Mullis 1 Kinetics Concept of rate of reaction Use of differential rate laws to determine order of...
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Transcript of Mullis 1 Kinetics Concept of rate of reaction Use of differential rate laws to determine order of...
Mullis1
Kinetics
Concept of rate of reaction Use of differential rate laws to determine
order of reaction and rate constant from experimental data
Effect of temperature change on rates Energy of activation; the role of catalysts The relationship between the rate-
determining step and a mechanism
Mullis2
Rate of Reaction
Rate = Δ[concentration] or d [product]
Δ time dt Rate of appearance of a product = rate of
disappearance of a reactant Rate of change for any species is inversely
proportional to its coefficient in a balanced equation.
Mullis3
Rate of Reaction
Assumes nonreversible forward reaction Rate of change for any species is inversely
proportional to its coefficient in a balanced equation.
2N2O5 4NO2 + O2 Rate of reaction = -Δ[N2O5] = Δ[NO2] = Δ[O2]
2 Δt 4 Δt Δtwhere [x] is concentration of x (M) and t is time (s)
Mullis4
Reaction of phenolphthalein in excess base
Use the data in the table to calculate the rate at which phenolphthalein reacts with the OH- ion during each of the following periods:
(a) During the first time interval, when the phenolphthalein concentration falls from 0.0050 M to 0.0045 M.
(b) During the second interval, when the concentration falls from 0.0045 M to 0.0040 M.
(c) During the third interval, when the concentration falls from 0.0040 M to 0.0035 M.
Conc. (M) Time (s)
0.0050 0
0.0045 10.5
0.0040 22.3
0.0035 35.7
0.0030 51.1
0.0025 69.3
0.0020 91.6
Mullis5
Reactant Concentration by Time
Phenolphthalein Concentation in Basic Solution Over Time
0
0.001
0.002
0.003
0.004
0.005
0.006
0 10 20 30 40 50 60 70 80 90 100
Time (s)
Co
nc
. (M
)
Mullis6
Finding k given time and concentration
Create a graph with time on x-axis. Plot each vs. time to determine the graph that gives
the best line:– [A]– ln[A]– 1/[A]– (Use LinReg and find the r value closest to 1)– k is detemined by the slope of best line (“a” in the linear
regression equation on TI-83) – 1st order (ln[A] vs. t): k is –slope– 2nd order (1/[A] vs t: k is slope)
Mullis7
Rate Law Expression
As concentrations of reactants change at constant temperature, the rate of reaction changes. According to this expression.
Rate = k[A]x[B]y… Where k is an experimentally determined rate
constant, [ ] is concentration of product and x and y are orders related to the concentration of A and B, respectively. These are determined by looking at measured rate values to determine the order of the reaction.
Mullis8
Finding Order of a Reactant - Example2ClO2 + 2OH- ClO3
- + ClO2- + H2O
Start with a table of experimental values:
To find effect of [OH-] compare change in rate to change in concentration.
When [OH-] doubles, rate doubles. Order is the power: 2x = 2. x is 1. This is 1st order for [OH-].
[ClO2] (M) [OH-] (M) Rate (mol/L-s)
0.010 0.030 6.00x10-4
0.010 0.060 1.20x10-3
0.030 0.060 1.08x10-2
2x 2x
Mullis9
Finding Order of a Reactant - Example2ClO2 + 2OH- ClO3
- + ClO2- + H2O
Start with a table of experimental values:
To find effect of [ClO2] compare change in rate to change in concentration.
When [ClO2] triples, rate increases 9 times. Order is the power: 3y = 9. y is 2. This is 2nd order for [ClO2].
[ClO2] (M) [OH-] (M) Rate (mol/L-s-1)
0.010 0.030 6.00x10-4
0.010 0.060 1.20x10-3
0.030 0.060 1.08x10-23x 9x
Mullis10
Finding Order of a Reactant - Example2ClO2 + 2OH- ClO3
- + ClO2- + H2O
Can use algebraic method instead. This is useful when there are not constant concentrations of one or more reactants. This example assumes you found that reaction is first order for [OH-] .
6.00 x 10-4=k(0.010)x(.030)1
1.08 x 10-2 = k (0.030)x(.060)1
0.0556 = .333x(.5)
For [ClO2]x , x = 2
[ClO2] (M) [OH-] (M) Rate (mol/L-s-1)
0.010 0.030 6.00x10-4
0.010 0.060 1.20x10-3
0.030 0.060 1.08x10-2
Mullis11
Rate Law:2ClO2 + 2OH- ClO3
- + ClO2- + H2O
Rate = k[ClO2]2[OH-]To find k, substitute in any one set of
experimental data from the table. For example, using the first row:
k = rate/[ClO2]2[OH-]k = 6.00x10-4Ms-1 = 200 M-2s-1
[0.010M]2[0.030M]Overall reaction order is 2+1=3. Note units of k.
