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Chemical Kinetics

Chemical Kinetics. Chemical Kinetics = study of the rates of chemical reactions = how fast reactants are converted to products. Factors affecting reaction rates : a. concentrations of reactants b. physical state, surface area of reactants c. temperature at which reaction occurs

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Chemical Kinetics

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  1. Chemical Kinetics • Chemical Kinetics = study of the rates of chemical reactions = how fast reactants are converted to products. • Factors affecting reaction rates : • a. concentrations of reactants • b. physical state, surface area of reactants • c. temperature at which reaction occurs • d. presence of a catalyst • Catalyst : substance that increases the rate of a reaction without being consumed in the overall reaction

  2. Reaction Rates • Reactants Products • Rate of reaction measured either as the rate of change of concentration of a reactant or as the rate of change of concentration of a product. • Therefore units of reaction rate usually M/s • Reaction Rate has to be a positive quantity • Reaction Rates are measured either as average rates or as instantaneous rates.

  3. Average Reaction Rates Average Reaction Rate in terms of reactant concentrations: Average Reaction Rate in Terms of Product Concentrations :

  4. Instantaneous Rates • Reactant concentrations decrease as the reaction progresses. • Reaction rates are dependant on reactant concentration. • Reaction rates therefore change with time as the reaction progresses • The reaction rate at any instant is called the instantaneous rate • The instantaneous rate at any time is the slope of the concentration-time curve at that time.

  5. Reaction Rates and Reaction Stoichiometry

  6. Molarity is represented by square brackets. For instance the molarity of a solution of a solute A = [A] = moles of A/L of solution [B] = moles of B/L of solution and so on. Change is represented by =final value – initial value [A] = final molarity of A – initial molarity of A If A is a reactant, to express the change as a positive number we use - [A] instead of [A]

  7. Consider the reaction: BrO3-(aq) +3SO32-(aq) Br-(aq) +3SO42-(aq). If the rate of formation of the sulfate ion 2.40Ms-1, calculate the rates of decomposition of the bromate ion and the sulfite ion and the rate of formation of the bromide ion. • The oxidation of iodide ions by arsenic acid in acidic aqueous solution occurs as follows: H3AsO4 + 3I¯ + 2 H3O+ H3AsO3 + I3¯ + H2O. The experimentally determined rate law for this reaction is • Rate = k [H3AsO4] [I¯] [H3O+]. • What is the order of the reaction with respect to I¯? • 3. The rate law for the decomposition of HI to H2 and I2 is • rate = k [HI]2. • The rate of the reaction 2.5  10-4 Ms-1 when the initial concentration of HI is 0.0558M. What is the value of the rate constant, k.?

  8. Concentration and Rate Rate Law : Equation relating instantaneous reaction rate and the reactant concentrations General form of a rate law for a general reaction aA+bBcC+dD : Rate Law : Rate = k[A]m[B]n k = rate constant m = order of the reaction with respect to reactant A n = order of reaction with respect to reactant B m + n = overall order of the reaction

  9. Rate Law is determined experimentally. One such experiment is to observe the effect of starting reactant concentrations on initial reaction rates . Reaction orders m, n can be positive or negative numbers or 0. They can be fractions or integers. They are constants for a reaction. A zeroth order reaction has an overall order of 0 A first order reaction has an overall order of 1 A second order reaction has an overall order of 2 and so on Rate constants, k, are temperature dependant. The value of k will change if a catalyst is introduced. However, the values of the orders, m, n, will not change if T is changed, the physical states are changed or if a catalyst is introduced. Rate Law

  10. Consider the hypothetical reaction A+2B + 3C  Products. • Calculate the rate law from the data in the table below: • Calculate the rate at which C is used when [A]=[B]=[C] = 0.5 M • Calculate the rate at which A is used under the same conditions • Calculate the rate at which B is used under the same conditions

  11. Units of Rate : Ms-1 Orders do not have units. But the rate constant does. Units of rate constant depend on the overall order of the reaction. For a reaction with overall order N, the units of rate constant = M(1-N)s-1 Units of k for a zeroth order reaction : Ms-1 Units of k for a 1st order reaction : s-1 Units of k for a 2nd order reaction : M-1 s-1 Units in Rate Law

  12. Rate Law from Initial Rates and Initial Reactant Concentrations For any reaction, the rate constant, the orders with respect to the reactants and the form of the rate law do not change with reactant concentrations. Thus the orders, rate constant and hence the rate law for a given reaction can be obtained by comparing the change in the initial instantaneous rates with changes in initial concentrations of one or more of the reactants. If in the rate law for a reaction, the order with respect to a reactant is zero, the rate of the reaction will not change with changes in the concentrations of that reactant. For example let the order with respect to reactant A = 0 for the reaction aA +bB  cC + dD; Rate = k[A]m[B]n = k[A]0[B]n = k[B]n Since [A] does not appear in the rate law, the rate won’t change if [A] alone is changed.

  13. Integrated Rate Laws: Concentration Change with Time First Order Reaction : Second Order Reaction : [A]t = Concentration of Reactant A after elapsed time t [A]0 = Initial concentration of Reactant A k = rate constant of the reaction

  14. Half-Life, t1/2 The half-life of a reaction is the time required for the concentration of a reactant, A, to drop to one half of the initial concentration. That is, at t = t1/2 , Half-life of a first order reaction: The half-life of a 1st order reaction is independent of the initial concentration of the reactant Half-life of a second order reaction : The half-life of a 2nd order reaction is inversely proportional to the initial reactant concentration.

  15. The half-life of a second order reaction is 180s. The initial reactant concentration is 0.100M. Calculate the value of the rate constant of the reaction. • 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) • Experiments conducted to study the rate of the reaction represented by the equation above. Initial concentrations and rates of reaction are given in the table below. • Determine the order for each of the reactants, NO and H2, from th data given and show your work.; Write the overall rate law for the reaction; Calculate the value of the rate constant, k, for the reaction. Include units.

  16. 4. The isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 minutes at 500 °C. a. What is the value of the rate constant? (include units) b. The time it takes for the concentration of cyclopropane to decrease from 1.00 M to 0.125 M at 500 °C 5. The rate constant for the second order decomposition of HI at 700K is k=0.012 M-1s-1. In a reaction, if the initial concentration of HI = 0.56M, what will be the concentration of HI at t= 2.00h? 6. The rate constant of a reaction 8.58 x 10-5 M-1s-1 at 374°C and it is 5.11 x 10-4 M-1s-1 at 410°C. What is the activation energy for the reaction? 1.82 x 105 J/mol

  17. 1. The activation energy for a certain reaction is 183kJ/mol. If the rate constant is 2.91 x 10-6 s-1 at 200C. Calculate the rate constant at 300C. Answer = 0.00980 s-1

  18. Temperature and Reaction Rate The rate constant, k, increases with increasing temperature : Arrhenius equation Therefore the rates of chemical reactions increase with increase in temperature. Collision Model : Molecules must collide to react Increasing the temperature increases molecular velocities which would result in harder and more frequent collisions. The reaction rate will increase if the number of collisions per second increases.

  19. Collision Model contd. • Not all collisions are productive because : • Reactions involve rearrangement of atoms which involves breaking and forming bonds which requires energy • The minimum energy required to initiate a reaction is called the Activation Energy, Ea. • The colliding reactant molecules must have a total kinetic energy greater than or equal to the activation energy for the reaction. The fraction, f, of molecules that has an energy equal to or greater than Ea at a temperature T (K) is . Larger the temperature, greater the fraction of energetic molecules

  20. Collision Model Contd. The arrangement of atoms at the top of the activation energy barrier is called the transition state of the reaction or the activated complex. 2. In addition to having enough kinetic energy, the colliding molecules must also have the right relative orientations during collisions to result in product formation. The frequency factor, A, of the reaction is a measure of the frequency of the collisions and the probability that the orientations of the reactant molecules during the collisions is favorable. Larger the value of A, more the number of productive collisions

