The study of. Kinetics - Chapter 14 reactions are reactions that will happen - but we can t tell how fast. - the steps by which a reaction takes place. Factors that Affect Rx Rates 1. The more readily molecules collide with each other, the more rapidly they react. Gases and liquid solutions are easily mixed Solids are limited to the surface area for RXs. Crushed powders react faster than large tablets 2. As concentration increases, the frequency with which the reactant molecules collide increases, leading to increased rates. 3. Rates increase as temp increases. Increases KE of molecules, move more rapidly, collide more frequently and also with higher energy Why we refrigerate foods 4. Increase rates without being used up. They affect the kinds of collisions (the mechanism). Enzymes are catalysts in biological systems. More on these at the end of Chapter! Reaction Rate The change in the concentration of reactants or products per unit time. (Molarity per second). Hypothetical reaction, A B A is disappearing as B is appearing Rate = Rate = For this reaction N 2 + 3H 2 2NH 3 As the reaction progresses the concentration H 2 goes down As the reaction progresses the concentration N 2 goes down 1/3 as fast As the reaction progresses the concentration NH 3 goes up. (Draw all three lines onto a set of axes)
Calculating Rates are taken over long intervals. Example: The initial concentration 1.00 M dropped to.90 M in 10 sec. Rate = - d[ ]/d t Solve for average rate. are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time Derivative. Average slope method Instantaneous slope method. Reaction Rates and Stoichiometry In general, for the reaction a A + b B c C + d D The rate is given by Rate = In our example: N 2 + 3H 2 2NH 3 You Try It How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction? 2 O3 (g) 3 O2 (g) If the rate at which oxygen appears is 6.0 x 10-5 M/s at a particular instant, at what rate is ozone disappearing at this same time?
More Practice The decomposition of dinitrogen pentoxide produces nitrogen dioxide and oxygen gas. Write the balanced equation. If the rate of decomposition of the reactant at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of the two products? 14.3 Concentration & Rate Reactions are. As products accumulate they can begin to turn back into reactants. Early on the rate will depend on only the amount of reactants present. We want to measure the reactants as soon as they are mixed. This is called the The method of Initial Rates This method requires that a reaction be run. The initial concentrations of the reactants are. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction. Determining Rate Laws shows how the rate depends on the concentrations of reactants. Must be determined from experimental data. For this reaction: 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won t play a role You will find that the rate will only depend on the concentration of the reactants. Rate = This is called a k is called the. (depends on temperature and the solvent used) n is the of the reactant -usually a positive integer.
Since the rate at twice the concentration is twice as fast the rate law must be Rate = We say this reaction is in N 2 O 5 The only way to determine order is to run the experiment. General Terms a A + b B c C + d D The rate law has the form: rate = We use the letters m, n, p, q, etc. Letter o not used because it may be confused as a zero. Reaction Orders The exponents m and n (etc.) are called. If the exponent is one, the rate is order with respect to that reactant. If the exponent is two, the rate is order with respect to that reactant. The reaction order is the of the orders with respect to each reactant. Reaction Orders The exponents in a rate law indicate how the rate is affected by the concentration of each reactant. the rate doubles when concentration doubles, triples when conc. triples, and so forth. if you double the concentration of a reactant, the rate will quadruple (2 2 = 4), tripling the concentration causes the rate to increase nine fold (3 2 =9) concentration has no effect on the rate You Try It The experimentally determined rate law for the reaction 2NO + 2H2 N2 + 2 H2O is rate = k[no] 2 [H2]. What are the reaction orders in this rate law? Does doubling the concentration of NO have the same effect on rate at which doubling the concentration of H 2?
Determine the order H 2 O 2 + 3I -1 + 2 H + I 3-1 + 2 H 2 O Given the following data, determine the rate law: Initial [H202] Initial [I-] Initial Rate (M/min).10.10 6.9 x 10-4.05.10 3.4 x 10-4.10.20 1.4 x 10-3 (no data given for Hydrogen ion means assume order is zero) Rate Constant, k Solve for the rate constant, k, and include its units. K units= [ ] (1-(total order)) time Another Example 2A + 2B C Initial [A] Initial [B] Initial rate (M/s).10.20 300.30.40 3600.30.80 14460 Find the rate law. Find k, with units. Another example For the reaction BrO - 3 + 5 Br - + 6H + 3Br 2 + 3 H 2 O The general form of the Rate Law is Rate = k[bro - 3 ] n [Br - ] m [H + ] p We use experimental data (next page) to determine the values of n,m,and p Solve for K, including units
Initial concentrations (M) BrO 3 - Br - H + Rate (M/s) 0.10 0.10 0.10 8.0 x 10-4 0.20 0.10 0.10 1.6 x 10-3 0.20 0.20 0.10 3.2 x 10-3 0.10 0.10 0.20 3.2 x 10-3 14.4 The Change of Concentration with Time Rate laws enable us to calculate from the rate constant and reactant concentrations. Rate laws can also be converted into equations that tell us what the concentrations of the reactants or products are at any during the course of a reaction. (Calculus) Types of Rate Laws l Rate law - describes how depends on. Rate Law - Describes how depends on. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms. Remember Differential Rate Laws Rate = An equation which shows how the. The concentration of the products does not appear in the rate law because this is an rate. The order must be determined experimentally, can t be obtained from the equation Integrated Rate Law If we integrate this equation, this relationship can be transformed into an equation that related concentration of A at the start of the reaction, [A]0 to its concentration at any other time, t, [A]t.
