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Chapter 14: Chemical Kinetics

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Chapter 14: Chemical Kinetics C h e m i c a l K i n e t i c s P a g e | 1 Chapter 14: Chemical Kinetics Homework: Read Chapter 14 Work out sample/practice exercises in the sections, Check for the MasteringChemistry.com assignment and complete before due date Introduction to Kinetics: Chemists generally want to know … What happens? (Balanced equations) To what extent does it happen? (Equilibrium) The rate it happens and by what mechanism? (Kinetics) Will the reaction spontaneously occur? (Thermodynamics) KINETICS is the area of chemistry concerned with the RATE of a reaction; the variables that affect rate and the REACTION MECHANISM ; the pathway by which a reaction occurs. Kinetic studies have environmental, biological and economic importance. Kinetics will not tell us the extent of the reaction (Equilibrium) or whether the reaction spontaneously occurs under specified conditions (Thermodynamics). Kinetic rates vary tremendously: Methane combustion: (very fast reaction after a spark initiates the reaction) CH4 (g) + 2 O2(g) CO2(g) + 2 H2O(g) H˚ = -802.3 kJ/mol Diamond combustion: (thankfully this does not occur at an observable rate) Cdiamond + O2(g) CO2 (g) H˚ = -395.4 kJ/mol (˚) represents a value under standard state conditions: 25˚C, 1 atm Try this #1: Predict some variables that may influence the rate of reactions? C h e m i c a l K i n e t i c s P a g e | 2 Factors that Affect Reaction Rates: 1. Nature of the reactants- periodic trends, physical state, surface area Examples… Li, Na, K react with water at increasing rates liquid gasoline burns slower that gas vapor powders react faster than large pieces dry solid reactants generally react slower than solutions of the same reactants Try this #2: Form an explanation for each of these trends. 2. The concentration/pressures of the reactants Examples… higher concentrations or partial pressures of reactants generally speed a reaction generally reaction rates slow over time Try this #3: Form an explanation for each of these trends. At any time an instantaneous rate can be calculated from the negative slope of the tangent to the curve of time (x axis) verses reactant concentration (y axis) or the positive tangent to the time verses product concentration. Figure 1 Kinetic Graph Rate of disappearance of the reactant and appearance of product 1.4 1.2 1 0.8 0.6 0.4 Reactant [M] 0.2 Product [M] 0 Concentration Concentration (M) 0 5 10 15 20 25 Time (minutes) C h e m i c a l K i n e t i c s P a g e | 3 3. The temperature at which the reaction occurs Example… as temperature increases the rate of reactants disappearing also increases as temperature drops the rate generally drops Try this #4: Form an explanation for this trend. 4. The presence of catalysts, enzymes, or inhibitors Examples… Catalysts alter the mechanism/pathway of the reaction. The activation energy is lowered and the steps in the reaction may vary from that without a catalyst. Enzymes are biological catalysts. Inhibitors alter the Figure 2 Reaction Coordinate mechanism sometimes Diagram blocking an active site with and without a catalyst to slow or stop a reaction 400 Try this #5: Form an explanation 300 for this trend. 200 reaction with catalyst Energy (kJ) Energy 100 0 Reactants → Products Reaction Rates: The speed of an event is its change over time. Similar to the speed of a car measured in miles per hour, a reaction rate is the concentration change over change in time, generally in Molarity/time Rate = [X]/ t 1. In slow reactions, samples can be taken of the reaction mixture after successive time intervals to determine rates. 2. In faster reactions, one can measure the change in a physical property, such as pressure if the number of moles of gas change or color by spectroscopy, some computer controlled spectroscopes can measure every picoseconds; 1 x 10-12 second C h e m i c a l K i n e t i c s P a g e | 4 Hypothetical Reaction: For the generic reaction… aA + bB cC + dD The amount is in Molarity [X] or partial pressure in atmospheres [Px]. Change, (), is always final minus initial Rate of disappearance of reactant =[reactant]/time (negative answer) Rate of formation of product =[product]/time (positive answer) Rate of Reaction needs to be the same rate for all species in a balanced reaction, (Finish the