C3H8
CO2 forms 3 times faster than C3H8 consumed
H2O forms 4/5 as fast as O2 consumed
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Rate=−Δ [O2 ]
Δt=−5
Δ [C 3 H8 ]
Δt
Rate=Δ [ CO2 ]
Δt=−3
Δ [C3 H8 ]
Δt
Δ[ H 2O ]
Δt=−
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Δ[O2 ]
Δt
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rates and CoefficientsIn general
αA + βB → γC + δD
Which one do I measure? Doesn't matter. All interrelated. Often determined by which one is easily
measured
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Rate=−1α
ΔAΔt
=−1β
ΔBΔt
=1γΔCΔt
=1δ
ΔDΔt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!In the reaction 2CO(g) + O2(g) 2CO→ 2(g), the rate of the reaction of CO is measured to be 2.0 M/s. What would be the rate of the reaction of O2?
A.the sameB.twice as greatC.half as largeD.you cannot tell from the given information
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Change of Reaction Rate with Time Generally reaction rate changes during
reaction i.e. Not constant
Often initially fast when lots of reactant present Slower at end when reactant depleted
Why? Rate depends on [reactants] Reactants being used up, so [reactant] is ↓ [A] vs. time is curve A is reactant ∴[A] is ↓ w/ time
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Measuring Rates Can measure rate using change in
concentration of any substance in reactionEx. 2A 3B→
Rates based on each substance are related to one another by the stoichiometric coefficients of reaction
Measured in three ways: Instantaneous rate Average rate Initial rate
Δ [B ]Δtime
=−32
Δ [A ]Δtime
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Instantaneous Reaction Rates Instantaneous rate
Slope of tangent to curve at any specific time Initial rate
Determined at initial time
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Average Rate of Reaction
Slope of line connecting starting and ending coordinates for specified time frame
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Δ[ Product ]Δtime
=rate
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Concentration vs. Time Curve for HI Decomposition at 508°C
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Rate=−slope=ΔyΔx
=riserun
Rate at any time t = negative slope (or tangent line) of curve at that point
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Table 14.1 Data at 508 °C
[HI] (mol/L) Time (s)
0.100 00.0716 500.0558 1000.0457 1500.0387 2000.0336 2500.0296 3000.0265 350
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2 HI(g) → H2(g) + I2(g)
rate=−(0 . 0716−0 .100 )M(50−0 )s
=−(−0 . 0284 M )
50 s =5 .68×10−4 M / s
Initial rate rate between first two data
points
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 22
Rate=−slope=ΔyΔx
=riserun
=−(0 . 044−0 . 068)M(150−50) s
=−−0 .024 M100 s
=2 . 4×10−4 M / s
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rate at 300 s 2 HI(g) → H2(g) + I2(g)
Rate = tangent of curve at 300 s
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Rate=−( 0. 0265−0 .0296)M(350−300 )s
=0 .0031 M50 s
=6 .20×10−5 M /s
[HI] (mol/L) Time (s)
0.100 00.0716 500.0558 1000.0457 1500.0387 2000.0336 2500.0296 3000.0265 350
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! A reaction was of NO2 decomposition was studied.
The concentration of NO2 was found to be 0.0258M at 5 minutes and at 10 minutes the concentration was 0.0097M. What is the average rate of the reaction between 5 min and 10 min?
A. 310 M/minB. 3.2 x 10-3 M/minC. 2.7 x 10-3 M/minD. 7.1 x 10-3 M/min
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( ) 30.0258 0.00973.2 x 10 / min
10min 5min
M MM−−
=−
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Concentration and Rate Rate Laws αA + βB → γC + δD Homogeneous reaction
Rate = k[A]m[B]n
Rate Law or Rate expression m and n = exponents found experimentally No necessary connection between stoichiometric
coefficients (α, β) and rate exponents (m, n) Usually small integers Sometimes simple fractions (½, ¾) or zero
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rate Laws
Rate = k[A]m[B]n
k = Rate Constant Dependence of rate on concentration goes as
some power (m) of concentration [A] All other factors (T, solvent) are included in k
Specific rate constant
k depends on T Must specify T at which you obtained k.
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Rate=−Δ [ A]
Δt= k[A]m[B]n
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
The rate law for the reaction 2A +B 3C is →rate= 0.045M-1s-1 [A][B]
If the concentration of A is 0.2M and that of B is 0.3M, what will be the reaction rate?
