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 ) v 0 /(10 -7 mol L -1 s -1 ) Determine the order of the reaction with respect to B and calculate the rate constant. Solution: Log([B] 0 ) Log(v 0 )

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 [N 2 O 5 ]/(mol L -1 ) 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 ln([A]/[A] 0 )

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 v = k [A] B → A v = k’ [B] the net rate change for A is therefore 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 the 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 = - (k a + k b )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 pKw = 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; [H + ] = [H + ] eq – x; [OH - ] = [OH - ] eq – x and obtain: 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 L mol -1 s -1 k 1 = 2.4 x 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; E a is the activation energy. The two quantities, A and E a, are called Arrhenius parameters. In an alternative expression lnk = lnA - 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 k(L mol -1 s -1 )7.9x x x x x10 8 Solution: 1/T (K -1 ) lnk (L mol -1 s -1 ) The slope of the above plotted straight line is –E a /R, so Ea = 23 kJ mol -1. The intersection of the straight line with y-axis is lnA, so A = 8x10 10 L mol -1 s Series1

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 and taking derivatives with respect to t: In order to satisfy = 0 t max = The maximum concentration of B can be calculated by plugging the t max into the equation.