equivalents for each reactant and product) Try this #6: Rate of Reaction = -[A]/at = = = 퐛 Try this: #7 for the reaction… H2 + 2 ICl I2 + 2HCl… a) Write the terms for the Rate of Reaction for each reactant/product Rate of Reaction = b) Draw the expected graph with concentrations verses time beginning with 1M H2 and 2 M ICl and no products initially. c) If the rate of disappearance H2 = -0.13 M/s at a specific time, what is the value for the rate of formation of HCl during the same interval? d) What is the generic reaction rate for that time interval? C h e m i c a l K i n e t i c s P a g e | 5 Rate Law Expression: For the generic reaction… a A + b B products Rate = k[A]x[B]y k = Specific rate constant x,y = orders, generally small positive integers, can be zero, fractions or negative k, x, y are experimentally determined Concentrations of Reactants and Catalysts are in the rate law expression. Rarely, Product concentration may be in a rate law expression. Specific rate constant : Information on k 1. Its value is for a specific reaction represented by a balance equation 2. Its value will change if a catalyst is added 3. It is for a specific temperature and will change when Temp changes 4. Its units depend on the overall order. M(1-overall order) time-1 5. It does not change with concentration 6. It does not change with time 7. It is experimentally determined Orders 1. Most common orders are 0, 1 and 2. 2. It is possible to have fractions such as ½ 3. On rare occasions the orders can be negative such as -1 4. Zero order reactant concentration has no affect on the rate as long as the reactant is present 5. First order reactant concentration is directly proportional to the rate. Doubling the concentration will double the rate. 6. The rate increases as a square of the second order reactant concentration. Doubling the concentration will quadruple the rate. 7. Overall order is the sum of all orders in the rate law expression. C h e m i c a l K i n e t i c s P a g e | 6 Rate Law Expression: Try this #8; Determine the orders with respect to each reactant and the overall order for each. Notice if rate law expression orders match the coefficients of the balanced reaction. a) 3NO (g) N2O (g) + NO2 (g) 2 Rate = k[NO] -1 -1 b) (CH3)3CBr (aq) + OH (aq) (CH3)3COH (aq) + Br Rate = k[(CH3)3CBr] c) 2NO (g) + O2 (g) 2 NO2 (g) 2 Rate = k[NO] [O2] Try this #9: Given: 2 NO (g) + 2 H2 (g) 2 H2O (g) + N2 (g) The reaction is second order with NO and first order with respect to H2. a) Write the rate law expression and determine the overall order. b) Specify the units of the specific rate constant using time in seconds. c) How does the reaction rate change if the NO concentration doubles? Triples? d) How does the reaction rate change if both reactant concentrations are doubled? C h e m i c a l K i n e t i c s P a g e | 7 Rate Graphs: Determining Rate Law Expression from Experimental Data: In order to determine orders and specific rate constant of a reaction, one may carry out a series of experiments (the minimum number is equal to the number of unknowns) in which initial rates are measured. It is best to design an experiment in which only one factor (concentration) is changed at a time. Do not confuse the rate of a reaction with the rate constant (k). Try this #10: The following rate data were obtained at 25°C for the reaction… x y 2NO (g) + 2H2 (g) N2 (g) + 2 H2O (g) Rate = k[NO] [H2] Experiment Initial [NO] Initial [H2] Rate forming N2 1 0.10 M 0.10 M 1.23 x 10-3 M/s 2 0.10 M 0.20 M 2.46 x 10-3 M/s 3 0.20 M 0.10 M 4.92 x 10-3 M/s 4 ? 0.30 M 8.00 x 10-3 M/s Solve for the orders, overall order, the Rate Law Expression and the value and units of k (specific rate constant). First try estimation, then verify through algebra and logarithms. Determine the initial concentration of [NO] in expt 4. C h e m i c a l K i n e t i c s P a g e | 8 Try this #11: The following rate data were obtained at 25°C for the reaction… x y z 2A + B + 2C D Rate = k[A] [B] [C] Experiment Initial [A] Initial [B] Initial [C] Rate forming D 1 0.34 M 0.16 M 0.12 M 1.4 x 10-2 M/min 2 0.34 M 0.32 M 0.24 M 2.8 x 10-2 M/min 3 0.34 M 0.16 M 0.35 M 1.4 x 10-2 M/min 4 0.68 M 0.32 M 0.49 M 5.6 x 10-2 M/min 5 0.50 M 0.50 M 0.20 M ? Solve for the orders, overall order, the Rate Law Expression and the value and units of k as well as the initial rate of formation of D in expt.
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