hydrophilic and hydrophobic colloids chapter 14, chemical ...mathews/chem122wi03/... · chapter 14,...
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1
Chapter 14, Chemical Kinetics
(continued)
Quizzes next week will be on Chap 14through section 14.5.
Last week we covered the following material:
Review Vapor Pressure with two volatile components
13.6 ColloidsHydrophilic and Hydrophobic ColloidsRemoval of Colloidal Particles
Chapter 14 Chemical Kinetics
14.1 Reaction RatesRates in Terms of ConcentrationsReaction Rates and Stoichiometry
14.2 The Dependence of Rate on ConcentrationsReaction OrderUnits of Rate ConstantsUsing Initial Rates to Determine Rate Laws
14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions
14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation
14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step
Week Five Chemical Kinetics (cont)
Results for a reaction of A → B
C4H9Cl(aq) + H2O (l) C4H9OH (aq) + HCl (aq)
tOHHC
tClHCRate
∆∆
+=∆
∆−=
][][ 9494
Note the signs!
2
In fact, the instantaneous rate corresponds to d[A]/dt
Consider the reaction 2 HI(g) H2(g) + I2(g)
It’s convenient to define the rate as
tI
tH
tHIrate
∆∆
+=∆
∆+=
∆∆
−=][][][
21 22
And, in general for
aA + bB cC + dD
tD
dtC
ctB
btA
aRate
∆∆
=∆∆
=∆∆
−=∆∆
−=][1][1][1][1
Sample exercise 14.2
The decomposition of N2O5 proceeds according to the equation
2 N2O5 (g) 4 NO2 (g) + O2 (g)
If the rate of decomposition of of N2O5 at a particular instant in a vessel is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 ?
tO
tNO
tONRate
∆∆
+=∆
∆+=
∆∆
−=][
11][
41][
21 2252
i.e. the rate of the reaction is 2.1 x 10-7 M/s
the rate of appearance of NO2 is 8.4 x 10-7 M/s
and the rate of appearance of O2 is 2.1 x 10-7 M/s
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆tmin min mol/L mol/L mol/L-min
0 2.33
184 2.08
319 1.91
526 1.67
867 1.35
1198 1.11
1877 0.72
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆tmin min mol/L mol/L mol/L-min
0 2.33184 0.25 1.36 x 10-3
184 2.08135 0.17 1.26 x 10-3
319 1.91207 0.24 1.16 x 10-3
526 1.67341 0.32 0.94 x 10-3
867 1.35331 0.24 0.72 x 10-3
1198 1.11679 0.39 0.57 x 10-3
1877 0.72
The information on the previous slide is a bit of a nuisance, since the instantaneous rate keeps changing—and you know how much we like constant values or linear relationships!
So let’s try something rather arbitrary at this point.
Let’s divide the instantaneous, average rate by
[N2O5] and/or [N2O5]2
3
2 N2O5 = 4 NO2 + O2 (g)
at T = 45 oC in carbon tetrachloride as a solvent
{N2O5] [N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t Avg rate Avg ratemol/L avg mol/L mol/L-min /[N2O5]av /[N2O5]av
2
2.332.21 0.25 1.36 x 10-3 6.2 x 10-4 2.8 x 10-4
2.082.00 0.17 1.26 x 10-3 6.3 x 10-4 3.2 x 10-4
1.911.79 0.24 1.16 x 10-3 6.5 x 10-4 3.6 x 10-4
1.671.51 0.32 0.94 x 10-3 6.2 x 10-4 4.1 x 10-4
1.351.23 0.24 0.72 x 10-3 5.9 x 10-4 4.8 x 10-4
1.110.92 0.39 0.57 x 10-3 6.2 x 10-4 6.7 x 10-4
0.72
Notice the nice constant value!!!
It’s convenient to write this result in symbolic form:
Rate = k [N2O5]
where the value of k is about 6.2 x 10-4
so that when [N2O5] = 0.221,
Rate = (6.2 x 10-4 )(0.221)
= 1.37 x 10-4 which is the ‘average rate’we started with
In fact, we really should take into account the 2 in front of theN2O5, in accordance with the rule we developed earlier.
This leads us to the general concept of Reaction Order
When Rate = k [reactant 1]m [reactant 2]n
we say the reaction is m-th order in reactant 1
n-th order in reactant 2
and (m + n)-th order overall.
Be careful—because these orders are NOT related necessarily
to the stoichiometry of the reaction!!!
