Chpt 12 - Chemical Kinetics Reaction Rates Rate Laws Reaction Mechanisms Collision Theory Catalysis HW set1: Chpt 12 - pg , # 22, 23, 28 Due Jan. 5 HW set2: Chpt 12 # 39, 48, 53, 56, 58, 60 due Jan. 10 HW set3: Chpt 12 #62, 69, 72 due Jan 12

Reaction Rate ? Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered.

Decomposition of NO 2 Graph Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point. 2NO 2 --> 2NO + O 2

Stoichiometry of Rates 2NO 2 --> 2NO + O 2 The rate of comsumption (disappearance) of NO 2 is same as rate of production of NO and twice the rate of production of O 2.

Calculate ave. rate from data

Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO 2 (g) → 2NO(g) + O 2 (g) Rate = k[NO 2 ] n :  k = rate constant  n = order of the reactant

Rate Law (cont) Rate = k[NO 2 ] n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation.

Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time.

Rate Laws (cont) typically consider reactions when the reverse reaction is unimportant, so our rate laws involve only [reactants]. differential and integrated rate laws for a given reaction are related in a well–defined way, can use either rate law. Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law.

Determining the form of the Rate Law Determine experimentally the power to which each reactant concentration must be raised in the rate law.

Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants.

Initial Rates examples Pg. 549 Table 12.4 discuss data Pg. 550 Table 12.5 discuss data

Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A] n [B] m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B

Integrated Rate Laws - zero first & second order Take actual data for each reactant and plot it… k is slope of line [A] vs t or ln[A] vs t or 1/[A] vs t straight line says O o, 1 o, 2 o for calculations at any time t t = time, [A] o = initial conc, [A] = conc at t, k = rate constant Table 12.6

Which Order? A plot of [A] vs time

Which order?

Exercise of Reaction Order Consider the reaction aA  Products. [A] 0 = 5.0 M and k = 1.0 x 10 –2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: a) Zero order b) First order c) Second order 4.7 M 3.7 M 2.0 M

Integrated rate from a graph Excel file AP Chem Chpt12problems.xls –Find it on wiki space

Half-life determination Half-life equation provided in Table 12.6 determined from integrated form. Can calculate yourself using the concentration when half remains as [A]=[A] o /2 which is half original conc and do the math. Answer has time related to k and [A] o

Reaction Mechanism Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction.

Mechanism example A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO NO 2 (g) + CO(g) → NO(g) + CO 2 (g)

Mechanisms Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Elementary Steps (Molecularity)

Mechanism Summary The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law.

Rate determining step A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction.

Mechanism Ex. Decomposition of N 2 O 5 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Step 1: 2( N 2 O 5 NO 2 + NO 3 )(fast) Step 2: NO 2 + NO 3 → NO + O 2 + NO 2 (slow) Step 3: NO 3 + NO → 2NO 2 (fast) Does this satisfy overall balanced eqn?

Elementary Reactions If you know that it is an elementary reaction, then the rate law just uses the coefficients from the balanced equation. If you have the mechanism in elementary steps you can figure out a plausible rate law

Reaction Mechanism Ex. The reaction A + 2B  C has the following proposed mechanism: A + B D(fast equilibrium) D + B  C(slow) Write the rate law for this mechanism. rate = k[A][B] 2 What if the fast and slow were reversed?

Collision Theory Molecules must collide to react. Main Factors:  Activation energy, E a  Temperature  Molecular orientations

Reaction Diagram

Arrhenius Model of rxns How fast a reaction happens is related by energy Number of collisions with Ea = (total collisions) x e -Ea/RT where R is gas constant and T is temp (K) But this value was still too high for observed rates so total collisions must not all react (orientation effect) so k = zpe -Ea/RT with z = collision frequency and p = steric factor (orientation)

Arrhenius Equation A=frequency factor (zp) E a =activation energy R=gas constant ( J/K·mol) T =temperature (in K)

Linear Form Arrhenius Eqn Slope = ?

Ex. Activation Energy, Ea Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? E a = 53 kJ

Catalysts A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy.

Rxn Diagram for catalyzed The catalyst provides a new pathway which has a lower Ea It does not lower the Ea for the original rxn

Effective collisions graphic Catalyzed Rxn has a lower Ea, so many more collisions are effective at making the product(s)

Heterogeneous Catalysts Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance.

Heterogeneous Catalysis Process 1. Adsorption and activation of the reactants. 2. Migration of the adsorbed reactants on the surface. 3. Reaction of the adsorbed substances. 4. Escape, or desorption, of the products.

Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts.