What is Chemical Kinetics?

1. Chemical Kinetics(4 lectures)Dr. Paul T. MaraghTue. 5:00 p.m. / Wed. 9:00 a.m.1 full question on C10K Paper 1

2. What is Chemical Kinetics? The study of the speed or RATE at which a chemical reaction occurs. What are some of the factors that affect the RATE of a chemical reaction? • The nature of the reactants and products • Temperature • Catalysts • The concentrations of the reacting species.

3. Homogeneous reactionsgas phase: H2(g) + I2(g) 2HI(g)liquid phase: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) Heterogeneous reactions Fe(s) + 2H+(aq) Fe2+(aq) + H2(g) Irreversible reaction Reversible reaction (equilibrium) e.g. N2O4(g) 2NO2(g)

4. The factors mentioned will affect the composition of the reaction mixture at any given time. Therefore The change in composition of the reaction mixture with time is the rate of reaction, denoted by R, r or . R is the same whether monitoring reactants or products Generally,

5. Example: 2H2 + O2 2H2O Then, Compare rate of loss of H2 and rate of loss of O2. • R has fixed dimensionality:ratio of concentration upon time, i.e. [(amount of material)(volume)-1] [time]-1common units:mol dm-3 s-1

6. Graphically [C] Conc. / M [A] Time / s The tangents to the curve are the slopes = Rate All reaction rates are positive

7. Rate laws, rate constants, reaction order • Consider the simple reaction A + B P R = ƒ ([A][B]) And, R  [A]m [B]n With the use of a proportionality constant k, which is the rate constant (independent of conc. but dependent on temp.), R = -d[A]/dt = k [A]m [B]n Such an equation is called the rate law

8. The exponents m and n are the orderof the reaction with respect to reactant A and the order of the reaction with respect to reactant B respectively. • The order of the reaction = m + n • If m = n =1, then the reaction is first-order in A and first-order in B, but second-orderoverall, therefore: R = k [A][B] Hence, Units for rate constant for 2nd order reaction If first-order overall???? Units for rate constant for 1st order reaction

9. Molecularity • Molecularity is the number of molecules coming together to react in an elementary step. • Elementary reactions are simple reactions (described by molecularity) (a) A Products UNI-molecular reaction e.g. (b) A + A Products or A + B Products BI-molecular e.g. CH3I + CH3CH2O- CH3OCH2CH3 + I- (c) 2A + B P or A + B + C P Ter-molecular

10. Molecularity = Order of reaction • Reaction order is determined by experiment only • Reaction order is an empirical quantity (values range -2 to 3). • Can be fractional – found mainly in gas phase • Can be negative, A is an inhibitor (decreases the rate)

11. Mechanism, Rate – determining step and Intermediates • Assembly of elementary steps (to give products(s)) is called the reaction mechanism. e.g. H2 + Cl2 2HCl. HCl is NOT formed in this one step, but proceeds by a series of elementary steps: Cl2 2Cl• Cl• + H2 HCl + H • Cl• + H • HCl H2 + Cl2 2HCl Overall reaction Mechanism – arrived at from theory and experiment

12. Rate-determining step (RDS) is the slowest elementary reaction in the mechanism and controls the overall rate of the reaction. e.g. A + 2B D + E mechanism: A + B C + E fast B + C D slow – rate determining step A + 2B D + E C is an intermediate – formed, and then used up in the reaction

13. Intermediates –A + B C Products Equilibrium is dynamic, this means Rf = Rr. Assume k << kr, then slow step is: A + B C C Prod. (Slow: RDS) R = k[C] Rate = kK[A][B] Rate = k’[A][B] k’ = kK

14. Deriving the Integrated Rate Expressions • First-order reactions – A B, then the rate of disappearance of A is: Rearranging gives: At time t = 0, [A] = [A]0 And when t = t, [A] = [A]t

15. Integrating: ln[A]t = ln[A]0 - kt y = c + mx Integrated form of the 1st order rate expression

16. Intercept = ln[A]0 -slope = -k ln[A]t t / s Other useful forms t / s -slope = -k ln([A]t/[A]0)

17. Recall ln[A]t = ln[A]o - kt Antilog gives: Intercept = [A]0 [A]t = [A]0 e-kt [A]t t / s

18. Second-order reactions – Two possible cases: Case I : A + A Products OR 2A Products Case II : A + B Products Rearranging gives: At time t = 0, [A] = [A]0 And when t = t, [A] = [A]t