  21. Arrhenius Equation It relates the rate constant, k, of a chemical reaction to the absolute reaction temperature, T : R = gas constant = 8.314 JK-1 mol-1 A = Frequency Factor; Ea = activation energy A, Ea do not change with T. They only change when a catalyst is introduced or physical states or surface areas of reactants are changed. k1 = rate constant at absolute temperature T1 k2 = rate constant at absolute temperature T2

  22. Reaction Mechanisms The series of single steps, called elementary steps, by which the reaction takes place is its reaction mechanism. Molecularity : The number of molecules that participate as reactants in an elementary step. Unimolecular Step : An elementary step involving 1 reactant molecule . Bimolecular step : An elementary step involving 2 reactant molecules . Termolecular Step : An elementary step involving 3 reactant molecules .

  23. A multi-step reaction mechanism will have more than one elementary step. The elementary steps in a multi-step mechanism must always add to give the chemical equation of the overall reaction. Intermediates in a reaction mechanism are substances formed in one elementary step that are consumed in a subsequent elementary step. Intermediates won’t appear in your overall equation because they cancel out when the equations for the elementary steps are added. Multi-step mechanisms involve one or more intermediates. The rate law of an elementary step can be written directly from its stoichiometry unlike an overall reaction. That is, given the balanced equation for an elementary step, you can write the rate law for that step. The orders = coefficients only for an elementary step. Therefore, the overall order of an elementary step is equal to its molecularity = sum of the coefficients of the reactants alone.

  24. Catalyst versus Intermediate in a Reaction Mechanism • Neither a catalyst nor an intermediate will appear in the overall reaction. • They will be both appear in elementary steps. • An intermediate will generated (a product) in an early step and will be consumed (a reactant) in a later step. Product first, reactant next: intermediate • In contrast, a catalyst will be consumed (a reactant) in an early step and will regenerated (a product) in a later step. Reactant first, product next: catalyst • Both the catalyst and intermediates will cancel out when the elementary steps are added to obtain the overall equation.

  25. Rate Laws of Multi-step Mechanisms Rate-determining step : the slow step in the reaction mechanism. The rate of the rate-determining step determines the rate of the overall reaction. The rate law of the slow step = rate law of the overall reaction. Mechanisms with an Initial Slow step : Overall Reaction : Experimentally determined rate law :

  26. Mechanism of the Reaction : The rate of the overall reaction is determined by the rate of the slow step ( step 1), the predicted rate law of the reaction is therefore the same as the experimentally determined rate law

  27. Mechanisms with an Initial Fast Step Overall Reaction : Experimentally Determined Rate Law : Reaction Mechanism :

  28. The rate law for the overall reaction is that for the rate-determining step( step 2). But the rate law then is in terms of the concentrations of an intermediate, NOBr2, which is unknown and hard to determine at anytime. We make the following assumption: rate of formation of intermediate = rate of decomposition of intermediate

  29. Consider the following reaction mechanism which occurs in 4 elementary steps. • 1. S2O8 2- + I- I S2O83- (slow) • 2. IS2O8 3- 2SO42- + I+ (fast) • 3. I+ + I- I2 (fast) • 4. I2 + I- I3- (fast) • Write the equation for the overall reaction. • b. Identify all intermediates in the reaction mechanism: I+, I2, IS2O8-3 • c. Identify the rate determining step and write the rate law for the overall reaction: step 1; rate =k[S2O8-2][I-]

  30. Catalysis Catalyst : Substance that changes the speed of a chemical reaction without undergoing a permanent chemical change itself. Homogeneous Catalysis : Catalyst in the same phase as the reactants. The bromide ion catalyses the decomposition of hydrogen peroxide to water and oxygen.

  31. Catalysis Contd. a.Heterogeneous Catalysis The catalyst is in a different phase from the reactants. In the above reaction, the reactants are adsorbed ( binding to the surface of catalyst ) onto the surface of the catalyst . The places on the catalyst surface where the reactant molecules bind to are called active sites.

  32. Answers to Slide 7 Problems 2. Order with respect to I-= 1 because power of I- = 1 in rate law

  33. Answers to Slide 14 Problems

  34. Answers to Slide 15 Problems

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