1 Chemical Kinetics (4 lectures) Dr. Paul T. Maragh Tue. 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 reactions gas phase: H 2 (g) + I 2 (g) 2HI(g) liquid phase: NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O(l) Heterogeneous reactions Fe(s) + 2H + (aq) Fe 2+ (aq) + H 2 (g) Irreversible reaction Reversible reaction (equilibrium) e.g. N 2 O 4 (g) 2NO 2 (g)

4 composition time. 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: 2H 2 + O 2 2H 2 O Then,  R has fixed dimensionality: ratio of concentration upon time, i.e. [(amount of material)(volume) -1 ] [time] -1 common units: mol dm -3 s -1 Compare rate of loss of H 2 and rate of loss of O 2.

6 Graphically [C] [A] Time / s Conc. / M 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 R = -d[A]/dt = k [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 order of 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-order overall, therefore: R = k [A][B] Hence, Units for rate constant for 2 nd order reaction If first-order overall???? Units for rate constant for 1 st 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. CH 3 I + CH 3 CH 2 O - CH 3 OCH 2 CH 3 + 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. H 2 + Cl 2 2HCl. HCl is NOT formed in this one step, but proceeds by a series of elementary steps: Cl 2 2Cl Cl + H 2 HCl + H Cl + H HCl H 2 + Cl 2 2HClOverall 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 R f = R r. Assume k << k r, 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 –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 Integrated form of the 1st order rate expression y = c + mx

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

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

18 Second-order reactions –Second-order reactions – Two possible cases: Case I Case I : A + A Products OR 2A Products Case II 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 2 nd order rate expression y = c + mx

20 (1/[A] t ) / dm 3 mol -1 t / s slope = 2k Intercept = 1/[A] 0 y = c + mx

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, t ½, 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 –First-order reactions – Remember that for a 1 st order reaction: ln[A] t = ln[A] 0 - kt At time t = 0, [A] = [A] 0 Then at time t = t ½ (half-life), [A] t ½ = [A] 0 /2 Substituting into above equation, ln([A] 0 /2) = ln[A] o – kt ½ ln([A] 0 /2) – ln[A] 0 = -kt ½ ln 1 – ln 2 = -kt ½, 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 1 st order reaction, the half-life is independent of reactant concentration but dependent on k.  The half-life is constant for a 1 st order reaction time concentration [A] 0 [A] 0 /2 [A] 0 /4 [A] 0 /8 Recall: [A] t = [A] 0 e -kt t 1/2

25 Second-order reactions –Second-order reactions – At time t = 0, [A] = [A] 0 And when t = t ½, [A] t½ = [A] 0 /2 So t 1/2 for 2 nd order reactions depends on initial concentration

26 Therefore, larger initial concentrations imply shorter half-lives (so faster the reaction). concentration [A] 0 [A] 0 /2 [A] 0 /4 [A] 0 /8 time t 1/2

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] t m [B] t n 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 r = k’[A] t m, where k’ = k[B] 0 n Since A is the reactant that changes, then the rate becomes dependent on A, and we can say Created a ‘false’ first-order (imitating first-order) PSEUDO-FIRST-ORDER, 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’ where k’ is the pseudo-first-order rate constant

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. [A] t t / s Initial rate Follow reaction to ~ 10% completion

32 Recall A + B P, Rate 0 = k[A] 0 a [B] 0 b Taking ‘logs’ log Rate 0 = log k + a log [A] 0 + b log[B] 0 y m xc ** Keep [A] 0 constant for varying values of [B] 0 to find b Log R o log[B] 0 slope = b Intercept = log k + a log[A] 0

33 ** Keep [B] 0 constant for varying values of [A] 0 to find a from the slope of the graph, log R 0 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 R 1 = k[A] a [B] b R 2 = k[nA] a [B] b Dividing R 2 by R 1 For these experiments, B is kept constant while A is varied and R 1 and R 2 are known.

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

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

36 COLLISION THEORY & ARRHENIUS EQUATION According to the Collision Theory Model: a bimolecular reaction properly oriented occurs when two properly oriented reactant molecules come sufficiently energetic collision 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 EaEa A---B---C A + BC AB + C Reactants Products Potential Energy Reaction Progress  The height of the barrier is called the activation energy, E a.  The configuration of atoms at the maximum in the P.E. profile is called the transition state.

38 **If the collision energy < E a, the reactant molecules cannot surmount the barrier and they simply bounce apart. **If the collision energy is  E a, 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 T 1, a certain fraction of the reactant molecules have sufficient K.E., i.e. K.E. > E a. At a higher temperature T 2, 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 o C.

41 Fraction of molecules Kinetic Energy EaEa T1T1 T2T2 T 2 > T 1 (i)The total area under the curve is proportional to the total # molecules present. (ii) Total area is the same at T 1 and T 2. (iii) The shaded areas represent the number of particles that exceed the energy of activation, E a. Maxwell-Boltzmann distribution curve

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. Arrhenius Equation This observation led to the development of the Arrhenius Equation: k = Ae -Ea/RT AE a Arrhenius parameters Collectively, A and E a are called the Arrhenius parameters of the reaction.  E a  E a = activation energy (kJ mol -1 ), and is the minimum kinetic energy required to allow reaction to occur

43 This fraction goes up when T is increased because of the negative sign in the exponential term. e -Ea/RT However, most of the collisions calculated by e -Ea/RT do not lead to products, and so e -Ea/RT The exponential term e -Ea/RT is simply the fraction of collisions that have sufficient energy to react.  A  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  T = Kelvin temperature R  R = ideal gas constant (8.314 J mol -1 K -1 )  k  k is the rate constant

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

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 1/T High activation energy Low activation energy

46 Manipulation of Arrhenius equation: (i)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 – E a /RT’ ln k = ln A – E a /RT Subtract these equations ln k’ – ln k = ln A – ln A – E a /RT’ – (-E a /RT) (ii) Can also find E a if k’, k, T’ and T are known.