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rate=0.0027 M/s ⇒ 0.003 M/s
rate=0.045 M-1 s-1 [0.2][0.3]
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rate Laws
Rate = k[A]m[B]n
Exponents tell Order of Reaction with respect to (WRT) each reactant
Order of Reaction m = 1 [A]1 1st order WRT reactant m = 2 [A]2 2nd order WRT reactant m = 3 [A]3 3rd order WRT reactant m = 0 [A]0 0th order WRT reactant [A]0 = 1 ⇒ means A doesn't affect rate
Overall order of reaction = sum of orders (m and n) of each reactant in rate law
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Example 1
5 Br− + BrO3− + 6H+ → 3Br2 + 3H2O
x = 1 y = 1 z = 2 1st order WRT BrO3
−
1st order WRT Br− 2nd order WRT H+
Overall order = 1 + 1 + 2 = 4
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−Δ [ BrO3
−]
Δt=k [BrO 3
−]x[Br−
]y[H +
]z
rate=k [ BrO3−]1[ Br− ]
1[ H+
]2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Example 2 Sometimes n and m are coincidentally the
same as stoichiometric coefficients2 HI (g) → H2 (g) + I2 (g)
2nd order WRT HI 2nd order overall
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rate=−Δ[ HI ]
Δt=k [ HI ]2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!The following rate law has been observed:Rate = k[H2SeO][I-]3[H+]2. The rate with
respect to I- and the overall reaction rate is:
A. 6, 2B. 2, 3C. 1, 6D. 3, 6
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Calculating k from Rate Law
If we know rate and concentrations, can use rate law to calculate k
Ex. 2 at 508 °C Rate= 2.5 x 10−4 M/s [HI] = 0.0558 M
rate=−Δ[ HI ]
Δt=k [ HI ]2
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k=rate
[HI ]2=
2. 5×10−4 M /s
(0 .0558 M )2=0 . 08029 M−1 s−1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How To Determine Exponents in Rate Law
Experiments Method of Initial Rates If reaction is sufficiently slow or have very fast technique
Can measure [A] vs. time at very beginning of reaction before it curves up very much, then
Set up series of experiments, where initial concentrations vary
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initial rate=−([ A]1−[ A ]0
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex 1. Method of Initial Rates3A + 2 B → productsRate = k[A]m[B]n Expt. # [A]0, M [B]0, M Initial Rate, M/s
1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4
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Convenient to set up experiments so [X] of one species is doubled or tripled while [X] of all other species are held constant
Tells us effect of [varied species] on initial rate
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Reaction Order and Rate If reaction is 1st order WRT given species X, Doubling [X]1 → 21
Rate doubles If reaction is 2nd order WRT X, Doubling [X]2 → 22
Rate quadruples If reaction is 0th order WRT X, Doubling [X]0 → 20
Rate doesn't change If reaction is nth order WRT X Doubling [X]n → 2n
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Back to our Example Expt. # [A]0, M [B]0, M Initial Rate, M/s
1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4
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Comparing 1 and 2 Doubling [A] Quadruples rate Reaction 2nd order in A [A]2
Rate 2Rate 1
=4. 8×10−4
1 . 2×10−4=4
4=Rate 2Rate 1
=k [A2 ]
m[B2]
n
k [ A1]m[B1 ]
n=
k [0 . 20 ]m [0 . 10 ]n
k [0. 10 ]m [0 .10 ]n=
[0 . 20 ]m
[0 . 10 ]m=2m