2 N2O5 = 4 NO2 + O2 (g) Rate = k [N2O5] !!!
CHCl3 (g) + Cl2 (g) CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2
H2 (g) + I2 (g) 2 HI (g) Rate = k[H2][I2]
Other reactions and their observed reaction orders
The order must be determined experimentally!!!
We’ll see later that it depends on the Reaction Mechanism,rather than the overall stoichiometry.
Let’s explore the results for the result Rate = k [N2O5]
This can be expressed as Rate = - (∆[N2O5] / ∆ t = - d[N2O5] / dt = k [N2O5]
or, in general for A products
Rate = - ∆[A] / ∆t = d[A] / dt = k [A]
rearrangement and integration some time = t and t =0 gives the result
ln[A]t - ln[A]o = -kt
or ln [A]t = -kt + ln [A]o
or ln ([A]t/[A]o = - kt
This is the expression of concentration vs time
for a First-Order Reaction
An example of the plots of concentration vs timefor a First-Order Reaction
4
The Change of Concentration with TimeThe Change of Concentration with TimeHalfHalf--LifeLife• Half-life is the time taken for the
concentration of a reactant to drop to half its original value.
• That is, half life, t1/2 is the time taken for [A]0to reach ½[A]0.
• Mathematically,
kkt 693.0ln 2
1
21 =−=
The Change of Concentration with TimeThe Change of Concentration with TimeFor a FirstFor a First--Order ReactionOrder Reaction
The identical length of thefirst and second half-life
is a SPECIFIC characteristicof First-Order reactions
Consider now Second-Order Reactions
2][][][ AkdtAd
tARate −==∆∆
−=
∫∫ −=tA
Adtk
AAdt
0
][
][ 20 ][
][
0][1
][1
Akt
A t
+=
Example of Second-Order Plots of conc vs time
5
SecondSecond--Order ReactionsOrder Reactions• We can show that the half life
• A reaction can have rate constant expression of the form
rate = k[A][B],i.e., is second order overall, but has first order dependence on A and B.
[ ]0A1
21 k
t =
t1/2 = 0.693/0.4 = 1.73 sec
t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1= 2.5 sec
t1/2 = [(0.4)(0.5)] -1= 5.0 sec
General Order of reaction
First Order reactions
Second Order reactions
Integrated form of each
Half lives of each
14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions
14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation
14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step
Week Five Chemical Kinetics (cont)
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MQ1 Results:
Hi 175/175
Lo 36/175
Mean: 125 (71.4 %)
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Chapter 14, Chemical Kinetics
(continued)
14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions
14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation
14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step
14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions
14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation
14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step
14.3 The Change of Concentration with Time
Mechanisms with and Initial Fast Step
Week Five Chemical Kinetics (cont)
Note the DRAMATIC effect of temperature on kTemperature and RateTemperature and Rate——Collision FrequencyCollision Frequency
The Collision Model The Collision Model egeg HH22 + I+ I22
The Collision ModelThe Collision Model• The greater the frequency of collision the faster
the rate for a given concentration.• Complication: not all collisions lead to products. In
fact, only a small fraction of collisions lead to product. Add an Orientation Factor.
• The higher the temperature, the more energyavailable to the molecules and the faster the rate.
• In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.
Activation EnergyActivation Energy• Arrhenius: molecules must posses a
minimum amount of energy to react. Why?– In order to form products, bonds must be broken
in the reactants.– Bond breakage requires energy.
• Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.
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Activation EnergyActivation Energy Activation EnergyActivation Energy• Consider the rearrangement of acetonitrile:
– In H3C-N≡C, the C-N≡C bond bends until the C-N bond breaks and the N≡C portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.
– The energy required for the above twist and break is the activation energy, Ea.
– Once the C-N bond is broken, the N≡C portion can continue to rotate forming a C-C≡N bond.
H3C N CC
NH3C H3C C N
Activation EnergyActivation Energy• The change in energy for the reaction is the
difference in energy between CH3NC and CH3CN.
• The activation energy is the difference in energy between reactants, CH3NC and transition state.
• The rate depends on Ea.• Notice that if a forward reaction is exothermic
(CH3NC → CH3CN), then the reverse reaction is endothermic (CH3CN → CH3NC).
‘‘Effective Collisions’Effective Collisions’• Consider the reaction between Cl and NOCl
to produce Cl2 and NO - maybe:– If the Cl collides with the Cl of NOCl then the
products are Cl2 and NO.– If the Cl collided with the O of NOCl then no
products are formed.• We need to quantify this effect.
Effective & Ineffective CollisionsEffective & Ineffective Collisions
8
The Arrhenius Equation The Arrhenius Equation ---- FinallyFinally• Arrhenius discovered most reaction-rate data
obeyed the Arrhenius equation:
– k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.
