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Chemistry

Chemistry. Chemical Kinetics - 2. Session Objectives. Methods of determining order of a reaction Theories of chemical kinetics Collision theory Transition state theory Temperature dependence of rate constant; Arrhenius equation Catalysis Reaction mechanism.

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Chemistry

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  1. Chemistry

  2. Chemical Kinetics - 2

  3. Session Objectives • Methods of determining order of a reaction • Theories of chemical kinetics • Collision theory • Transition state theory • Temperature dependence of rate constant; Arrhenius equation • Catalysis • Reaction mechanism

  4. Methods of determining the order of a reaction Integrated method The equation which gives a constant value of k decides the order of reaction Graphical method The data are plotted acc to different integrated rate equations so as to yield a straight line .Slope gives the value of rate constant Initial rate method Concentration of one of the reactant is varied Half life method In this method we plot half life of the reactant versus concentration of the reactant.

  5. Reaction Order Differential rate law Integrated rate law Characteristic kinetics Plot Slope of kinetics plot Units of rate constant Zero [A]=[Ao]-kt [A] vs t -k Mole l-1 sec-1 First [A]=[Ao]e-kt In[A] vs t -k sec-1 Second 1/[A] vs t k L mole-1 sec-1 Methods of determining the order of a reaction

  6. Graphical Representation

  7. Graphical Representation Conc. [A] log [A] 1/ [A] t t t zero order first order second order Graphical representation for concentration of integrated rate equation versus time

  8. Illustrative Example For the thermal decomposition of CH3CHO; CH3CHO = CH4(g) + CO(g) The following rate data are obtained. Experiment Initial pressure Initial rate of increase in total pressure 1. 300 0.61 2. 200 0.27 Predict the order of the reaction.

  9. or n = 2.0 Solution Let the expression for the rate of the reaction is r = k[CH3CHO]n Putting the values for both experiments separately Thus, 0.61 = k (300)n 0.27 = k (200)n Dividing both equations and taking log of the resultant equation

  10. Initial rate method • Complications occurs when multiple reactant are involved in the reaction. • This method involves the determination of the order of each reactant separately. • To determine the order of a particular reactant, its concentration is varied keeping the concentrations of other reactants constant. • In every experiment, we determine the initial rate of the reaction and observe the dependence of rate on that particular reactant. Keeping [B] constant

  11. Illustrative Example Rate of the reaction; A + B = products; is given below as a function of the initial concentration of ‘A’ and ‘B’. [A](mol L-1) [B](mol L-1) rate(mol L-1min-1) 0.01 0.01 0.005 0.02 0.01 0.010 0.01 0.02 0.005 Determine the order of the reaction with respect to ‘A’ and ‘B’. Solution: For first two experiments, the concentration of the reactant ‘B’ is constant. Rate of the reaction depend linearly on reactant ‘A’. Now, taking experiments first and third, the concentration of the reactant ‘A’ is constant. Therefore, rate of the reaction is independent of ‘B’. Thus, order of the reaction with respect to ‘A’ = one. Order of the reaction with respect to ‘B’ = zero.

  12. Half life method

  13. Illustrative Example The half-life of a particular chemical reaction at two different initial concentrations 5 x 10-4 and 25 x10-5 M are 2 and 16 hours. Calculate the order of reaction. Solution:

  14. Integrated method In this method, we put the data into the integrated form of the rate laws and calculate the values of the rate constants for different kinetics of the reaction. The order of the reaction is that one for which the value of rate constant is constant.

  15. Some first order reactions Decomposition of H2O2 Let Vo & Vt be volume of KMnO4 used during zero & time t respectively

  16. Let V0 , Vt be the volume of alkali required for titration during 0 ,t & infinite time Some first order reactions Hydrolysis of ester

  17. Theories of chemical kinetics 1. Collision theory 2. Transition state theory

  18. Collision theory Reaction occurs when reacting species have sufficient energy to collide and proper orientation in space. Energy barrier: The minimum energy which the colliding particles possess in order to bring about the chemical reaction is called threshold energy(activation energy). Orientation barrier: Colliding molecules should be in their proper orientation at the time of collision.

