CHEMICAL KINETICS

1. CHEMICAL KINETICS In thermochemistry we studied the question of whether or not a reaction “can” occur under given conditions. is G +ve or -ve In chemical kinetics we are concerned with two primary characteristics of the reaction. What are these properties?

2. A) The rate at which a reaction occurs How can this be measured? The increase in concentration of a product or the decrease in concentration of a reactant per unit time. Properties to be monitored can be conductivity, absorbance, pH or any other physical property which is convenient.

3. B) The mechanism by which a reaction occurs the pathway or series of steps involved. Why would the pathway or series of steps effect the rate? There are many cases in which G is –ve but the reactions occur so slowly that the rate is immeasurable at room temperature.

4. Example: Diamond in air C (diamond) + O2 (g)  CO2 (g) G = -396 kJ Why does diamond not oxidise to cardon dioxide in air? This can be explained by kinetics.

5. What factors will affect the rate of a chemical reaction? (1) nature of the reactants (2) concentration of the reactants (3) temperature (4) catalysis

6. What must atoms do before they can react with one another? Do all collisions result in reactions? WHY? What factors are involved?

7. Collision theory of reaction rates For a collision to be effective the reacting species must (a) possess a minimum (kinetic energy) KE to overcome repulsion between their electron clouds and break existing bonds (b) have the proper orientations toward each other

8. Example: Consider the reaction (Whitten) NO + N2O  NO2 + N2 Bond breaking N N N N N N O O O N N O O N O Bond forming EFFECTIVE Collision

9. N N N N N O N O O O O N N N O Collision INEFFECTIVE Unfavourably placed for NO bond formation.

10. BE is Potential energy What property involves the measure of potential energy? manifests itself as H.

11. A C B D Ea (forward reaction) Ea (reverse reaction) Reactants well separated H -ve Products well separated TIME

12. Products PE Reactants H +ve TIME Transition state

13. The first section of the curve shows... Reactants well separated and having no effect on each other. As they approach each other there is an increase in PE due to ... Electrostatic repulsion between their electron clouds, including those of the bonds which are to be broken.

14. If the reactants have enough KE to overcome this repulsion they form a high-energy, short lived transition complex which then breaks up into the products. If H is -ve the reaction is said to be exothermic (heat released) if H is +ve the reaction is endothermic (heat absorbed).

15. The amount of KE that must be converted to PE for the transition state to be reached is called the activation energy, Ea. Ea is then released as the products move apart.

16. Example: aA + bB  cC + dD Rate = - [A] = - [B] = [C] = [D] at bt ct dt “Reaction rate” is defined as the change in concentration of a reactant or product per unit time. If a moles of A that disappears, b moles of B appears. If a = 1 say and b = 3, then 3 mol of B disappear for every 1 mol of A.

17.    [ ]          t0 t1 t2 .. .. .. Determination of reaction rate This is done by measuring [ ] at t0, t1, t2, ……etc. and plotting as follows. The slope of the tangent to the curve at any given time gives the reaction rate at that time.

18. Rate decreases with time due to fewer collisions between reacting species. Concentration of reactants is decreasing by convention the rate is +ve because of the method used for the determination of the slope of the tangent.

19. That is tangent equals [ ] at t2 - [ ] at t1 t2 - t1 Since [ ] at t2 < [ ] at t1, the value of the tangent is –ve and hence Tangent = -(- conc./t) = conc./t

20. Rate equations and reaction order These “instantaneous” rates can be used to find the mathematical relationship between reaction rate and concentration of each reactant, that is the rate equation. Rate = k [A]a [B]b [C]c for the reaction aA + bB + cC  k = rate constant, which is characteristic of a particular reaction.

21. k reflects the fraction of collisions that are effective in producing a reaction at a given temperature. For example for Rate = k [A]2 [B] A + B  no dependence on products The reaction is described as “second order in A” and “first order in B”. The “overall reaction order” is the sum of the exponents, for example in the above reaction, overall reaction order = 3

22. Example: • If one reactant is present in large excess, so that its concentration does not change appreciably as the reaction proceeds • HCl + H2O  H3O+ + Cl- • (solvent) • That is a one-step reaction but [H2O] not significant • HCl dissolved in H2O

23. Determination of rate equations For the simple case: Reactions that are known to be elementary (ie occur in one step) and do not involve the solvent rate equations can be written based solely on the chemical equation for the reaction. eg A + 2B  C Rate = k [A] [B]2

24. What if a two step reaction process is involved? Whats the so called rate determining step? The rate determining step will be the slower of the two steps. eg For the overall reaction 2NO2 (g) + F2 (g)  2NO2F (g) Experimentally determined rate equation is Rate = k [NO2] [F2]

25. This is because the reaction takes place in 2 steps as follows. NO2 (g) + F2 (g)  2NO2F + F NO2 (g) + F (g)  2NO2F (g) Rate depends on the rate at which F is produced in the first reaction since it is used in the second reaction immediately as it is formed. Slow Fast

26. INITIAL RATES FOR FINDING RATE EQUATIONS Values of concentration vs time data from a series of separate experiments. Each must have a different initial concentration of one or more reactants A + 2B  AB2 Trial run Initial [A] Initial [B] Initial rate of AB2 formation (mol L-1) (mol L-1 s-1 ) 1 1.0 x 10-2 1.0 x 10-2 1.5 x 10-4 2 1.0 x 10-2 2.0 x 10-2 1.5 x 10-4 3 2.0 x 10-2 3.0 x 10-2 6.0 x 10-4

27. For trial 1,

28. Equations for first order reactions • Rate = k [A] Integrating gives: k t = - [A]/[A] log [A] = log [A0] - kt/2.303 where [A0] = concentration at t = 0 [A] = concentration at time t

29. This equation gives us a lot of information on the progress of a first order reaction. If k is known, the time needed to reach a given concentration of reactant can be calculated, or can calculate the conc of the reactant after a given amount of time.

