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CHEMICAL KINETICS

CHEMICAL KINETICS. Reaction Mechanism Rate Law Order of Reactions. Chemical Kinetics  by Dr. Akinyugha  is licensed under a  Creative Commons Attribution- NonCommercial 4.0 International License. Reaction Mechanism Most reactions occur in a series of steps at a molecular level

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CHEMICAL KINETICS

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  1. CHEMICAL KINETICS • Reaction Mechanism • Rate Law • Order of Reactions Chemical Kinetics by Dr. Akinyugha is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

  2. Reaction Mechanism • Most reactions occur in a series of steps at a molecular level • Each step is called an elementary step or reaction • Consequently, reaction mechanism is the sequence of elementary steps that lead to the formation of the product • The mechanism shows the order in which bonds are broken and formed and the rate of each of the elementary steps • The slowest step or reaction is called the rate determining step or reaction

  3. Reaction Mechanism Rate determining step or reaction • The rate determining step is known by calculating the rate of reaction for each elementary step • The rate determining step is not necessarily the first elementary step of the reaction.

  4. Reaction Mechanism Two requirements for balanced reaction mechanism: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The mechanism must agree with the experimentally determined rate law. • Intermediates formed, if any, must be detected experimentally, the presence of which is used to arrive at the mechanism of the reaction

  5. 2NO (g) + O2(g) 2NO2(g) Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Reaction Mechanism The sum of the elementary steps must give the overall balanced equation for the reaction For the production of Nitrogen Dioxide – However, N2O2is detected during the reaction N2O2 is an intermediate in the reaction Intermediatesare species that appear in a reaction mechanism but not in the overall balanced equation They can be detected using spectroscopy

  6. Reaction Mechanism 1. The decomposition of Hydrogen peroxide is catalysed by iodide ion thus – H2O2 + I- H2O + IO- H2O2 + IO- H2O + O2 + I- • Write the overall net reaction for the equations • Identify the intermediate(s) in the equation • Ozone decomposes into Oxygen in the following manner – O3⇌ O2 + O O3 + O  2O2 • Identify the intermediate • Write the overall chemical reaction 3. Tetrachloromethane is formed from the chlorination of chloroform thus – Cl2⇌ 2Cl Cl + CHCl3 HCl + CCl3 Cl + CCl3  CCl4 Write out the net or overall reaction of the process.

  7. Reaction Mechanism Molecularity Molecularity is the number of molecules that react in each elementary step or reaction Rate Law for Elementary Processes: Elementary Reactions Molecularity Rate Law  A  product Unimolecular Rate = k[A] 2A  product Bimolecular Rate = k[A]2 A + B  product Bimolecular Rate = k[A][B] 2A + B  product Termolecular Rate = k[A]2[B] 

  8. Rate Law The Rate Law expresses the relationship between • The rate of reaction • The rate constant • The concentration of reactants Rate Law cannot be determined from the knowledge of stoichiometry alone The Rate Law is usually experimentally determined

  9. Rate Law For a given reaction where A B Rate of reaction, R is proportional to concentration of A R ∝ A To get rid of the proportionality sign, a constant, k is introduced R = k [A] Where k = Rate Constant, [A] is the concentration of A in mol/dm3 • Direct proportionality means that if you increase the concentration of A, the Rate of Reaction increases • So, if you double the concentration of A, the Rate of Reaction will double

  10. Rate Law: Order of Reaction For the rate law for a reaction between A and B: R = k[A]x [B]y x and y are the order of the reaction, i.e., • The reaction is said to be of xth order with respect to A and yth order with respect to B • The overall order of the reaction is the sum of the individual orders, i.e., Overall Order = x + y • The order of a reaction is experimentally determined • Note the order of the reaction is NOT the stoichiometric coefficient of the reaction

  11. Rate Law: Order of Reaction Examples 1. For the reaction: H2 + O2 H2O The Rate Law is given as follows: Rate = k[H2] [O2] The order of the reaction is 1 with respect to H2 and 1 with respect to O2 2. For the reaction: N2 + 3H2  2NH3 The Rate Law is given as follows:Rate = k[N2]2[H2] The order of the reaction is 2 with respect to N2 and 1 with respect to H2 3. For a reaction: 3A + B + 2C  4D + E The Rate Law is given as: Rate = k[A]1[B]3[C]1 The Order of the reaction is 1 with respect to A, 3 with respect to B and 1 with respect to C

