1. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) ERT 108/3 : PHYSICAL CHEMISTRY Chemical Kinetics By; Mrs Haf iza Bint i Shu kor

2. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) TOPIC COVERED…; Experimental Chemical and Kinetics Reactions First Order Reactions Second Order Reactions Reaction Rates and Reaction Mechanisms Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions).

3. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) CHEMICAL KINETICS?? Also called reaction kinetics Study of the rates & mechanisms of chemical reactions 2 types of reaction; a)homogeneous – reaction occurs in 1 phase (gas @liquid phase) b)heterogeneous – reaction occurs in 2 @ > phase

4. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Experimental Chemical and Kinetics Reactions Rates of chemical Reactions:  the rate of speed with which a reactant disappears or a product appears.  the rate at which the concentration of one of the reactants decreases or of one of the products increases with time.  mol L -1 s -1. 4

5. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 1 The decomposition of dinitrogen pentoxide (N 2 O 5 ) in an inert solvent (carbon tetrachloride) at 45 0 C: The data of the formation of O 2 (g) and the disappearance of N 2 O 5 is shown in Table 1. The initial concentration [N 2 O 5 ] = 1.40M. What is the concentration, [N 2 O 5 ] at time, t=423s? 5

6. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Rate of Reaction: A variable quantity Rate of reaction is expressed as either: or 6 [ Negative value ] [ Positive value ] Disappearance of reactant Formation of products

7. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 1) At t=0, Initial [N 2 O 5 ] = 1.40M At t = ∞, Final [N 2 O 5 ] = 0M [decomposes completely] 5.93cm 3 O 2 (g) is obtained at STP. After 423s, the volume of O 2 (g) collected is 1.32cm 3 of a possible 5.93cm 3. The fraction of the N 2 O 5 decomposed is 1.32/5.93. The decrease in concentration of N 2 O 5 at this point = (1.32/5.93) x 1.40M = 0.312 M. After 423s, [N 2 O 5 ] remaining undecomposed = 1.40-0.31 = 1.09M. 7

8. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) From the figure, determine the rate of decomposition of N 2 O 5 at 1900s. What is the initial rate of reaction? 8 Example 2: Note: the rate of reaction can be expressed as the slope of a tangent line.

9. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 2) 9 Based on the graph of concentration of reactant vs time,  the slope of a tangent line at t=1900s,

10. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) 10 The initial rate = = 8.0 x 10 -4 mol N 2 O 5 L -1 s -1

11. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) The Rate Law for Chemical Reactions The rate law or rate equation – mathematical equation. Reaction rate, r = k[A] m [B] n …..  The rate, r at time t is experimentally found to be related to the concentrations of species present at that time, t.  The exponents in the rate reaction are called the order of the reaction.  The term k in the equation is called the rate constant.  it is a proportionality constant that is characteristic of the particular reaction & is significantly dependent only on temperature. 11

12. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Method of Initial Rates This simple method of establishing the exponents in a rate equation involves measuring the initial rate of reaction for different sets of initial concentration. 12

13. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 3 The data of three reactions involving S 2 O 8 2- and I - were given in the below table. (i) Use the data to establish the order of reaction with respect to S 2 O 8 2-, the order with respect to I - & the overall order. 13

14. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Cont…Example 3 (ii) Determine the value of k for the above reaction. (iii) What is the initial rate of disappearance of S 2 O 8 2- reaction in which the initial concentrations are [S 2 O 8 2- ] =0.050M & [I-]=0.025M? (iv) What is the rate of formation of SO 4 2- in Experiment 1? 14

15. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 3) 15 (i) In the experiments 1 & 2, [I - ] is held constant & [S 2 O 8 2- ] is increased by a factor of 2, from 0.038 to 0.076M.  The reaction rate, R increased by a factor of 2 also. R 2 = k (0.076) m (0.060) n = k (2x0.038) m (0.060) n = k (2) m (0.038) m (0.060) n = 2.8 x 10 -5 mol L -1 s -1 R 1 = k (0.038) m (0.060) n =1.4 x 10 -5 mol L -1 s -1 If 2 m =2, then m =1. The reaction is first order in S 2 O 8 2-.

16. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) 16 R 2 = k (0.076) m (0.060) n = k (0.076) m (2x0.030) n = k (0.076) m (2) n (0.030) n = 2.8 x 10 -5 mol L -1 s -1 R 3 = k (0.076) m (0.030) n =1.4 x 10 -5 mol L -1 s -1 If 2 n =2, then n =1. The reaction is first order in I -. The overall order of the reaction is m + n = 1+1 = 2 (second order).

17. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) TRT401 Physical Chemistry BBLee@UniMAP 17 (ii) Use any one of the three experiments: k = 6.1 x 10 -3 L mol -1 s -1. (iii) Once the k value is determined, the rate law can be used to predict the rate of reaction. Reaction rate, R = 6.1 x 10 -3 L mol -1 s -1 x 0.050 mol L -1 x 0.025 mol L -1 = 7.6 x 10-6 mol L -1 s -1.

18. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) 18 Based on the stoichiometry, 2 moles of SO 4 2- are produced for every mole of S 2 O 8 2- consumed. = 2.8 x 10 -5 mol SO 4 2- (L -1 s -1 ).

19. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Zero-order, First-order, Second-order Reactions 19

20. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Cont……… 20

21. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Zero-order, First-order, Second- order Reactions 21 Zero order First orderSecond order

22. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 4 (a) When [N2O5] =0.44M, the rate of decomposition of N2O5 is 2.6 x 10 -4 mol L -1 s -1.  what is the value of k for this first-order reaction? (b) N2O5 initially at a concentration of 1.0 mol/L in CCl4, is allowed to decompose at 45 0 C. At what time will [N2O5] be reduced to 0.50M? 22

23. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 4) 23 (a) Rate of disappearance of N 2 O 5 (R) = k [N 2 O 5 ] = 5.9 x 10 -4 s -1 (b) For 1 st order of reaction, to determine t, we can use: log [A] 0 = log [N 2 O 5 ] 0 = log 1.0 = 0. log [A] t = log [N 2 O 5 ] t = log 0.50 = -0.30. use, k = 5.9 x 10 -4 s -1.

24. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 5 The data of the above table were obtained for the decomposition reaction: A → 2B + C. (a)Establish the order of the reaction. (b)What is the rate constant, k? 24 Time, min[A], Mlog [A]1/[A] 01.000.001.00 50.63-0.201.59 100.46-0.342.17 150.36-0.442.78 250.25-0.604.00

25. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 5) 25 (a) Plot graph based on the data given in the Table. (b) The slope of the 3 rd graph: Not Straight line – Not Zero order Not Straight line – Not First order Straight line – 2 nd order

26. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Reaction rates: Effect of temperature Chemical reactions tend to go faster at higher temperature.  slow down some reactions by lowering the temperature. Increasing the temperature increases the fraction of the molecules that have energies in excess of the activation energy.  this factor is so important that for many chemical reactions it can lead to a doubling or tripling of the reaction rate for a temperature increase of only 10 0 C. 26

27. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Cont…. In 1889, Arrhenius noted that the k data for many reactions fit the equation: where, A & E a are constants characteristics of the reaction R = the gas constant. E a = the Arrhenius activation energy (kJ/mol or kcal/mol) A = the pre-exponential factor (Arrhenius factor). ( the unit of A is the same as those of k.) Taking log of the above equation: 27

28. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) If the Arrhenius equation is obeyed:  a plot of log 10 k versus 1/T is a straight line with slope: -E a /2.303 R and intercept log 10 A.  This enables E a and A to be found. Another useful equation: (eliminate the constant A).  T 2 and T 1 - two kelvin temperatures.  k 2 and k 1 - the rate constants at these temperatures.  E a – the activation energy (J/mol)  R – the gas constant (8.314 Jmol -1 K -1 ). 28 Cont….

29. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 6 (a) Use the figure given to find A and E a for: (b) Calculate E a for a reaction where rate constant at room temperature is doubled by a 10Kelvin increase in T. 29 Figure: Rate constant versus temperature for the gas-phase first order decomposition reaction

30. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 6a) 30 Tabulate the data as follows. Construct the Arrhenius plot of log 10 k versus 1/T for the reaction.  Intercept (log 10 A)=13.5 A = 3x10 13 s -1  Slope=-5500K, E a =25kcal/mol =105 kJ/mol Temp, 0 CTemp, K1/Temp, 1/Kk, s -1 log 10 k 252980.00340.001-3 Figure: Arrhenius plot of log 10 k versus 1/T for this reaction. Note: the long extrapolation needed to find A.

31. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 6b) 31 Based on the given info:  k 2 = 2k 1,  T1 = room temperature (298K),  T 2 =298+10 = 308K, The Arrhenius equation: Substitute: E a = 53 kJ/mol

32. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Reaction Mechanisms Each molecular event that significantly alters a molecule’s energy or geometry is called an elementary process (reaction). The mechanism of a reaction:  the sequence of elementary reactions that add up to give the overall reaction.  A mechanism is a hypothesis about the elementary steps through which chemical change occurs. 32

33. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Reaction Mechanisms Elementary processes in which a single molecule dissociates (unimolecular) or two molecules collide (bimolecular) much more probable than a process requiring the simultaneous collision of three bodies (termolecular). All elementary processes are reversible and may reach a steady-state condition. In the steady state the rates of the forward & reverse processes become equal. One elementary process may occur much more slower than all the others. In this case, it determines the rate at which the overall reaction proceeds & is called the rate-determining/ limiting step. 33

34. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) The Hydrogen-Iodine Reaction H2 (g) + I2 (g) → 2HI (g) Rate of formation of HI = k [H2][I2] The hydrogen-iodine reaction is proposed to be a two-step mechanism [Sullivan J. (1967). J.Chem.Phys.46:73].  1 st step: iodine molecules are believed to dissociate into iodine atoms.  2 nd step: simultaneous collision of two iodine atoms and a hydrogen molecule. (this termolecular step is expected to occur much more slowly – the rate-determining step). 34

35. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) The Hydrogen-Iodine Reaction 1 st step: [Fast] 2 nd step: [Slow] Net: If the reversible step reaches a steady state condition:  rate of disappearance of I 2 = rate of formation of I 2 35

36. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) The Hydrogen-Iodine Reaction For the rate-determining step: Rate of formation of HI = k 3 [I] 2 [H 2 ] where = K[H 2 ][I 2 ] where (K=k 1 k 3 /k 2 ) 36

37. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Example 7 The thermal decomposition of ozone to oxygen: 2O 3 (g) → 3O 2 (g) The observed rate law: Rate of disappearance of O 3 = Show that the following mechanism is consistent with this experiment rate law. 1 st : 2 nd : 37

38. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Answer (Example 7) 38 Assume the 1 st step reaches the steady state condition: Rate of formation of O = Rate of disappearance of O k 1 [O 3 ] = k 2 [O 2 ] [O] Assume the 2 nd step is the rate-determining step: Rate of disappearance of O 3 = k 3 [O][O 3 ] (where k = k 1 k 3 /k 2 )

39. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) Experimental methods for fast reactions Many reactions are too fast to follow by the classical methods. Several ways to study fast reactions : 39 1. Rapid flow methods:(i)Continuous flow (ii)Stopped flow 2. Relaxation methods:(i)Temperature jump (T-jump) method (ii)Pressure jump method (iii)Electric field jump method 3. Flash photolysis 4. Shock tube 5. Nuclear-magnetic-resonance (NMR) spectroscopy

40. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) ASSIGNMENT 1 Write a short note for the following fast reaction: a)Rapid flow methods b)Relaxation methods c)Flash photolysis d)Shock tube e)Nuclear-magnetic-resonance (NMR) spectroscopy 40

41. ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) 41 The End