Mullis12
Determining k Given Overall Reaction Order
Rate(M/s) = k[A]x
x = overall order of reaction
[A] = the reactant concentration (M)
Overall reaction order Example Units of k
1 Rate=k[A] (M/s)/M = s-1
2 Rate=k[A]2 (M/s)/M2 = M-1s-1
3 Rate=k[A]3 (M/s)/M3 = M-2s-1
1.5 Rate=k[A]1.5 (M/s)/M1.5 = M-0.5s-1
Mullis13
Integrated Rate LawUse when time is given or requested
Relates concentration and time to rate 1st order: ln[A] = -kt + ln[A]0 or [A]=[A]0e-kT
2nd order: 1__ = kt + 1__
[B] [B]0
Wow! y = mx + b
Both equations can be used with linear regression to solve for slope, or k.
Mullis14
Half life for 1st vs 2nd Order Reactions
1st order: t1/2 = 0.693
k
2nd order: t1/2 = 1__
k[A]0
Mullis15
The Arrhenius Equation and Finding Ea
k=Ae-Ea/RT
Where A is the frequency factor– Related to requency of collisions and favorable
orientation of collisions Ea is activation energy in J R = 8.314 J/mol-K T is temp in K k is the rate constant
Mullis16
Using the Arrhenius equation
As Ea increases, rate decreases. – Fewer molecules have the needed energy to react.
As temp increases, rate increases– More collisions occur and increased kinetic energy
means more collisions have enough energy to react.– Mathematically, T is in denominator of the power –
Ea/T.
ln k = -Ea + lnA
T
Mullis17
Activation Energy: Energy vs. Reaction Progress
Ea is lowered with the addition of a catalyst.
Peak is where collisions of reactants have achieved enough energy to react
The arrangement of atoms at the peak is activated complex or the transition state.
A+B --> C+D
0
100
200
300
400
0 2 4 6
Reaction path
En
erg
y (
kJ
)
no catalyst
with catalyst
Ea
ΔH
A + B
C + D
Mullis18
Temperature effects on a reaction
Two factors account for increased rate of reaction.1. Energy factor: When enough energy is in collision for
formation of activation complex, bonds break to begin reaction. With higher temp, more collisions have this energy.
2. Frequency of collision: Particles move faster and collide more frequently with higher temp, increasing chance of reaction.
Increasing temperature increases the rate of a reaction more if the reaction is endothermic to start with.
K increases according to k=Ae-Ea/RT
Mullis19
Finding Ea given info at two temperatures
(or to find a temperature given conditions at one temp and Ea)
ln k1 = Ea [ 1_ - 1_ ]
k2 R T2 T1
Mullis20
Mechanisms: Multistep Reactions
The following reaction occurs in a single step. CH3Br (aq) + OH-(aq) CH3OH(aq) + Br-(aq)
– Rate = k(CH3Br)(OH-) This one occurs in several steps: (CH3) 3CBr(aq) + OH-(aq) (CH3) 3COH (aq) + Br- (aq)
1. (CH3)3CBr (CH3) 3C+ + Br- Slow step
2. (CH3)3C+ + H2O (CH3)COH2+ Fast step
3. (CH3)3COH2+ + OH- (CH3)3COH + H2O Fast step
– Rate = k((CH3)3CBr)
Mullis21
Mechanisms: Multistep Reactions
The overall rate of reaction is more or less equal to the rate of the slowest step. (rate-limiting step)
If only one reagent is involved in the rate-limiting step, the overall rate of reaction is proportional to the concentration of only this reagent.
Ex. For the reaction with Rate = k((CH3) 3CBr)
Although the reaction consumes both (CH3) 3CBr and OH-, the rate of the reaction is only proportional to the concentration of (CH3)3CBr.
The rate laws for chemical reactions can be explained by the following general rules.– The rate of any step in a reaction is directly proportional to the concentrations of
the reagents consumed in that step.– The overall rate law for a reaction is determined by the sequence of steps, or the
mechanism, by which the reactants are converted into the products of the reaction.
– The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch22/rateframe.html
Mullis22
Substituting for an IntermediateStep 1 : N2H2O2 N2HO2- + H + fast equilibriumStep 2: N2HO2- N2O + OH- slowStep 3: H+ + OH- H2O fast
Requirement: A fast equilibrium prior to the rate determining (slow) step that contains the intermediate for which you wish to substitute.
1. N2H2O2 N2HO2- + H+ fast equilibrium
2. For the fast equilibrium, write the rate law (leaving out the k and R) for the reactants and set it equal to the rate law for the products. This can be done because in an equilibrium reaction the forward rate must be equal to the reverse rate.
[N2H2O2] = [N2HO2-] [H+]3. Algebraically solve for the intermediate, N2HO2-
[N2H2O2] / [H+] = [N2HO2-]4. Algebraically substitute into the rate law for N2HO2-
Rate law with intermediate is: R = k [N2HO2-] , so R = k [N2H2O2] / [H+]