  12. Zero Order Reaction For A  Product The Rate Law for the Zero Order Reaction is given as Rate = k [A]0 = k The Rate of Reaction of a Zero Order Reaction is independent of the Concentration of the Reactant A That is, as you increase the concentration of A, the rate of reaction increases. Graphically, first order reactions are: Slope = 0 Rate [A]

  13. Zero Order Reaction: Example Decomposition of Nitrogen Dioxide 2N2O  2N2 + O2 (Pt catalyst) The reaction is Zero Order provided the pressure is high enough to saturate the surface of the metal catalyst Unit of Zero Order Rate Constant Zero Order Rate Equation: Rate = k = Moldm-3 s -1

  14. First Order Reactions For A  Product The Rate Law for the First Order Reaction is given as Rate = k [A] The Rate of Reaction of a First Order Reaction is directly proportional to the Concentration of the Reactant A That is, as you increase the concentration of A, the rate of reaction increases. Graphically, first order reactions are: Rate [A]

  15. First Order Reactions: Examples Decomposition of DinitrogenPentoxide 2N2O5 4NO2 + O2 Rate = k [N2O5] Radioactive decay reactions are essentially First Order reactions 226 88 Ra  222 86 Rn + 4 2 He Unit of First Order Rate Constant First Order Rate Equation: Rate = k [A] Rate is Concentration divided by time; that is, M/secs or mol/dm3/seconds or moldm-3 s -1 A is concentration in mol dm-3. Therefore, k = s-1

  16. Second Order Reactions There are two types of Second Order Reactions • Second Order Reaction depending on one reactant A  Product The Rate Law for the Second Order Reaction is given as Rate = k [A]2 Here, the Rate of Reaction is directly proportional to the square of the Concentration of the Reactant A Rate [A]2

  17. Second Order Reactions • Second Order Reaction depending on two reactants A + B  Product The Rate Law for the Second Order Reaction is given as Rate = k [A][B] Here, the Rate of Reaction is directly proportional to [A] when [B] is constant and directly proportional to [B] when [A] is constant Rate Rate When [B] is constant When [A] is constant [A] [B]

  18. Unit of Second Order Rate Constant First Order Rate Equation: Rate = k [A] [B] Rate is Concentration divided by time; that is, M/secs or mol/dm3/seconds or moldm-3 s -1 [A] = [B] = Concentration in mol dm-3

  19. Determination of Rate Equations by Initial Rate Method • Rate Equations consist of the order of reaction and the rate constant • The rate equation can be determined by measuring experimental changes in the initial rate of reaction as the initial concentration of the reactants changes • Consequently, the order of reaction can be calculated and the rate constant determined

  20. Example The initial rate of the reaction between A and B is given in the table below. From the information, calculate • The order of the reaction with respect to A and B • The rate constant of Experiment I • The rate law of the reaction

  21. Answer Let the hypothetical rate equation be Rate = k [A] x [B] y Where k – rate constant, [A] is Conc of A, [B] is Conc of B, x is the order with respect to A and y is the order with respect to B Rate Equation for the three experiments is thus – 0.0010 = k [0.02] x[0.04] y I 0.0020 = k [0.04] x[0.04] y II 0.0040 = k [0.02] x[0.08] y III On inspection, y is the same for Equation I and II. So, divide I by II Rate (I) = 0.0010 = k [0.02] x [0.04] y Rate (II) = 0.0020 = k [0.04] x [0.04] y 0.5 = (0.5)x Therefore, x = 1 Divide I by III and solve likewise, y = 2. Reaction is First Order with respect to A and second order with respect to B ii. Rate = k [0.02] x [0.04] y =k [0.04] 1[0.04] 2 = iii. Rate = k [A] 1[B] 2

  22. Example An experiment was carried out to find out the rate of decomposition of Ozone into Oxygen. The following data was recorded at 350K. From the table, determine • The order of the reaction with respect to Ozone • The rate constant and state its unit • The rate law of the reaction Chemical Kinetics by Dr. Akinyugha is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

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