2m = 4 or m = 2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Back to our Example Expt. # [A]0, M [B]0, M Initial Rate, M/s
1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4 3 0.20 0.20 4.8 × 10−4
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Comparing 2 and 3 Doubling [B] Rate does not change Reaction 0th order in B [B]0
Rate 3Rate 2
=4. 8×10−4
4. 8×10−4=1
1=Rate 3Rate 2
=k [ A3]
m[B3]
n
k [ A2]m[B2]
n=
k [0. 20 ]m [0 . 20 ]n
k [0 . 20]m [0. 10]n=
[0 . 20 ]n
[0 .10 ]n=2n
2n = 1 or n = 0
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 1 Method of Initial Rates
Conclusion: rate = k[A]2
Can use data from any experiment to determine k Let’s choose experiment 1
Expt. # [A]0, M [B]0, M Initial Rate, M/s
1 0.10 0.10 1.2 × 10−4 2 0.20 0.10 4.8 × 10−4
3 0.20 0.20 4.8 × 10−4
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k=rate
[ A ]2=
1 . 2×10−4 M/s
(0. 10 M )2=1 . 2×10−2 M−1 s−1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2. Method of Initial Rates
2 SO2 + O2 → 2 SO3
Rate =
Expt #
[SO2], M
[O2], M
Initial Rate of SO3 formation, M·s−1
1 0.25 0.30 2.5 × 10−3
2 0.50 0.30 1.0 × 10−2
3 0.75 0.60 4.5 × 10−2
4 0.50 0.90 3.0 × 10−2
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2 SO2 + O2 → 2 SO3
Rate = k[SO2]m[O2]n
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2 Compare 1 and 2
[SO2] doubles, [O2] constant, Rate quadruples, 22
40
Rate 2Rate 1
=1 .0×10−2
2 . 5×10−3=4
4=Rate 2Rate 1
=k [SO2 ]2
m[O2 ]2
n
k [SO2 ]1m[O2 ]1
n=k [0 . 50 ]m [0. 30 ]n
k [0 . 25 ]m
[0 .30 ]n
=[0. 50 ]
m
[0. 25 ]m=2m
2m = 4 or m = 2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2 Compare 2 and 4
[O2] triples, [SO2] constant Rate triples, 31
41
Rate 4Rate 2
=3 . 0×10−2
1 . 0×10−2=3
3=Rate 4Rate 2
=k [SO2]4
m[O2]4
n
k [SO2]2m[O2]2
n=k [0.50 ]m [0 . 90 ]n
k [0.50 ]m
[0 . 30 ]n
=[0. 90 ]
n
[0. 30 ]n=3n
3n = 3 or n = 1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2
Rate = k[SO2]2[O2]1
1st order WRT O2
2nd order WRT SO2
3rd order overall Can use any experiment to find k
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k=rate
[ SO2 ]2 [O2 ]1
=3. 0×10−2 M / s
(0 . 50 M )2( 0. 90 M )=0 . 13 M−2 s−1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 3. Method of Initial Rates
BrO3− + 5 Br− + 6H+ → 3Br2 + 3H2O
Rate=−Δ [BrO3
−]
Δt=k [BrO3
−]m[Br−
]n[H +
]p
Expt #
[BrO3−],
mol/L[Br−], mol/L
[H+], mol/L
Initial Rate, mol/(L·s)
1 0.10 0.10 0.10 8.0 × 10−4
2 0.20 0.10 0.10 1.6 × 10−3
3 0.20 0.20 0.10 3.2 × 10−3
4 0.10 0.10 0.20 3.2 × 10−3
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 3 Compare 1 and 2
44
Rate 2Rate 1
=1 .6×10−3 M / s
8 .0×10−4 M / s=
k (0 . 20 M )m(0. 10 M )
n(0 .10 M )
p
k (0. 10 M )m(0 . 10 M )n (0. 10 M )p
2 . 0=(0 .20 M0 .10 M )
m
=(2 . 0 )m ∴m=1
Rate 3Rate 2
=3 . 2×10−3 M / s
1 .6×10−3 M / s=
k (0 . 20 M )m( 0.20 M )
n(0 .10 M )
p
k (0. 20 M )m(0 .10 M )n(0 .10 M )p
2 . 0=(0 .20 M0 .10 M )
n
=(2 .0)n ∴n=1
Compare 2 and 3
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 3 Compare 1 and 4
First order in [BrO3−] and [Br−]
Second order in [H+] Overall order = m + n + p = 1 + 1 + 2 = 4 Rate Law is: Rate = k[BrO3
−][Br−][H+]2
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Rate 4Rate 1
=3 .2×10−3 M /s
8 .0×10−4 M /s=
k ( 0. 10 M )m(0 . 10 M )
n(0 .20 M )
p
k (0 .10 M )m(0 . 10 M )n (0.10 M )p
4 . 0=(0 .20 M0.10 M )
p
=(2 .0 ) p ∴ p=2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Using the following experimental data, determine the
order with respect to NO and O2 .
