Topic 16 Kinetics
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Transcript of Topic 16 Kinetics
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Topic 16 Kinetics• Rate expressions• Reaction mechanism• Activation energy
A + B → C + D
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16.1 Rate expression
• Change in concentration usually affects the rate of reaction
• The change in rate isn’t the same for all reactants (A and B)
• Must be determined by experiment. (Change the concentrations of one reactant and hold the others constant)
A + B → C + D rate =k*[A]a*[B]b
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If a reaction involves the reactants A, B etc =>
The Rate expressionRate of reaction = - d[A]/dt = k[A]a[B]b k = rate constant
• Order of reaction: “a” in substance A and “b” in substance B
• Overall order of reaction: a+b
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What happens to the rate in the reactionA + B C + …
[A] [B] Reaction rate
Order
Double the concentration
Keep constant No change
20 = 1 Zero order
Double the concentration
Keep constant X 2 21 = 2 First order
Double the concentration
Keep constant X 4 22 = 4 Second order
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Reactants A, B and C in four experiments with altering concentrations:A + B + C => ……
Experiment [A]mol*dm-3
[B]mol*dm-3
[C]mol*dm-3
Initial ratemol*dm-3*s-1
1 0.400 1.600 0.0600 4.86
2 0.800 1.600 0.0600 9.72
3 0.400 0.800 0.0600 4.86
4 0.800 1.600 0.1800 87.5
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Compare Experiment 1 and 2:
Initial rate: [A]: 2x[B] and [C] : constant Þ2X rate => [A]1
The reaction is first order in [A]
Experiment [A]mol*dm-3
[B]mol*dm-3
[C]mol*dm-3
Initial ratemol*dm-3*s-1
1 0.400 1.600 0.0600 4.86
2 0.800 1.600 0.0600 9.72
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Compare Experiment 1 and 3:
Initial rate: [A]: constant [B]: ½ [C] : constant Þsame rate => [B]0
The reaction is zero order in [B]
Experiment [A]mol*dm-3
[B]mol*dm-3
[C]mol*dm-3
Initial ratemol*dm-3*s-1
1 0.400 1.600 0.0600 4.86
3 0.400 0.800 0.0600 4.86
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Compare Experiment 2 and 4:
Initial rate: [A]: constant [B]: constant [C] : 3XÞ32 = 9X=> [C]2
The reaction is second order in [C]
Experiment [A]mol*dm-3
[B]mol*dm-3
[C]mol*dm-3
Initial ratemol*dm-3*s-1
2 0.800 1.600 0.0600 9.72
4 0.800 1.600 0.1800 87.5
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Conclusion
• Rate = k*[A]1’*[B]0*[C]2 = k*[A]1*[C]2
• Overall order 1+2 = 3• k can be calculated using the data from one of
the experiments above
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Exercises
• 1-2 on page 120
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The order can also be found in a graph where initial concentration is set against initial rate.
The gradient of the graph => rate of the reaction.
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First order reactionsThey show an exponential decrease:
the time to half the concentration is equal to go from ½ to 1/4
Half life, t½ = 0.693/k
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16.2 Reaction mechanism
Types of reactions: Molecularity• A Products
Unimolecular• A + B Products Bimolecular
• In a Bimolecular reaction the reactants collide and form an activated complex
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Nucleophilic Substitution bimolecular, SN2- topic 10
2 molecules
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If we study the reaction: CH3COCH3 + I2 CH3COCH2I + HI
• It could be a bimolecular process with a rate expression rate = k*[CH3COCH3] *[I2]
• The rate is independent of [I2], but first-order in propanone and acid => rate = k*[CH3COCH3]*[H+]
• The reaction must proceed through a series of steps, a mechanism must be found:
H+
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CH3
C
OH
H2C CH3
CH2C
O
I
CH3
C
O
H3C
H+
CH3
C
OH
H3C
+
Propanone
Fast equilibrium
CH3
C
OH
H3CCH3
C
OH
H2C
++ H+
Slow-Rate DeterminingStep
+ I2 + HI Fast
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• CH3C(OH+)CH3 is known as a intermediate, not an activated complex, though it occur at an energy minimum. In the mechanism there will be several activated complexes
CH3
C
OH
H3C
+
Intermediate
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Activated complex = Transition state, T
CH3
C
OH
H3C
+
CH3
C
O
H3C
CH3
C
OH
H2C
CH3
CH2C
O
I
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Exercises
• 1 and 2 on page 122
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16.3 Activation energyRecall: Maxwell-Boltzmann energy distribution curve.
Temperature Average speed
Higher temperature =>More particles with higher speed => Greater proportion of particles with energy enough
to react
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The Arrhenius equation
• The rate constant, k, can be given if collision rate and orientation is given
• Ea = activation energy• T = temperature, K• R = Gas constant
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The equation can also be given in a logarithmic form:
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Exersize: Consider the following graph of ln k against (temperature in Kelvin) for the second order decomposition of N2O into N2 and O
N2O → N2 + O
(a) State how the rate constant, k varies with temperature, T(b) Determine the activation energy, Ea, for this reaction.(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3 mol–1 s–1 at 750 °C.A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted.
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Solution:(a) State how the rate constant, k varies with temperature, TThe Arrhenius equation (it’s in the Data booklet): k=Ae(-Ea/RT)
Logaritming the equation on both sides:lnk=lnA –Ea/RT In the graph we see that the gradient is negativeAnswer: when k increases T decreases
(b) Determine the activation energy, Ea, for this reactionThe gradient (-Ea/R) can be calculated from the graph:DY/DX= -3/0,1*10-3= -3*104= -30000
Therefore:Ea=gradient*R=-30000*8.31= 2,49*105
Answer: Ea= 2,49*105
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Solution:(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted.
10 % has reacted → 90 % left → 0.200*0.9= 0.180 mol/dm3
Rate= 0.244 * [0.180]2 = 7.91*10-3