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 v0 = k [A0]a taking logarithms gives: logv0 = log k + a log[A0] 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 v0/(10-7 mol L-1s-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(v0) -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]0e-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 N2O5 in liquid bromine varied with time as follows: t/s 0 200 400 600 1000 [N2O5]/(mol L-1) 0.110 0.073 0.048 0.032 0.014 confirm that the reaction is first-order in N2O5 and determine the rate constant. Solution: To confirm whether a reaction is first order, plot ln([A]/[A]0) against time. First order reaction should yield 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 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 rate is equal to the reverse: 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 = - (ka + kb)x therefore is called the relaxation time

Example 22.4: The H2O(l) ↔ H+(aq) + OH-(aq) equilibrium relaxes in 37 μs at 298 K and pKw = 14.0. Calculate the rate constants for the forward and backward reactions. Solution: the net rate of ionization of H2O is we write [H2O] = [H2O]eq + x; [H+] = [H+]eq – x; [OH-] = [OH-]eq – x and obtain: Because x is small, k2x2 can be ignored, so Because k1[H2O]eq = k2[H+]eq[OH-]eq at equilibrium condition = = hence k2 = 1.4 x 1011 L mol-1 s-1 k1 = 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.

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 = 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 300 350 400 450 500 k(L mol-1s-1) 7.9x106 3.0x107 7.9x107 1.7x108 3.2x108 Solution: 1/T (K-1) 0.00333 0.00286 0.0025 0.00222 0.002 lnk (L mol-1s-1) 15.88 17.22 18.19 18.95 19.58 The slope of the above plotted straight line is –Ea/R, so Ea = 23 kJ mol-1. The intersection of the straight line with y-axis is lnA, so A = 8x1010 L mol-1s-1 5 10 15 20 25 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 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;