A. 2, 0B. 3,1C. 2, 1D. 1, 1
46
Expt #
[NO], M
[O2], M
Initial Rate of NO2 formation, M·s−1
1 0.12 0.25 1.5 × 10−3
2 0.24 0.25 6.0 × 10−3
3 0.50 0.50 5.2 × 10−2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
47
3 12
3 11
22 13
3 1 21
0.24 0.256.0 x 10 1.5 x 10 0.12 0.25
2
0.50 0.505.2 x 10 1.5 x 10 0.12 0.25
1
x y
x y
y
y
M MR M sR M s M M
x
M MR M sR M s M M
y
− −
− −
− −
− −
= =
=
= =
=Pers
onal
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Concentration and Time Rate law tells us how speed of reaction varies
with [X]'s. Sometimes want to know [reactants] and [products] at given time during
reaction How long for [reactants] to drop below some
minimum optimal value Need dependence of Rate on Time
48
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Concentration vs. Time for 1st Order Reactions
Corresponding to reactions A → products
Integrating we get
Rearranging gives
Equation of line y = mx + b 49
Rate=−Δ [ A ]
Δt=k [ A ]
ln[ A ]0[ A ]t
=kt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Plot ln[A]t (y axis) vs. t (x axis)
Yields straight line Indicative of 1st
order kinetics slope = − k intercept = ln[A]0
If we don't know already
ln [ A ]t=−kt+ ln [ A ]0
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Slope = − k
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
First Order Kinetics Graph Plot of [A] vs. time gives an exponential decay
51
[ A]t=[ A ]o e−kt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Concentration vs. Time for 2nd Order Reactions
Corresponding to reactions 2B → products
Integrating we get
Rearranging gives
Equation of line y = mx + b52
1[ B ]t
−1
[B ]0=kt
1[ B ]t
=kt+1
[ B ]0
Rate=k [B ]2=−Δ [ B ]
Δt
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Plot 1/[B]t (y axis) vs. t (x axis)
Yields straight line indicative of 2nd
order kinetics slope = + k intercept = 1/[B]0
1[ B ]t
=kt+1
[ B ]0
53
Graph of [B] vs. time is still ↓ function, which curves up
Not an exponential
Slope= +k
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How to Determine Reaction Order Using Graphs
Make 2 plots1.ln [A] vs. time2.1/[A] vs. time If ln [A] is linear and 1/[A] is curved, then
reaction is 1st order WRT [A] If 1/[A] plot is linear and ln [A] is curved,
then reaction is 2nd order WRT [A] If both horizontal lines, then 0th order WRT
[A]
54
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex 4. SO2Cl2 → SO2 + Cl2
Time, min [SO2Cl2], M ln[SO2Cl2] 1/[SO2Cl2] (L/mol)
0 0.1000 -2.3026 10.000
100 0.0876 -2.4350 11.416
200 0.0768 -2.5666 13.021
300 0.0673 -2.6986 14.859
400 0.0590 -2.8302 16.949
500 0.0517 -2.9623 19.342
600 0.0453 -3.0944 22.075
700 0.0397 -3.2264 25.189
800 0.0348 -3.3581 28.736
900 0.0305 -3.4900 32.787
1000 0.0267 -3.6231 37.453
1100 0.0234 -3.7550 42.73555
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex 4. SO2Cl2 → SO2 + Cl2
56
Clearly reaction is 1st order in SO2Cl2.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2 HI (g) → H2 (g) + I2 (g)
Time (s)
[HI] (mol/L)
ln[HI] 1/[HI] (L/mol)
0 0.1000 -2.3026 10.000
50 0.0716 -2.6367 13.9665
100 0.0558 -2.8860 17.9211
150 0.0457 -3.0857 21.8818
200 0.0387 -3.2519 25.840
250 0.0336 -3.3932 29.7619
300 0.0296 -3.5200 33.7838
350 0.0265 -3.6306 37.735857
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 5 HI (g) → H2 (g) + I2 (g)
58
Clearly reaction is 2nd order in HI.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! A plot for a zeroth order
reaction is shown. What is the proper label for the y-axis in the plot ?
A. ConcentrationB. ln of ConcentrationC. 1/ConcentrationD. 1/ ln Concentration
59
0 200 400 600 800 1000 1200
Zeroth Order Plot
time (min)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Half-lives for 1st Order Reactions Half-life = t½ First Order Reactions Set
Substituting into
Gives
Canceling gives ln 2 = kt½
Rearranging gives
60
[ A]t=12[ A]0
ln[ A ]0[ A ]t
=kt
ln[ A ]0
12 [ A]0
=kt 12
t 12
=ln 2k 1
=0 . 693k 1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Half-life for 1st Order ReactionsObserve: 1. t½ is independent of [A]o
For given reaction (and T) Takes same time for concentration to fall from
2 M to 1 M as from 5.0 × 10-3 M to 2.5 × 10-3 M
1. k1 has units (time)-1, so t½ has units (time) t½ called half-life
Time for ½ of sample to decay
61
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Half-life for 1st Order ReactionsDoes this mean that all of sample is gone
in two half-lives (2 x t½)? No! In 1st t½, it goes to ½[A]o
In 2nd t½, it goes to ½(½[A]o) = ¼[A]o
In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o
In nth t½, it goes to [A]o/2n
Existence of [X] independent half-life is property of exponential function Property of 1st order kinetics
62
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Half-Life Graph
63
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. Using Half-Life 131I is used as a metabolic tracer in hospitals.
It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?
64
N=N oe−kt
lnNN o
=−kt=−t ln 2
τ 12
t=−
τ 12
lnNN o
ln 2=−
(8. 07days ) ln( 1100 )
ln 2=53. 6 days
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s-1?
ln(10033
)=k ( 21da )
65
ln( A0
A )=kt
k = 6.11×10-7 s-1
k = 0.0528 da-1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
66
The half-life of I-132 is 2.295h. What percentage remains after 24 hours?
ln( 2)
k=t 1
20.302 h–1 = k
A = .0711 % Ao
k=ln 22 . 295h
ln( Ao
A )=kt
ln( Ao
A )=0 . 302h−1×24h=7 . 248
A=A o e−kt
=Aoe−7. 248
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which order has a half-life that is independent of the original amount?