– A is called the frequency factor, and is a measure of the probability of a favorable collision (including the collision frequency).
– Both A and Ea are specific to a given reaction.
RTaE
Aek−
=
RTaE
Aek−
=
The Arrhenius EquationThe Arrhenius Equation• If we have a lot of data, we can determine Ea
and A graphically by rearranging the Arrhenius equation:
• If we do not have a lot of data, or if we don’t know A, then we can use
ATEk a lnR
ln +−=
−
−=
121
2 11R
lnTT
Ekk a
Sample Exercise 14.8 For CH3NC CH3CN
Temp. /oC k / (s-1 )
189.7 2.52 x 10-5
198.9 5.25 x 10-5
230.3 6.30 x 10-5
251.2 3.16 x10-5
Sample Exercise 14.8 For CH3NC CH3CN
Temp. (oC) T (K) 1/T (K-1) k (s-1 ) ln k
189.7 2.52 x 10-5
198.9 5.25 x 10-5
230.3 6.30 x 10-5
251.2 3.16 x10-5
Sample Exercise 14.8 For CH3NC CH3CN
Temp. (oC) T (K) 1/T (K-1 x 103) k (s-1 ) ln k
189.7 462.9 2.160 2.52 x 10-5
198.9 472.1 2.118 5.25 x 10-5
230.3 503.5 1.986 6.30 x 10-5
251.2 524.4 1.907 3.16 x10-5
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Sample Exercise 14.8 For CH3NC CH3CN
Temp. (oC) T (K) 1/T (K-1 x 103) k (s-1 ) ln k
189.7 462.9 2.160 2.52 x 10-5 -10.589
198.9 472.1 2.118 5.25 x 10-5 -9.855
230.3 503.5 1.986 6.30 x 10-5 -7.370
251.2 524.4 1.907 3.16 x10-5 -5.757
Fig 14.18 For CH3NC CH3CN reaction
From the graph we find the slope = -1.9 x 104 K
But this is also equal to - Ea/R
Or Ea = -(slope)(R)
= - ( - 1.9x104 )(8.314 J mol-1 K-1)(1 kJ / 1000 J)
= 1.62 x 102 kJ/mol
or 162 kJ/mol
We can now use these results to calculate the rate constant at any temperature.
−
−=
121
2 11R
lnTT
Ekk a
To calculate k1 for a temperature of 430.0 K, make substitutionsfor all other parameters, after choosing a starting point:
k2 = 2.52 x 10-5 s-1
T2 = 462.9 KAnd T1 = 430.0 K
to obtain k1 = 1.0 x 10-6 s -1
RTaE
Aek−
= Catalysts serve to reduce the activation energy, thus increasing k.
10
Reaction Mechanisms
Involve Elementary Steps – or Reactions
When we know these, we CAN write downdirectly the rate expressions!!!
Furthermore, these elementary steps represent fundamental molecular processes—
and lead to the term molecularity of the elementary reaction.
Elementary StepsElementary Steps• Elementary steps must add to give the
balanced chemical equation.• Intermediate: a species which appears in an
elementary step which is not a reactant or product.
Rate Laws of Elementary StepsRate Laws of Elementary Steps• The rate law of an elementary step is
determined by its molecularity:– Unimolecular processes are first order,– Bimolecular processes are second order, and– Termolecular processes are third order.
Rate Laws of Multistep MechanismsRate Laws of Multistep Mechanisms• Rate-determining step: is the slowest of the
elementary steps.• Therefore, the rate-determining step governs
the overall rate law for the reaction.Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• It is possible for an intermediate to be a
reactant.• Consider
2NO(g) + Br2(g) → 2NOBr(g)
Reaction MechanismsReaction MechanismsMechanisms with an Initial Fast StepMechanisms with an Initial Fast Step
2NO(g) + Br2(g) → 2NOBr(g)• The experimentally determined rate law is
Rate = k[NO]2[Br2]• Consider the following mechanism
for which the rate law is (based on Step 2):
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• The rate law is (based on Step 2):
Rate = k2[NOBr2][NO]• The rate law should not depend on the
concentration of an intermediate because intermediates are usually unstable.
• Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBrand Br2 assuming there is an equilibrium in step 1 we have
[ ] ]Br][NO[NOBr 21
12
−=
kk
Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• By definition of equilibrium (see Chap 15):
k1[NO][Br2] = k-1[NOBr2]• Therefore, the overall rate law becomes
• Note the final rate law is consistent with the experimentally observed rate law.
]Br[]NO[Rate 22
11
2−
=kkk