Unit #10 Chemical Kinetics Zumdahl Chapter 12 College Board Performance Objectives: Express the rate of a reaction in terms of changes in the concentration of a reactant or a product per time. Understand how to change from one to the other. Understand the difference graphically between average rate and instantaneous rate. Be able to calculate both. Explain the meaning of the reaction rate law and the rate law constant. Be able to determine a reaction rate law for a reaction from experimental data. Calculate the rate law constant (including units) after finding the rate law constant from experimental data. After this, calculate the rate of another experiment not included in the data. Understand what is meant by order in terms of a reactant as well as the overall order. Explain graphically the concept of activation energy and how temperature affects reaction rate. Understand how temperature affects the rate law constant for a reaction. Be able to relate the collision model to all of the above. Explain what is meant by a reaction mechanism and know the meaning of elementary steps, rate-determining step, and intermediate species. Be able to explain and show how a rate law is derived from a certain reaction mechanism. Describe the theory of how a catalyst works.
I. Reaction Rates A. Chemical Kinetics The Rate of a Chemical Reaction depends on 4 things: 1. 2. 3. 4. B. Reaction Rate 1. Average Rate = [reactant] or [product] t t 2. Instantaneous Rate 3. Sample Exercise 14.1- Using the following data, calculate the average rate of disappearance of C 4 H 9 Cl over the time interval from 50.0 to 150.0 s C 4 H 9 Cl (l) + H 2 0 (l) C 4 H 9 OH (aq) + HCl (aq) Time (s) [C 4 H 9 Cl] (M) 0.0 0.1000 50.0 0.0905 100.0 0.0820 150.0 0.0741 200.0 0.0671 C. Reaction Rates and Stoichiometry aa + bb cc + dd Rate = 1 [A] = 1 [B] = 1 [C] = 1 [D] a t b t c t d t 2
D. Sample Exercise 14.2 (a) How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2 O 3 (g) 3 O 2 (g)? (b) If the rate of appearance of O 2 is 6.0 x 10 5 M/s at a particular instant, what is the value of the rate of disappearance of O 3 at this same time? II. The dependance of rate of concentration A. Rate Law An equation (expression) that relates the reaction rate to the [conc] of reactants (or sometimes products). Reaction Constant k a constant of proportionality between the reaction rate and [conc] of reactants that appear in the rate law. General Form = Rate = k[a] x [B] y [C] z. B. Reaction orders and Overall Reaction Order The exponents in the rate law expression (x,y,z) are called the rate order for those reactants. The sum of all the rate orders in the rate law expression is called the overall reaction order. Reaction orders are determined experimentally, not by the coefficients from the balanced chemical equation. C. Units of Rate Constants Units of rate constants are important in helping to solve rate order problems Units of Rate = (Units of Rate Constant) (Units of Concentration) x D. Sample Exercise 14-3 (a) What are the overall reaction orders for the following reactions? #1 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Rate = k[n 2 O 5 ] #2 CHCl 3 (g) + Cl 2 (g) CCl 4 (g) + HCl Rate = k[chcl 3 ] [Cl 2 ] 1/2 (b) What are the usual units of the rate constant for the rate law for equation #1? 3
E. Using Initial Rates to Determine Rate Laws Consider the reaction below between bromate and bromide ions in acid solution. The reactions kinetics can be studied and some results are given in the table. BrO 3 -(aq) + 5Br - (aq) + 6H + (aq) 3Br 2(aq) + 3H 2 O (l) Initial concentrations Mixture [BrO - 3 ] / M [Br - ] / M [H + ] / M Relative Rate A 0.05 0.25 0.3 1 B 0.1 0.25 0.3 2 C 0.1 0.5 0.3 4 D 0.1 0.5 0.6 16 By comparing the results from mixtures A & B; Doubling [BrO 3 - ] doubles the rate. (Note the other concentrations are kept constant to ensure a fair test). Therefore the rate α [BrO 3 - ] and the rate is said to be first order with respect to, (w.r.t), [BrO 3 - ]. By comparing the results from mixtures B & C; Doubling the [Br - ] doubles the rate. (Note the other concentrations are kept constant to ensure a fair test). Therefore the rate α [Br - ] and the rate is said to be first order w.r.t [Br - ]. By comparing the results from mixtures C & D; Doubling the [H + ] quadruples the rate. (Note the other concentrations are kept constant to ensure a fair test). Therefore the rate α [H + ] 2 and the rate is said to be second order w.r.t [H + ]. Combining these results gives Rate [BrO 3 - ][Br - ][H + ] 2 And introducing the rate constant (k), the equation becomes Rate = k[bro 3 - ][Br - ][H + ] 2 F. Sample Exercise 14.4 The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, with the results given below: Experiment # [A] (M) [B] (M) Initial Rate (M/s) 1 0.100 0.100 4.0 x 10 5 2 0.100 0.200 4.0 x 10 5 3 0.200 0.100 16.0 x 10 5 Determine (a) the rate law for the reaction; (b) the magnitude of the rate constant; (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M. 4