A. ZeroB. FirstC. SecondD. None depend on the original quantity
67
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Half-lives of Second Order Reactions
How long before [A] = ½[A]o?
t½, depends on ½[A]o
t½, not useful quantity for 2nd order reaction
68
t 12
=1
k∗[ A ]0
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
The rate constant for the second order reaction 2A B is 5.3×10→ -5 M-1s-1. What is the original amount present if, after 2 hours, there is 0.35M available?
1[ 0.35 ]
−1
[ A 0 ]=
5 .3×10−5
M s×7200 s
69
1[ A ]
−1
[ A 0 ]=kt
A0=0.40 M
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! Plutomium-239 has a
half life of 24,100 yrs. How many years will it
take for 1.0 grams of Pu-239 to decay to 0.025 g ?
A. 9.6 x 105 yrsB. 2.4 x 105 yrsC. 1.3 x 105 yrsD. 4.8 x 105 yrs
70
5
The number of half-lifes if given by:1.00 g
0.025 g = 2 402
log 2 = log 40 = 5.32
5.32 x 24,100 yrs = 1.3 x 10 yrs
nn
n n
=
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Theories about Reaction Rates Reaction rate depends on [reactants] and T Collision Theory Based on Kinetic Molecular Theory Accounts for both effects on molecular level Central Idea Molecules must collide to react Greater number of collision/sec = greater reaction
rate
71
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Theories about Reaction Rates
Collision Theory As [reactants] ↑ number of Collisions ↑ Reaction rate ↑
As T ↑ Molecular speed ↑ Molecules collide with more force (energy) Reaction rate ↑
72
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Collision Theory Rate of reaction proportional to number of effective
collisions/sec among reactant molecules Effective collision
1 that gives rise to product
Ex. At Room Temperature and Pressure H2 and I2 molecules undergoing 1010 collisions/sec
Yet reaction takes a long time Not all collisions lead to reaction
Only very small % of all collisions lead to net change
73
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
1. Molecular Orientation
Molecules must be oriented in a certain way during collisions for reaction to occur
Ex. NO2Cl + Cl → NO2 + Cl2 Cl must come in pointing directly at another Cl atom
for Cl2 to form
74
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
1. Molecular Orientation
75
Wrong Orientation
Correct Orientation
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
2. Temperature Greatly Affects Rates
Over moderate T range, Ea unchanged
As ↑ T, More molecules have Ea
So more molecules undergo reaction
Reaction rate ↑ as T↑ 76
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
3. Activation Energy, Ea Molecules must possess certain amount of kinetic
energy (KE) in order to reactActivation Energy, Ea
Minimum KE needed for reaction to occur Get energy from collision with other molecules Upon collision, KE converted to potential energy (PE),
used to stretch, bend, and break bonds leading to chemical reaction
If molecules move too slowly, too little KE, they just bounce off each other
Without this minimum amount, reaction will not occur even when correctly oriented
Major reason all collisions do not lead to reaction77
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Transition State Theory Used to explain details of reactions What happens when reactant molecules collidePotential Energy Diagram To visualize what actually happens during
successful collision Relationship between Ea and developing Total
PE
78
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Potential Energy Diagram
79
Reaction Coordinate (progress of reaction)
Po
ten
tia
l E
ner
gy
Activation energy (Ea) = hill or barrier
between reactants and products
heat of reaction (∆H) = difference in PE between
products and reactants
∆Hreaction = Hproducts – Hreactants ProductsPers
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Potential Energy Diagram (Exothermic)
80
Reaction Coordinate (progress of reaction)
Po
ten
tia
l E
ner
gy Exothermic reaction
• Products lower PE than reactants
Exothermic Reaction
∆H = −
Products
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Exothermic Reaction
∆Hreaction < 0 (–) ↓ in PE of system Appears as ↑ in KE So T system ↑
Reaction gives off heat Can’t say anything about Ea from size of ∆H Ea could be high and reaction slow even if ∆Hrxn
large and negative Ea could be low and reaction rapid
81
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Potential Energy Diagram (Endothermic)
82
Endothermic Reaction
∆H = +
∆Hreaction = Hproducts – Hreactants
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Endothermic Reaction
∆Hreaction > 0 (+) ↑ in PE Appears as ↓ in KE So T system ↓
Have to add E to get reaction to go Ea ≥ ∆Hrxn as Ea includes ∆Hrxn If ∆Hrxn large and + Ea must be high Reaction very slow
83
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Transition State or Activated Complex
Arrangement of atoms at top of activation barrier
Brief moment during successful collision when bond to be broken is partially broken and bond to be formed is partially formed
Ex.
