Reaction order The rate law can be written

Reaction order • The rate law can be written in a generalized form: v = k [A]a[B]b…. where a is the order of the reaction with respect to the species A, and b is the order of the reaction with respect to the reagent B. • The reaction order is (a + b +…. ). • The order of a chemical reaction needs not to be an integral. Certain reactions do not have an overall order !!! Example 1: v = k [A]1/2[B]1 Example 2: v = k (zero order reaction, such as ……) • How to determine the unit of k?

Determination of the rate law • Isolation method: v = k [A]a[B]b -----> v = k’[B]b • Method of initial rates (often used in conjunction with the isolation method): v = k [A]a at the beginning of the reaction v 0 = k [A 0]a taking logarithms gives: logv 0 = log k + a log[A 0] therefore the plot of the logarithms of the initial rates against the logarithms of the initial concentrations of A should be a straight line with the slope a (the order of the reaction).

Self-test 22. 3: The initial rate of a reaction depended on the concentration of a substance B as follows: [B]0/(mmol L-1) 5. 0 8. 2 17 30 v 0/(10 -7 mol L-1 s-1) 3. 6 9. 6 41 130 Determine the order of the reaction with respect to B and calculate the rate constant. Solution: Log([B]0) -2. 30 -2. 086 -1. 770 -1. 523 Log(v 0) -6. 444 -6. 018 -5. 387 -4. 886

22. 3 Integrated rate law • First order reaction: A Product The solution of the above differential equation is: or: [A] = [A]0 e-kt • In a first order reaction, the concentration of reactants decreases exponentially in time.

Self-test 22. 4: In a particular experiment, it was found that the concentration of N 2 O 5 in liquid bromine varied with time as follows: t/s 0 200 400 600 1000 [N 2 O 5]/(mol L-1) 0. 110 0. 073 0. 048 0. 032 0. 014 confirm that the reaction is first-order in N 2 O 5 and determine the rate constant. Solution: To confirm that a reaction is first order, plot ln([A]/[A]0) against time and expect a straight line: t/s 0 200 400 600 1000 ln([A]/[A]0) 0 -0. 410 -0. 829 -1. 23 -2. 06

Half-lives and time constant • For the first order reaction, the half-live equals: therefore, is independent of the initial concentration. • Time constant, , the time required for the concentration of a reactant to fall to 1/e of its initial value. for the first order reaction.

Second order reactions • Case 1: second-order rate law: (e. g. A → P) • Can one use A + A → P to represent the above process? • The integrated solution for the above function is: or • The plot of 1/[A] against t is a straight line with the slope k.

• Case 2: The rate law (e. g. A + B → Product) • The integrated solution (to be derived on chalk board) is :

22. 4 Reactions approaching equilibrium Case 1: First order reactions: A → B B → A the net rate change for A is therefore v = k [A] v = k’ [B] if [B]0 = 0, one has [A] + [B] = [A]0 at all time. the integrated solution for the above equation is [A] = As t → ∞, the concentrations reach their equilibrium values: [A]eq = [B]eq = [A]0 – [A]eq =

• The equilibrium constant can be calculated as K = thus: • In a simple way, at the equilibrium point there will be no net change and thus the forward reaction will be equal to the reverse reaction: k[A]eq = k’ [B]eq thus the above equation bridges thermodynamic quantities and reaction rates through equilibrium constant. • For a general reaction scheme with multiple reversible steps:

Determining rate constants with relaxation method • After applying a perturbation, the system (A ↔ B) may have a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A]eq + x; [B] = [B]eq - x; Because one gets dx/dt = - (ka + kb)x therefore is called the relaxation time

Example 22. 4: The H 2 O(l) ↔ H+(aq) + OH-(aq) equilibrium relaxes in 37 μs at 298 K and p. Kw = 14. 0. Calculate the rate constants for the forward and backward reactions. Solution: the net rate of ionization of H 2 O is we write [H 2 O] = [H 2 O]eq + x; and obtain: [H+] = [H+]eq – x; [OH-] = [OH-]eq – x Because x is small, k 2 x 2 can be ignored, so Because k 1[H 2 O]eq = k 2[H+]eq[OH-]eq at equilibrium condition = hence k 2 = 1. 4 x 1011 L mol-1 s-1 k 1 = 2. 4 x 10 -5 s-1 =

• Self-test 22. 5: Derive an expression for the relaxation time of a concentration when the reaction A + B ↔ C + D is second-order in both directions. To be demonstrated on in class

22. 5 The temperature dependence of reaction rates • Arrhenius equation: A is the pre-exponential factor; Ea is the activation energy. The two quantities, A and Ea, are called Arrhenius parameters. • In an alternative expression lnk = ln. A one can see that the plot of lnk against 1/T gives a straight line.

Example: Determining the Arrhenius parameters from the following data: T/K 300 350 400 450 500 k(L mol-1 s-1) 7. 9 x 106 3. 0 x 107 7. 9 x 107 1. 7 x 108 3. 2 x 108 Solution: 1/T (K-1) lnk (L mol-1 s-1) 0. 00333 15. 88 0. 00286 17. 22 0. 0025 18. 19 0. 00222 18. 95 0. 002 19. 58 25 20 15 Series 1 10 5 0 0 0. 0005 0. 0015 0. 0025 0. 0035 The slope of the above plotted straight line is –Ea/R, so Ea = 23 k. J mol-1. The intersection of the straight line with y-axis is ln. A, so A = 8 x 1010 L mol-1 s-1

The interpretation of the Arrhenius parameters • Reaction coordinate: the collection of motions such as changes in interatomic distance, bond angles, etc. • Activated complex • Transition state • For bimolecular reactions, the activation energy is the minimum kinetic energy that reactants must have in order to form products.

Applications of the Arrhenius principle Temperature jump-relaxation method: consider a simple first order reaction: A↔B at equilibrium: After the temperature jump the system has a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A]eq + x; [B] = [B]eq - x;

22. 6 Elementary reactions • Elementary reactions: reactions which involves only a small number of molecules or ions. A typical example: H + Br 2 → HBr + Br • Molecularity: the number of molecules coming together to react in an elementary reaction. • Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism. It must be an integral.

• An elementary bimolecular reaction has a second-order rate law: A + B → P • If the reaction is an elementary bimolecular process, then it has second-order kinetics; However, if the kinetics are second-order, then the reaction might be complex.

22. 7 Consecutive elementary reactions • An example: 239 U → 239 Np → 239 Pu • Consecutive unimolecular reaction A → B → C The rate of decomposition of A is: • The intermediate B is formed from A, but also decays to C. The net rate of formation of B is therefore: • The reagent C is produced from the unimolecular decay of B:

• Integrated solution for the first order reaction (A) is: • Then one gets a new expression for the reactant B: the integrated solution for the above equation is: when assuming [B] 0 = 0. • Based on the conservation law [A] + [B] + [C] = [A]0

Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration? Solution: At the maximum value of B Using the equation 25. 7. 6 and taking derivatives with respect to t: In order to satisfy =0 tmax = The maximum concentration of B can be calculated by plugging the tmax into the equation.