  19. Transition State Theory In the activated complex theory, we consider two reactants approaching and their potential energy rising and reaching a maximum. Activation energy - the energy needed to start a chemical reaction. It is very low for some reactions and very high for others.

  20. Some Points about Ea 1. Ea is always positive. 2. The larger the value of Ea, the slower the rate of a reaction at a given temperature. 3. The larger the value of Ea, the steeper the slope of (ln k) vs (1/T). A high activation energy corresponds to a reaction rate that is very sensitive to temperature. 4. The value of Ea itself DOES NOT CHANGE with temperature.

  21. The ratio is called the temperature coefficient and its value is 2 or 3 Effect of temperature on rate of chemical reaction A is frequency factor or Arhenius constant,Ea is activation energy Plot of log k vs 1/T is a straight line & slope =-Ea/2.303R

  22. Temperature Dependence Plot of log k vs 1/T is a straight line & slope =-Ea/2.303R

  23. Illustrative Example If we plot log k versus 1/T,we get a straight line. What is the slope and intercept of the line. Solution: Taking log of both sides intercept = log k.

  24. Effect of temperature on rate of chemical reaction

  25. Illustrative Example The activation energy of one of the reactions in the Krebs citric acid cycle is 87 kJ/mol. What is the change in the rate constant when the temperature falls from 37oC to 15oC? Solution:

  26. Illustrative Example The specific reaction rate for a reaction increases by a factor 4 if the temperature is changed from 27o C to 47o C. Find the activation energy for the reaction. Solution: = 13222.98 cals Ea = 13.311 K cals

  27. Catalysis • Catalyst: A substance that changes the rate of a reaction without being consumed in the reaction. • Provides an easier way to react. • Lower activation energy. • Still make the same products. • Enzymes are biological catalysts. • Inhibitor: A substance that decreases the rate of reaction (a negative catalyst).

  28. How catalyst change reaction rate • Catalysts are the one way to lower the energy of activation for a particular reaction. • The presence of the catalyst alters the path of the reaction. • The lower activation energy allows the reaction to proceed faster.

  29. Rate Law and Mechanism A mechanism is a collection of elementary steps devised to explain the reaction in view of the observed rate law. For the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), the rate law is, rate = k [NO2] [F2] . Can the elementary reaction be the same as the overall reaction? If they were the same the rate law would have been rate = k [NO2]2 [F2], Therefore, the overall reaction is not an elementary reaction.

  30. Rate-determining Step in a Mechanism Slowest step: The rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction. Steady-state approximation: The steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified.It is based on the assumption that the concentration of the intermediate is constant.

  31. The hypothetical reaction follows the following mechanism The order of the overall reaction is (a) 0 (b) 1 (c) 2 (d) 3/2 Illustrative Problem

  32. Solution The order depends on slowest step Hence, the answer is (d).

  33. The decomposition of ozone is believed to occur by mechanism O3 O2 + O (fast), O + O3 2O2 (slow) When the concentration of O2 is increased, the rate (a) increases (b) decreases (c) remains the same (d) Cannot say Illustrative Problem

  34. Solution \When concentration of O2 is increased rate decreases.

  35. Illustrative Example The decomposition of H2O2 in the presence of I– follow this mechanism, i H2O2 + I–  H2O + IO– (k1 )slow ii H2O2 + IO–  H2O + O2 + I– (k2 )fast What is the rate law? Solution The slow step determines the rate, and the rate law is: rate = k1 [H2O2] [I –] Since both [H2O2] and [I –] are measurable in the system, this is the rate law.

  36. Illustrative Example The (determined) rate law is, rate = k [NO2] [F2], for the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), and a two-step mechanism is proposed:i NO2 (g) + F2 (g)  NO2F (g) + F (g)ii NO2 (g) + F (g)  NO2F (g)Which is the rate determining step? Solution: The rate for step i is rate = k [NO2] [F2], which is the rate law, this suggests that step i is the rate-determining or the slow step.

  37. Illustrative Problem At 380° C, the half-life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ mole-1. Calculate the time required for 75% decomposition at 450° C.

  38. Solution

  39. Thank you

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