19. Integrating: OR Integrated form of the 2nd order rate expression y = c + mx

20. y = c + mx slope = 2k (1/[A]t) / dm3 mol-1 Intercept = 1/[A]0 t / s

21. What can we conclude about RATE LAWS versus INTEGRATED RATE EXPRESSSIONS?? • a rate law can tell us the rate of a reaction, once the composition of the reaction mixture is known • An integrated rate expression can give us the concentration of a species as a function of time. It can also give us the rate constant and order of the reaction by plotting the appropriate graph

22. The Study of Half-Lives • The half-life, t½, of a reaction is the time taken for the concentration of a reactant to fall to half its initial value. • It is a useful indication of the rate of a chemical reaction.

23. First-order reactions – Remember that for a 1st order reaction: ln[A]t = ln[A]0 - kt At time t = 0, [A] = [A]0 Then at time t = t½ (half-life), [A]t½ = [A]0/2 Substituting into above equation, ln([A]0/2) = ln[A]o – kt½ ln([A]0/2) – ln[A]0 = -kt½ ln 1 – ln 2 = -kt½, where ln 1 = 0 Therefore, ln 2 = kt ½

24. Hence, or What is/are the main point(s) to note from this expression?? • For a 1st order reaction, the half-life is independent of reactant • concentration butdependent on k. • The half-life is constant for a 1st order reaction [A]0 Recall: [A]t = [A]0e-kt concentration t1/2 [A]0/2 t1/2 [A]0/4 t1/2 [A]0/8 time

25. Second-order reactions – At time t = 0, [A] = [A]0 And when t = t½, [A]t½ = [A]0/2 So t1/2 for 2nd order reactions depends on initial concentration

26. Therefore, larger initial concentrations imply shorter half-lives (so faster the reaction). [A]0 concentration t1/2 [A]0/2 t1/2 [A]0/4 t1/2 [A]0/8 time

27. Determining Rate Laws Rate laws have to be determined experimentally. Techniques for monitoring the progress of a reaction include: • Absorption measurements (using a spectrophotometer) • Conductivity (reaction between ions in solution) • Polarimetry (if reactants/products are optically active, e.g. glucose) • Aliquot method (employing titration technique) Recall A + B P, r = k[A]m[B]n

28. (A) Isolation Method:This technique simplifies the rate law by making all the reactants except one, in large excess. Therefore, The dependence of the rate on each reactant can be found by isolating each reactant in turn and keeping all other substances (reactants) in large excess. Using as example: r = k[A]tm [B]tn Make B in excess, so [B]>>[A]. Hence, by the end of the reaction [B] would not have changed that much, although all of A has been used up And we can say, [B]  [B]0

29. Since A is the reactant that changes, then the rate becomes dependent on A, and we can say r = k’[A]tm , where k’ = k[B]0n Created a ‘false’ first-order (imitating first-order) PSEUDO-FIRST-ORDER, where k’ is the pseudo-first-order rate constant Logging both sides gives: log r = log k’ + m log [A]t y = c + m x A plot of log r vs log [A]t gives a straight line with slope = m, and intercept log k’

30. If m = 1, the reaction is said to be pseudo-first-order With the roles of A and B reversed, n can be found in a similar manner k can then be evaluated using any data set along with the known values of m and n

31. (B) Initial Rate Method: - often used in conjunction with the isolation method, -The rate is measured at the beginning of the reaction for several different initial concentrations of reactants. Initial rate Follow reaction to ~ 10% completion [A]t t / s

32. Recall A + B P, Rate0 = k[A]0a[B]0b Taking ‘logs’ log Rate0 = log k + a log [A]0 + b log[B]0 y c x m ** Keep [A]0 constant for varying values of [B]0 to find b slope = b Log Ro Intercept = log k + a log[A]0 log[B]0

