DIFFERENTIAL RATES OF REACTION

2. aA + bB cD + dD DIFFERENTIAL RATES OF REACTION Rate = d[B]/dt Rate’ = -d[A]/dt -d[A]/dt = (1/2)d[B]/dt Net rate = forward rate – reverse rate

3. THE EMPIRICAL RATE LAW (EXPERIMENTAL) aA + bB ----> cC + dD It is often (but not always) possible to express the rate of a reaction in the form • k is the rate constant; independent of [A],[B],[C],[D], but dependent on T • m,n are the reaction “orders” with respect to the reagents; usually integers • m + n is the “overall order” • In general, the orders bear no relationship to the stoichiometry of the reaction. That is, m, n are not related to a and b. The orders m, n must be experimentally determined.

4. DETERMINING THE EMPIRICAL RATE LAW: method of initial rates A + B ----------> C INITIAL RATE = Ri = C/t  k[A0]m[B0]n where t is chosen sufficiently small near t = 0 so that [A]  [A0] and [B]  [B0] RATE = k [A][B]2

5. DETERMINING THE EMPIRICAL RATE LAW: integrated rate laws First order A------> Products (1) k has units of time-1 First order reactions have some interesting and unique properties. For example, write the rate law in the form: (2) A constant fraction of A disappears per unit time = k Equation (1) may be directly integrated between time 0 and t with the result (3) (4)

6. First order rates can be characterized by a "half-life": By definition, the half-life, t1/2, is the time at which [A] = [A]o /2. Substituting this condition in the integrated rate law, we find (5) Thus the time it takes the reaction to proceed 50% is independent of the initial amount of material! 1Kg or 1mg will react 50% in the same time

7. Second order processes A------> Products (6) k has units conc-1time-1 -d[A]/dt = k [A]2 The rate constant depends on the units of concentration, and the probability of reaction per unit time is not a constant throughout the reaction, but depends on the amount of [A] present at any time -d[A]/[A] 1/dt = k [A] Equation (6) can be directly integrated from [A]o to [A] and t=0 to t to give (7) The half-life may be derived in the same way as for first order (not so useful; depends on [A]o)

8. For a second order process

9. Consider a more complex type of second order, A + B ------> products (first order in each, second order overall) -d[B]/dt = k[A][B] If [A]o = [B]o integration gives the same result as above, ie, for either A or B, 1/[A] - 1/[A]o = kt If [A]o ≠ [B]o, the result is Where x is the amount of A or B reacted at any time.

10. 4HBr + O2 2Br2 + 2H2O N2O5 2NO2 + 1/2O2 Rate = -d[ N2O5]/dt = k’[ N2O5] N2O5 NO2 + NO3 NO2 + NO3 NO + O2 + NO2 NO + NO3 2NO2 STOICHIOMETRY AND MECHANISMS OF CHEMICAL REACTIONS Stoichiometric equations Stoichiometric equation Stoichiometric equations do not imply the mechanism of the reaction; the above equation does not imply that the reaction occurs by the simultaneous collision of 4 molecules of HBr and 1 of O2. That would be most improbable due to the low probability of a simultaneous collision of 5 molecules! Reaction mechanisms Consider the “simple” reaction Stoichiometric equation Empirical rate law “elementary steps” Mechanism

11. Rate = -d[ N2O5]/dt = k[ N2O5] PROPERTIES OF ELEMENTARY REACTIONS (STEPS) 1. Elementary reactions are classified according to “molecularity”, the number of molecules involved in the reaction • unimolecular: one molecule of reactant • bimolecular: two molecules of reactant • termolecular: three molecules of reactant 2. The pathway of an elementary step is known; it occurs as written. Thus, the rate law can be derived theoretically unimolecular N2O5 NO2 + NO3 (forward rate only; reverse is bimolecular) A unimolecular step is first order. However, a first order reaction is not necessarily unimolecular

12. NO2 + NO3 NO + O2 + NO2 Bimolecular A direct collision between NO2 and NO3 is required Rate = (# of collisions per second/cm3)x(probability of a reaction per collision) # of collision per second/cm3 = a[NO2][NO3] probability of a reaction per collision = b (a constant) Rate = k[NO2][NO3] A bimolecular elementary step is second order, first order in each reactant Termolecular Similarly, for a termolecular process A + B + C products Rate = k[A][B][C] A termolecular process is third order, first order in each reactant (very rare).

13. DEDUCING MECHANISMS FROM RATE DATA

14. COMPLEX REACTIONS (combinations of elementary steps) Reversible unimolecular Parallel unimolecular Consecutive unimolecular Consecutive reversible unimolecular

15. Reversible unimolecular 1. A connection between kinetics and thermodynamics The rate law (1) Net rate = forward rate - reverse rate At equilibrium d[A]/dt = 0 k1[A]eqm = k-1[B]eqm (2)

16. With an intermediate X For any number of intermediates

17. 2. The principle of microscopic reversibility (PMR) Consider an overall reversible process A possible mechanism The PMR specifically disallows this mechanism "Each and every elementary step in a reversible reaction must have equal forward and reverse rates at equilibrium" Another possible mechanism

18. An example of the PMR Uncatalyzed pathway Catalyzed pathway (C is the catalyst) From thermodynamics, it is clear that the catalyst cannot change the equilibrium constant K = [B]eqm/[A]eqm In the presence of the catalyst, the reaction can proceed by both pathways, and the overall rate is At equilibrium, -d[A]/dt = 0

19. The equilibrium constant should not involve C; what’s wrong? We have stipulated only that the overall rate at equilibrium is = 0, but the PMR requires that the rate of each pathway be = 0 at equilibrium! For each step having equal forward and reverse rates: Uncatalyzed pathway Catalyzed pathway

20. PARALLEL UNIMOLECULAR 1. The rate law A disappears with first order kinetics, with a rate constant k = k1 + k2. The overall rate is determined by the fastest step. At the completion of the reaction, the ratio of products is