84
N CH3C C NH3CH3CN
C
Transition State (TS)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. NO2Cl + Cl → NO2 + Cl2 As NO2Cl and Cl come
together Start to form Cl····Cl bond Start to break N····Cl bond
Requires E, as must bring 2 things together
In TS N····Cl bond ½ broken Cl····Cl bond ½ formed
After TS Cl—Cl bond forms N····Cl breaks
Releases E as products more stable
85
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Examine the potential energy diagram. Which is the Slowest (Rate Determining) Step?A. Step 1B. Step 2C. Can’t tell from the given information
Reaction ProgressPo
ten
tial
En
erg
y
Has greatest Ea
1 2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Measuring Activation Energy
Generally we find k ↑ as T ↑ Usually magnitude of effect ↓ as T ↑ Arrhenius Equation Equation expressing T dependence of k
A = Frequency factor—has same units as k R = gas constant in energy units = 8.314 J·mol−1·K−1
Ea = Activation Energy—has units of J/mol T = Temperature in K
87
k=Ae−E
a/RT
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How To Calculate Activation Energy
Method 1. Graphically Take natural logarithm of both sides
Rearranging
Equation for a line y = b + mx
Arrhenius Plot Plot ln k (y axis) vs. 1/T (x axis)
88
ln k= ln A−( Ea
R )∗( 1T )
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2N2O5(g) → 4NO2(g) + O2(g)
T (°C) T (K) 1/T (K−1) k (s−1) ln k
20 293 3.41 × 10−3 2.00 × 10−5 −10.82
30 303 3.30 × 10−3 7.30 × 10−5 −9.53
40 313 3.19 × 10−3 2.70 × 10−4 −8.22
50 323 3.09 × 10−3 9.10 × 10−4 −7.00
60 333 3.00 × 10−3 2.90 × 10−3 −5.84
89
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Arrhenius Plot
3.0E-03 3.1E-03 3.2E-03 3.3E-03 3.4E-03 3.5E-03-11
-10
-9
-8
-7
-6
-5
1/T (1/K)
ln k
90
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Yields straight line slope = − Ea/R intercept = A
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How To Determine Ea w/ Arrhenius Plot
Pick 2 points on line Point 1 (3.030 × 10−3, −6.20) Point 2 (3.375 × 10−3, −10.4)
91
slope=Δ (ln k )
Δ(1T )
=(−10 . 4−(−6 . 20))(3 . 375−3 . 030)×10−3 K−1
=−4 . 20
3 . 45×10−4K
slope=−1 . 22×104 K=−E a
R
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Determine Ea from Arrhenius Plot
Ea = − (−1.22 × 105 K) * 8.314 J·mol−1K−1
Ea = 1.01 × 105 J/mol
Ea=−slope∗R
92
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Working With Arrhenius Equation
Given the following data, predict k at 75°C using the graphical approach
k (M/s) T °C T, K0.000886 25 2980.000894 50 3480.000908 100 3980.000918 150 448
? 75 348
ln (k) = -36.025/T – 6.908
ln k=−EaR
×1T
+lnA
ln (k) = – 36.025/(348) – 6.908 = – 7.01152
k=e−7 . 012=9 . 01×10−4
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Working with Arrhenius Equation
94
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Method 2. van't Hoff Equation
Sometimes a graph is not needed Only have 2 k s at 2 Ts
Here use van't Hoff Equation Derived from Arrhenius equation
95
ln( k 2
k 1)=−Ea
R ( 1T 2
−1T 1)Pers
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Using van't Hoff EquationEx. CH4 + 2 S2 → CS2 + 2 H2S
k (L/mol·s) T (°C) T (K)1.1 = k1 550 823 = T1
6.4 = k2 625 898 = T2
96
ln(6 . 41. 1)=
−E a
8 .3145 J /K⋅mol ( 1898 K
−1
823 K )
Ea=
−(8 . 314 J /K⋅mol ) ln(6 . 41. 1)
( 1898K
−1
823 )=1 . 4×105 J /mol
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckGiven that k at 25°C is 4.61×10-1 M/s and that at 50°C it is 4.64×10-1 M/s, what is the activation energy for the reaction?
ln(k 2
k 1)=
−E a
R ( 1T 2
−1T 1)
Ea = 208 J/mol
ln(4 . 64×10−1 M/s
4 . 61×10−1 M/s)=
−E a
8. 314J/ ( mol⋅K )( 1323K
−1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70 oC to 80 0C?