III. Change of Concentration with Time A. First Order Reactions Rxn whose rate depends on the [conc] of a single reactant (raised to the first power) Rate = [A] = k [A] ln[a] t - ln[a] 0 = - k t t ln[a] 0 = - k t + ln[a] t y = m x + b B. Half Life - t 1/2 - the time required for the reactant concentration to drop to one half of its initial value. For a 1 st Order Rxn t 1/2 = 0.693 k C. Second Order Reactions Rate depends on the reactant concentration raised to the second power. For Rate = k [A] 2 1 = k t + 1 [A] t [A] 0 D. Sample Exercises 14.5 The first-order rate constant for the decomposition of a certain insecticide in water at 12 C is 1.45 yr 1, A quantity of this insecticide is washed into a lake in June, leading to a [ ] of 5.0 x 10 7 g/cm 3 of water. Assume that the effective temperature of the lake is 12 C, (a) What is the concentration of the insecticide in June of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 x 10 7 g/cm 3? 14.7 The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300 C, 2NO 2 (g) 2 NO(g) + O 2 (g) Is the reaction 1 st or 2 nd Order? Time (s) [NO 2 ] (M) Time (s) [NO 2 ] (M) 0.0 0.0100 150.0 0.0048 50.0 0.0079 200.0 0.0038 100.0 0.0065 5
IV. Temperature and Rate A. The Collision Model As a guideline in many reactions a 10 o C rise will result in an approximate doubling of the rate. Consider the Maxwell- Boltzman distribution graph of molecular energies. The area underneath the graph represents the total number of molecules. At higher temperatures, (T high), the average energy of the molecules is greater and a greater number of molecules will have energies > E a. (i.e. the area to the right of Eact underneath the T high curve in the graph above is much larger than that of the area to the right of Eact underneath the T low curve). A reaction will always be much faster at a higher temperature because many more molecules possess the activation energy on collision. In addition, as the temperature increase so will the number of collisions. 6
B. Activation Energy (i) The activation energy is always endothermic regardless of the overall heat change. (ii) H is positive for an endothermic reaction and negative for an exothermic reaction. (iii) The maximum energy state (TS) is called the transition state or activated complex and can be regarded as the point at which the bonds in the reactants have (or are about to be) broken and the bonds in the products have (or are about to be) formed. (iv) A comparison of (Y) and (Z) shows that H is the same in each reaction. It shows that the products in both reactions are thermodynamically more stable (and to the same extent) than the reactants. (Y) however has a lower activation energy meaning that the reactants in this reaction are less kinetically stable than those in (Z). i.e. the reaction in (Y) will proceed at a faster rate. C. The Arrhenius Equation V. Reaction Mechanisms A. Elementary Steps B. Intermediate Steps 7
C. Rate Laws of Elementary Processes D. Rate Laws of Multistep Mechanism 1. Rate-determining Step 2. Mechanism with an initial fast step 8
VI. Catalyst A. Definition A. Homogeneous Catalyst B. Heterogeneous Catalyst C. Enzymes 9
1. For each table of data below, deduce the orders of reaction, the rate equation, the rate constant and the units of the rate constant. Experiment Starting [A] / M Starting [T] / M Rate / Ms -1 1 0.05 0.25 0.16 2 0.10 0.25 0.64 3 0.05 0.125 0.16 Experiment Starting [C] / M Starting [D] / M Rate / Mhr -1 1 0.05 0.25 0.0083 2 0.10 0.25 0.0166 3 0.05 0.125 0.00415 Experiment Starting [X] / M Starting [Y] / M Rate / Mmin -1 1 0.05 0.25 0.044 2 0.10 0.25 0.044 3 0.05 0.125 0.022 3. The table below shows the results of an investigation into the rate of reaction between hydroxide ions and phosphinate ions (PH 2 O 2 - ) at 80 o C. Experiment [phosphinate ion] / moldm -3 [hydroxide ion] / moldm -3 Rate / cm3 min -1 1 0.6 1 2.4 2 0.6 2 9.6 3 0.6 3 21.5 4 0.1 6 14.4 5 0.2 6 28.8 6 0.3 6 43.2 (i) (ii) What is the rate expression for this reaction? Calculate the rate constant k, and give its units. 10
D. Summary of graphical interpretations of orders of reaction Order Zero First Second Rate Law Rate = k Rate = k [A] Rate = k [A] 2 Integrated rate law [A] = -kt + [A] 0 ln [A] = -kt + ln [A] 0 Plot to give straight line [A] versus t ln [A] versus t versus t Slope (gradient) Slope = -k Slope = -k Slope =k Half-life 11