33. ** Keep [B]0 constant for varying values of [A]0 to find a from the slope of the graph, log R0 vs log [A]0 ** Substitute values of a, b, [A]0, [B]0 to find k. However, in some cases, there may be no need to use the plots as shown previously. EXAMPLE R1 = k[A]a[B]b R2 = k[nA]a[B]b For these experiments, B is kept constant while A is varied and R1 and R2 are known. Dividing R2 by R1

34. (a) If R2 = 2R1, and n=2, then a = 1, so 1st order with respect to A (b) If R2 = 4R1, and n=2, then a = 2, so 2nd order with respect to A

35. Concluding: if n=2, and Rate doubles 1st order Rate increases by a factor of 4 2nd order Rate increases by a factor of 9 3rd order

36. COLLISION THEORY & ARRHENIUS EQUATION According to the Collision Theory Model: a bimolecular reaction occurs when two properly oriented reactant molecules come together in a sufficiently energetic collision. i.e. for a reaction to occur, molecules, atoms or ions must first collide. Consider the hypothetical reaction: A + BC AB + C A + BC A----B----C AB + C

37. Potential Energy Profile A---B---C • The height of the barrier is called the • activation energy, Ea. • Theconfiguration of atoms at the maximum in the P.E. profile is called the transition state. Ea Potential Energy A + BC Reactants AB + C Products Reaction Progress

38. **If the collision energy < Ea, the reactant molecules cannot surmount the barrier and they simply bounce apart. **If the collision energy is  Ea, the reactants will be able to surmount the barrier and be converted to products.

39. Very few collisions are productive because very few occur with a collision energy as large as the activation energy. Also, proper orientation is necessary for product formation. There must be some effect by Temperature on reaction systems. Temperature can result in an increase in energy. This leads us to say: The average kinetic energy of a collection of molecules is proportional to the absolute temperature.

40. At a temperature T1, a certain fraction of the reactant molecules have sufficient K.E., i.e. K.E. > Ea. At a higher temperature T2, a greater fraction of the molecules possess the necessary activation energy, and the reaction proceeds at a faster rate. **In fact it has been found that reaction rates tend to double when the temperature is increased by 10 oC.

41. Maxwell-Boltzmann distribution curve T2 > T1 Fraction of molecules T2 T1 Ea Kinetic Energy • The total area under the curve is proportional to • the total # molecules present. (ii) Total area is the same at T1 and T2. • The shaded areas represent the number of particles • that exceed the energy of activation, Ea.

42. It was observed by Svante Arrhenius that almost all of the reaction rates (obtained from experiments) accumulated over a period showed similar dependence on temperature. This observation led to the development of the Arrhenius Equation: k = Ae-Ea/RT Collectively, A and Ea are called the Arrhenius parameters of the reaction. • Ea = activation energy (kJ mol-1), and is the minimum kinetic energy required to allow reaction to occur

43. The exponential term e-Ea/RTis simply the fraction of collisions that have sufficient energy to react. This fraction goes up when T is increased because of the negative sign in the exponential term. However, most of the collisions calculated by e-Ea/RTdo not lead to products, and so • A = the frequency factor or pre-exponential factor (same units as k), is the fraction of sufficiently energetic collisions that actually lead to reaction. • T = Kelvin temperature • R = ideal gas constant (8.314 J mol-1 K-1) • k is the rate constant

44. Logarithmic form of the Arrhenius equation: Recall : k = Ae-Ea/RT y c mx A plot of ln k versus 1/T gives slope= –Ea/R and intercept= ln A Cannot extrapolate for intercept. Obtain A by substituting one of the data values along with value of Ea into equation. ln k y x 1/T

45. High activation energy corresponds to a reaction rate that is very sensitive to temperature (the Arrhenius plot has a steep slope). Converse also applies. ln k Low activation energy High activation energy 1/T

46. Manipulation of Arrhenius equation: • Once the activation energy of a reaction is known, it is a simple • matter to predict the value of a rate constant k’ at a temperature, • T’ from another value of k at another temperature, T. ln k’ = ln A – Ea/RT’ Subtract these equations ln k = ln A – Ea/RT ln k’ – ln k = ln A – ln A – Ea/RT’ – (-Ea/RT) (ii) Can also find Ea if k’, k, T’ and T are known.