1. Chemistry

2. Chemical Kinetics - 2

3. Session Objectives 1.Methods of determining order of a reaction 2.Theories of chemical kinetics 3.Collision theory 4.Transition state theory 5.Temperature dependence of rate constant; Arrhenius equation 6.Catalysis 7.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. Methods of determining the order of a reaction L mole -1 sec -1 k1/[A] vs tSecond sec -1 -kIn[A] vs t[A]=[Ao]e -kt First Mole l -1 sec -1 -k[A] vs t[A]=[Ao]-ktZero Units of rate constant Slope of kineti cs plot Characteristi c kinetics Plot Integrated rate law Differential rate law Reaction Order

6. Graphical Representation

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

8. Illustrative Example For the thermal decomposition of CH 3 CHO; CH 3 CHO = CH 4 (g) + CO(g) The following rate data are obtained. ExperimentInitial pressureInitial rate of increase in total pressure 1. 300 0.61 2. 200 0.27 Predict the order of the reaction.

9. Solution Let the expression for the rate of the reaction is r = k[CH 3 CHO] 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 or n = 2.0

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 -1 min -1 ) 0.010.010.005 0.020.010.010 0.010.020.005 Determine the order of the reaction with respect to ‘A’ and ‘B’. 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. Solution:

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 H 2 O 2 Let V o & V t be volume of KMnO 4 used during zero & time t respectively

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

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 E a 1. E a is always positive. 2. The larger the value of E a, the slower the rate of a reaction at a given temperature. 3. The larger the value of E a, 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 E a itself DOES NOT CHANGE with temperature.

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

22. Temperature Dependence Plot of log k vs 1/T is a straight line & slope =-E a /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. intercept = log k. Taking log of both sides Solution:

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 37 o C to 15 o C? Solution:

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

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 NO 2 (g) + F 2 (g)  2 NO 2 F (g), the rate law is, rate = k [NO 2 ] [F 2 ]. Can the elementary reaction be the same as the overall reaction? If they were the same the rate law would have been rate = k [NO 2 ] 2 [F 2 ], 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. Illustrative Problem The hypothetical reaction follows the following mechanism The order of the overall reaction is (a) 0(b) 1 (c) 2(d) 3/2

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

33. Illustrative Problem The decomposition of ozone is believed to occur by mechanism O 3 O 2 + O (fast), O + O 3 2O 2 (slow) When the concentration of O 2 is increased, the rate (a) increases(b) decreases (c) remains the same(d) Cannot say

34. Solution  When concentration of O 2 is increased rate decreases.

35. Illustrative Example The decomposition of H 2 O 2 in the presence of I – follow this mechanism, i H 2 O 2 + I –  H 2 O + IO – (k 1 )slow ii H 2 O 2 + IO –  H 2 O + O 2 + I – (k 2 )fast What is the rate law? The slow step determines the rate, and the rate law is: rate = k1 [H 2 O 2 ] [I – ] Since both [H 2 O 2 ] and [I – ] are measurable in the system, this is the rate law. Solution

36. Illustrative Example The (determined) rate law is, rate = k [NO 2 ] [F 2 ], for the reaction, 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g), and a two-step mechanism is proposed: i NO 2 (g) + F 2 (g)  NO 2 F (g) + F (g) ii NO 2 (g) + F (g)  NO 2 F (g) Which is the rate determining step? The rate for step i is rate = k [NO 2 ] [F 2 ], which is the rate law, this suggests that step i is the rate- determining or the slow step. Solution:

37. Illustrative Problem At 380° C, the half-life period for the first order decomposition of H 2 O 2 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