30. Example: For the decomposition of N2O5 in CHCl3 2N2O5 4NO2 + O2 Reaction is 1st order, k = 6.32 x 10-4 s-1 Initial [N2O5] = 0.40 mol L-1 Find [N2O5] after 1 hr (3600 s) Use: log [A] = log [A0] - kt/2.303

31. log [N2O5] = log 0.40 - (6.32x10-4)(3600)/2.303 = -1.39 Taking antilog [N2O5] = 0.041 mol L-1

32. Example: The reaction N2O5 (g) N2O4 (g) + ½ O2 (g) obeys the rate law Rate = k [N2O5] where k = 1.68 x 10-2 s-1 Initial [N2O5] = 2.50 moles container volume = 5.0 L How many moles of N2O5 would remain after 1 minute?

34. Taking antilog [A] = 1.82 x 10-1 mol L-1 Moles A remaining = 0.182 mol L-1 x 5L = 0.910 mol

35. Half-life of first order reactions The Half-life or t ½ of a reactant is the time it takes for one-half of a reactant to be converted into product. Same as that calculated for nuclear decay. Solve previous equation for t Since [A] = ½ [A]0 at t = t ½ (by definition)

36. Example For a solution originally containing 1.30 x 10-6 mol L-1 of AmCl3. Only 1.27 x 10-6 mol L-1 of this radioactive substance remained after 2 hrs. Given that radioactive decay is a first-order process, determine the half life of 240Am.

38. What happens if our reaction is second order or greater?

39. Rate equations for second order reactions The rate of a second order reaction can be dependent upon the concentration of either one or two reactants Therefore Rate = k[A]2 or Rate = k[A][B] If Rate = k [A]2 Integrate 1/[A] = 1/[A0] + kt

40. The reaction A + B C + D is second order in A and second order overall. And if k = 0.622 L mol-1 min-1 What is the half life of A if [A]o=4.10 x 10-2 mol L-1? What is the equation for the half-life of a second order reaction?

41. For t½ [A] = ½[A0] Thus 1/[A] = 1/[A0] + kt becomes 2/[A] - 1/[A0] = kt½ t½ = 1/k[A0] for the question then where k = 0.622 L mol-1 min-1and [A]o=4.10 x 10-2 mol L-1. t½ = 1/(0.622 )(4.10 x 10-2 ) = 39.2 min

42. Rate equations for zero-order reactions We have looked at first and second order reactions, what is a zero order? What controls the rate of this reaction? The rate is not dependent on any reactant conc. i.e. Rate = k [A]0[B]0 = k i.e. rate is constant. That means that the rate is controlled by something other than collisions Like What?

43. The amount of Light Photochemical reactions. Rate = k

44. Graphic method for finding rate equations Type of reactionEqn for straight linelinear plotslope Zero order [A] = -kt +[A0] [A] vs t -k (rate = k) first order log[A] = -kt/2.303 + log[A0] log[A] vs t -k/2.303 (rate =k[A] ) second order 1/[A] = 1/[A0] + kt 1/[A] vs t k (rate = k[A]2)

45. Reaction mechanisms How the reaction occurs. Example: The overall third order reaction of NO with H2 2NO(g) + 2H2(g) N2 (g) + 2H2O(g) Rate = k[NO]2[H2] (experimentally determined)

46. 2NO(g) + H2(g) N2(g) + H2O2(g) H2O2(g) + H2(g) 2H2O(g) Slow Fast How could we envisage the reaction occurring?

47. The factor is always less than 1 and is the fraction of molecules which have the minimum energy required for reaction The Arrhenius eqn. from this equation • Reactions with large activation energies have smaller values of k and are slower. • For a given Ea, k will increase as temperature increases, reaction proceeds at a fast rate.,

48. Example: The rate constant for the decomposition of N2O5 in CHCl3 (chloroform) 2N2O5 4NO2 + O2 was measured at T1 = 25°C (k1=5.54 x10-5 S-1) and T2 = 67°C (k2 = 9.30 x 10-3 S-1). Find Ea for this reaction. Ans 1.0x10-5 J mol-1

49. Homogeneous and heterogeneous reactions Dependent on the opportunities for effective contact between reactants. Liq – Liq good contact Liq – Solid good contact Solid – Solid difficult contact Catalysis A catalyst is a substance which increases the rate of a chemical reaction and is not itself consumed in that reaction. A catalyst works by providing an alternative and easier pathway from reactants to products.