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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 expression Rate 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 reaction A + B C + … [A][B] Reaction rate Order Double the concentration Keep constant No change 2 0 = 1 Zero order Double the concentration Keep constant X 22 1 = 2 First order Double the concentration Keep constant X 42 2 = 4Second 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 rate mol*dm -3 *s

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Compare Experiment 1 and 2: Initial rate: [A]: 2x [B] and [C] : constant 2X rate => [A] 1 The reaction is first order in [A] Experiment [A] mol*dm -3 [B] mol*dm -3 [C] mol*dm -3 Initial rate mol*dm -3 *s

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Compare Experiment 1 and 3: Initial rate: [A]: constant [B]: ½ [C] : constant same rate => [B] 0 The reaction is zero order in [B] Experiment [A] mol*dm -3 [B] mol*dm -3 [C] mol*dm -3 Initial rate mol*dm -3 *s

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Compare Experiment 2 and 4: Initial rate: [A]: constant [B]: constant [C] : 3X 3 2 = 9X=> [C] 2 The reaction is second order in [C] Experiment [A] mol*dm -3 [B] mol*dm -3 [C] mol*dm -3 Initial rate mol*dm -3 *s

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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 reactions They show an exponential decrease: the time to half the concentration is equal to go from ½ to 1/4 Half life, t ½ = 0.693/k

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16.2 Reaction mechanism Types of reactions:Molecularity A ProductsUnimolecular A + B ProductsBimolecular In a Bimolecular reaction the reactants collide and form an activated complex

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Nucleophilic Substitution bimolecular, S N 2- topic 10 2 molecules

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If we study the reaction: CH 3 COCH 3 + I 2 CH 3 COCH 2 I + HI It could be a bimolecular process with a rate expression rate = k*[CH 3 COCH 3 ] *[I 2 ] The rate is independent of [I 2 ], but first-order in propanone and acid => rate = k*[CH 3 COCH 3 ]*[H + ] The reaction must proceed through a series of steps, a mechanism must be found: H+H+

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CH 3 C(OH + )CH 3 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 Intermediate

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Activated complex = Transition state, T

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Exercises 1 and 2 on page 122

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16.3 Activation energy Recall: 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 E a = 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 N 2 O → N 2 + 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 [N 2 O] 2 and the rate constant is dm 3 mol –1 s –1 at 750 °C. A sample of N 2 O of concentration mol dm –3 is allowed to decompose. Calculate the rate when 10 % of the N 2 O has reacted.

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Solution: (a) State how the rate constant, k varies with temperature, T The Arrhenius equation (it’s in the Data booklet): k=Ae (-Ea/RT) Logaritming the equation on both sides: lnk=lnA –E a /RT In the graph we see that the gradient is negative Answer: when k increases T decreases (b) Determine the activation energy, E a, for this reaction The gradient (-E a /R) can be calculated from the graph: Y/ X= -3/0,1*10 -3 = -3*10 4 = Therefore: E a =gradient*R=-30000*8.31= 2,49*10 5 Answer: E a = 2,49*10 5

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Solution: (c) The rate expression for this reaction is rate = k [N 2 O] 2 and the rate constant is dm 3 mol –1 s –1 at 750 °C. A sample of N 2 O of concentration mol dm –3 is allowed to decompose. Calculate the rate when 10 % of the N 2 O has reacted. 10 % has reacted → 90 % left → 0.200*0.9= mol/dm3 Rate= * [0.180] 2 = 7.91*10 -3

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