A. Rate increases approximately 1.5 timesB. Rate increases approximately 5000 timesC. Rate does not increaseD. Rate increases approximately 3 times
98
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
99
40000 JJ
8.314 x(80 273)Kmol K
2
1 40000 JJ
8.314 x(70 273)Kmol K
Rate is proportional to the rate constant
e 1.49
e
kk
÷
− ÷ ÷+ ÷
÷
− ÷ ÷+ ÷
= =
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Collision Theory and Reaction Mechanisms Sometimes rate law has simple form N2O5 → NO2 + NO3
NO2 + NO3 → N2O5
But others are complex H2 + Br2 → 2 HBr
100
Rate=−d [ N 2O5 ]
dt=k 1[ N 2O5 ]
Rate=−d [ NO2 ]
dt=k 2 [NO 2 ][ NO3 ]
Rate =−d [ H 2 ]
dt=
k [ H 2 ][ Br2 ]1
2
1+k ' [HBr ]
[ Br2 ]
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Why?
Some reactions occur in a single step, as written NO2 bumps into NO3
Bond forms Others involve a sequence of steps Reaction Mechanism Entire sequence of steps
Elementary Process Each individual step in sequence (mechanism) Single step that occurs as written
101
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
What makes an elementary step elementary?
Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process
Rate laws for elementary processes are directly related to stoichiometry
Number of molecules that participate in elementary process defines molecularity of step
102
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Unimolecular Process Only one molecule as reactant H3C—N≡C → H3C—C≡N
Rate = k[CH3NC]
1st order overall As number of molecules ↑, number that rearrange
in given time interval ↑'s proportionally
103
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Bimolecular Process Elementary step with 2 reactants NO (g) + O3 (g) → NO2 (g) + O2 (g)
Rate = k[NO][O3]
2nd order overall From collision theory: If [A] doubles, number of collisions between A and
B will double If [B] doubles, number of collisions between A and
B will double Thus, process is 1st order in A, 1st order in B, and
2nd order overall104
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Termolecular Process Elementary reaction with 3 molecules Very rare
Why? Very low probability that 3 molecules will collide
simultaneously 3rd order overall
105
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Summary of Elementary Processes
Molecularity Elementary Step Rate Law
Unimolecular A → products Rate = k[A]
Bimolecular A + A → products Rate = k[A]2
Bimolecular A + B → products Rate = k[A][B]Termolecular A + A + A → products Rate = k[A]3
Termolecular A + A + B → products Rate = k[A]2[B]
Termolecular A + B + C → products Rate = k[A][B][C]
106
Significance of elementary steps: • If we know that reaction is elementary step• Then we know its rate law
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Multi-Step Mechanisms Contains 2 or more steps to yield net reaction Elementary processes in multi-step mechanism
must always add up to give chemical equation of overall process
Any mechanism we propose must be consistent with experimentally observed rate law
107
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Multi-Step Mechanisms
Ex. Net reaction is:NO2(g) + CO(g) → NO(g) + CO2(g)
Proposed mechanism is:NO2(g) + NO2(g) → NO3(g) + NO(g)
NO3(g) + CO(g) → NO2(g) + CO2(g)
2NO2(g) + NO3(g) + CO(g) → NO2(g) + NO3(g) + NO(g) + CO2(g)
or NO2(g) + CO(g) → NO(g) + CO2(g)
108
1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Intermediates Species which are formed in one step and used
up in subsequent steps Species which are neither reactant nor product
in overall reaction NO3 in this reaction
Product in one step Reactant in later step
Mechanisms may involve one or more intermediates
109
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How Observed Rate Law Relates to Mechanism
Can we deduce mechanism based on rate law?
Complicated problem Sometimes we can make simplifying
assumptions Most important: concept of Rate-determining
or Rate-limiting step Slowest step in mechanism
110
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How Observed Rate Law Relates to Mechanism
If process follows sequence of steps, slow step determines rate.
Think of an assembly line Fast earlier steps may cause intermediates to pile
up Fast later steps may have to wait for slower initial
steps Rate-determining step governs rate law
for overall reaction Can only measure up to rate determining step
—too fast to see111
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Mechanism Examples
(CH3)3CC l(aq) + OH− (aq) → (CH3)3COH (aq) + Cl− (aq)
chlorotrimethylmethane trimethylmethanol
Observed rate = k[(CH3)3CCl]
If reaction was elementary Rate would depend on both reactants Frequency of collisions depends on both
concentrations ∴Mechanism is more complex than single step What is mechanism? Evidence that it is a two step process
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Mechanisms with Slow Initial Step
Step 1: (CH3)3CCl(aq) → (CH3)3C+(aq) + Cl−(aq) (slow)
Step 2: (CH3)3C+(aq) + OH−(aq) → (CH3)3COH(aq) (fast)
Two steps Go at different rates
Each step in multiple step mechanism is elementary process, so Has its own rate constant Has its own rate law
Hence only for each step can we write rate law directly
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Mechanisms with Slow Initial Step
For above reaction Observed rate law says that step 1 is very slow
compared to step 2 i.e. Carbonium ion is formed very slowly
Once it forms, it reacts immediately Thus, rate determining step of overall
reaction is controlled by this slow step ≡ Rate-determining step (RDS)
In this case, RDS = step 1 Overall rate = k1[(CH3)3CCl] Consistent with observed rate
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Mechanisms with Fast Initial Step
1st step involves fast, reversible reactionEx. Decomposition of Ozone (No catalysts)Net reaction: 2 O3 (g) → 3 O2 (g)
Proposed mechanism:
O3 (g) O2 (g) + O (g) (fast)
O (g) + O3 (g) → 2 O2 (g) (slow)115
Observed Rate =k [O3 ]
2
[O2 ]
kf
k2
kr
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! What is the activated complex in the reaction
3O2 (g) ⇌ 2O3(g) ? The * indicates the species is in a high energy (activated) state.
A. [O2]*
B [O3]*
C. [O]*
D. [O4]* The activated complex is formed during the
slow, rate determining step.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Consistent with Observed Rate Law? Rate of formation of O2 = Rate of rxn 2
= k2[O][O3] But O is intermediate Need rate law in terms of reactants and
products and possibly catalysts
Rate (forward) = kf[O3] Rate (reverse) = kr[O2][O] When step 1 comes to equilibrium Rate (forward) = Rate (reverse)
kf[O3] = kr[O2][O]117
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Consistent with Observed Rate Law?
Solving this for intermediate O gives:
Substitution into rate law for step 2 gives:
Rate of RXN 2 = k2[O][O3] = where
This is observed rate law Yes, mechanism consistent
Observed Rate =k [O3 ]
2
[O2 ]
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[O ]=k f [O3 ]
k r [O2 ]
k 2k f [O3 ]2
k r [O2 ]k obs=k 2k f
k r
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
????? Questions ?????1. How do you determine if a proposed
mechanism is consistent with the observed rate law?
2. What do you do if the first step is rate determining?
3. What if the first step is a fast equilibrium and the second step is slow?
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
The reaction mechanism that has been proposed for the decomposition of H2O2 is
1. H2O2 + I– → H2O + IO– (slow)
2. H2O2 + IO– → H2O + O2 + I– (fast)
What is the expected rate law?
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First step is slow so RDSrate=k[H2O2][I–]Pers
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckThe reaction: A + 3 B → D + F was studied and the following mechanism was finally determined:
1. A + B C (fast)2. C + B → D + E (slow)3. E + B → F (very fast)
What is the expected rate law?
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Rate Step 2=k2[C][B] Rate forward = kf[A][B]Rate reverse = kr[C]kf[A][B] = kr[C][C]= kf[A][B]/kr
Rate = kobs[A][B]2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Catalyst Substance that changes rate of chemical
reaction without itself being used up Speeds up reaction, but not consumed by
reaction Appears in mechanism, but not in overall
reaction Does not undergo permanent chemical change Regenerated at end of reaction mechanism May appear in rate law May be heterogeneous or homogeneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
How Does A Catalyst Work?
By providing alternate mechanism One with lower Ea
Because Ea lower, more reactants and collisions have minimum KE, so reaction proceeds faster
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Homogeneous Catalyst Same phase as reactantsConsider : S (g) + O2 (g) + H2O (g) → H2SO4 (g)
S (g) + O2 (g) → SO2 (g) NO2 (g) + SO2 (g) → NO (g) + SO3 (g) Catalytic pathway
SO3 (g) + H2O (g) → H2SO4 (g) NO (g) + ½O2 (g) → NO2 (g) Regeneration of catalyst
Net: S (g) + O2 (g) + H2O (g) → H2SO4 (g)
What is Catalyst? Reactant (used up) in early step Product (regenerated) in later step
Which are Intermediates? 124
NO2 (g)
NO and SO2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Heterogeneous Catalyst
Exists in separate phase from reactants Usually a solid Many industrial catalysts are heterogeneous Reaction takes place on solid catalyst
125
Ex. 3 H2 (g) + N2 (g) → 2 NH3 (g)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Heterogeneous Catalyst
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H2 & N2 approach Fe catalyst
H2 & N2 bind to Fe& bonds break
N—H bonds forming
N—H bonds forming
NH3 formation complete
NH3 dissociates
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Other Examples of Heterogeneous Catalysts
Catalytic converters in cars to remove CO, NO, N2O, and unburned hydrocarbons
Enzymes Proteins that catalyze reactions in living
systems. Insulin, digestive enzymes, nitrogenase, etc.
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
????? Question ????? What is the difference between a catalyst and
an intermediate?
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