**Expresses the reaction. Form of the equation depends on the order of the rate law (differential). Changes Rate = We will only work with n=0, 1, and 2 First Order For the reaction 2N 2 O 5 4NO 2 + O 2 We found the Rate = k[n2 O 5 ] 1 If concentration doubles rate doubles. If we integrate this equation with respect to time we get the ln is the natural log [N 2 O 5 ] 0 is the initial concentration. Differential form Rate = Integrated form In the form y = mx + b (straight line) y = ln[a] m = -k x = t b = ln[a] 0 A graph of ln[a] vs time is a. By getting the straight line you can prove it is first order Often expressed in a ratio Let s Practice The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr -1 at 12 C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm 3. Assume that the average temperature of the lake is 12 C. What is the concentration of the insecticide on June 1 of the following year? How long will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g/cm 3?
Another Try The decomposition of dimethyl ether, (CH3)2O at 510 C is a first-order process with a rate constant of 6.8 x 10-4 s-1. (CH3)2O (g) CH4 (g) + H2 (g) + CO (g) If the initial pressure of dimethyl ether is 135 torr, what is its partial pressure after 1420 seconds? Half Life The time required to. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 ln(2) = kt 1/2 t 1/2 = 0.693/k The time to reach half the original concentration does not depend on the starting concentration. First Order Half-Life Does not depend on the. Half-life remains throughout the reaction The concentration of the reactant decreases by ½ in each of a series of regularly spaced time intervals, namely, t1/2 is a first-order process. Practice with Half-Life The decomposition of N2O5 into 2NO2 and ½ O2 in CCl4 solvent is first order with a rate constant of 6.2 x 10-4 min -1 at a given temperature. What is the half-life? What is the time required for 20% of the N2O5 to decompose?
Second Order Differential Rate Law: Rate = Integrated Rate Law y= 1/[A] m = k x= t b = 1/[A] 0 A straight line if 1/[A] vs t is graphed Knowing k and [A] 0 you can calculate [A] at any time t Important One way to distinguish between first and second order rate laws is to graph both ln[a]t and 1/[A]t against t. If the ln[a]t plot is linear, the reaction is order. If the 1/[A]t plot is linear, the reaction is order. Try It The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300C, NO2 NO + ½ O2 Time(s) [NO2] (M) 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 Is this reaction first or second order in NO2? Practice this Consider again the decomposition of NO2. The reaction is second order in NO2 with k = 0.543 M -1 s -1. If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the remaining concentration after 0.500 hours?
Second Order Half Life [A] = [A] 0 /2 at t = t 1/2 In contrast to the behavior of first-order reactions, the half-life for second order and other reactions Zero Order Rate Law Rate = k[a] 0 = k Rate does. Integrated [A] = When [A] = [A] 0 /2 t = t 1/2 t 1/2 = Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry. Temperature and Rate The rates of most chemical reactions as the temperature rises. Why? Molecules must collide in order to react. The greater the number of collision occurring per second, the greater the reaction rate. Greater Concentration and Higher Temperature both lead to an increased reaction rate according to Collision Model Orientation Factor Molecules must be oriented in a certain way during collisions in order for a reaction to occur. Atoms must be suitably positioned for bonds to break and form new bonds. Review Molecules must to react. affects rates because collisions are more likely. Must collide hard enough. Temperature and rate are related. Only a small number of collisions produce reactions. Activation Energy, Ea In order to react, colliding molecules must have a total kinetic energy equal to or greater than some minimum value. - the minimum energy needed to make a reaction happen.
Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier. Arrhenius Said that reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier. The number of collisions with the necessary energy increases. The fraction of molecules that has an energy equal to or greater than Ea is given by the expression f = e-e a /RT e is (opposite of ln) E a = R = T is Magnitude of f Suppose that Ea is 100 kj/mol, a typical value of many reactions, and that T is 300 K, around room temp. Calculate the value of the fraction of molecules that has energy equal to or greater than Ea. Calculate the f at 310 K. Thus, a 10-degree change increase in temperature produces a 3.7 fold increase in the fraction of molecules possessing at least 100 kj/mol of energy Arrhenius Arrhenius found that most obeyed an equation based on three factors The fraction of molecules possessing an energy of Ea or greater. The number of collisions occurring per second The fraction of collisions that have the appropriate orientation Arrhenius Equation Incorporating these 3 factors into his equation: k = Ae Ea/RT k is the
Ea is R is T is A =, is constant as temperature is varied. As the magnitude of Ea, k because the fraction of molecules that possess the required energy is smaller. Activation Energy Reaction rates decrease as activation energy increases. Calculating Activation Energy: Taking the natural log of both sides of Arrhenius Equation we have Straight line equation predicts that a graph of versus will be a line with a slope equal to, y intercept will be. Non-graphical Way to Calculate Ea We need to know the rate constant of a reaction at two or more temperatures. Practice Problem The following table shows the rate constants for the rearrangement of methyl isonitirle at various temperatures. Temperature C k (s -1 ) 189.7 2.52 x 10-5 198.9 5.25x 10-5 230.3 6.30x 10-4 251.2 3.16x 10-3 From these data, calculate the Ea for the reaction. What is the value of the rate constant at 430.0 K?
14.6 Reaction Mechanisms The process by which a reaction occurs is called a. There is activation energy for each elementary step. Activation energy determines k. k = Ae - (E a /RT) k determines rate Slowest step (rate determining) must have the highest activation energy. Reaction Mechanisms The series of steps that actually occur in a chemical reaction. Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products. 2NO 2 + F 2 2NO 2 F Rate = k[no 2 ][F 2 ] The proposed mechanism is NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F (fast) F is called an It is formed then consumed in the reaction Reaction Mechanisms Each of the two reactions is called an. The rate for a reaction can be written from its. Molecularity is the number of pieces that must come together. step involves one molecule - Rate is first order. - requires two molecules - Rate is second order step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule. Rate Law for Elementary Reactions
Elementary reactions are significant in a very important way: It is based directly on its molecularity. A products Rate = k[a] A+A products Rate= k[a] 2 2A products Rate= k[a] 2 A+B products Rate= k[a][b] A+A+B Products Rate= k[a] 2 [B] 2A+B Products Rate= k[a] 2 [B] A+B+C Products Rate= k[a][b][c] Let s Practice It has been proposed that the conversion of ozone gas into oxygen gas proceeds by a two-step mechanism: O3 O2 + O O3 + O 2 O2 Describe the molecularity of each elementary reaction Write the equation for the overall reaction Identify the intermediate(s) More Practice If the following reaction occurs in a single elementary reaction predict the rate law: H2 + Br2 2 HBr Consider the following reaction: 2 NO + Br2 2NOBr Write the rate law for the reaction, assuming it involves a single elementary reaction Is a single-step mechanism likely? Remember Multi-Step Mechanisms most common
Most chemical reactions occur by mechanisms that involve two or more elementary reactions. Each step has its own Often one of the steps is much slower than the others. The overall rate of a reaction cannot exceed the rate of the slowest elementary step = The slowest step in a multi-step reaction limits the overall rate. The rate-determining step governs the rate law for the overall reaction. Mechanisms with a Slow Initial Step Step 1: Step 2: Overall: Step 2 is much faster, k2 >>k1 Rate = Practice The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism N2O N2 + O (slow) N2O + O N2 + O2 (fast) Write the equation for the overall reaction Write the rate law for the overall reaction Mechanisms with a Fast Initial Step It is difficult to derive the rate law for a mechanism in which an intermediate is a reactant in the rate determining step. Example: 2NO + Br2 2 NOBr Experimental rate law: Rate = k [NO] 2 [Br2] How could this happen? Predict a reaction mechanism for this rate law: One possibility is that the reaction occurs in a single termolecular step: NO + NO + Br2 2 NOBr Not likely Let s consider an alternative
Alternative Mechanism Step 1: Step 2: Problem: Rate is dependent on the concentration of the intermediate too unstable to measure. Solution Express the concentration of the intermediate in terms of the concentrations of the starting reactants. Because step 1, both forward and reverse, occur rapidly with respect to step 2, a dynamic equilibrium is established. Their rates are equal. Rate of forward reaction = rate of reverse reaction Solve for [NOBr 2 ] [ Substituting this relationship into the rate law for the rate-determining step: Rate = Rate = Consistent with experimental rate law In general Whenever a fast step precedes a slow one, we can solve for the concentration of an intermediate by assuming that equilibrium is established in the fast step. Practice Show that the following mechanism for the reaction 2NO + Br2 2 NOBr also produces a rate law consistent with the experimentally observed one. Step 1: NO + NO N2O2 (fast) Step 2: N2O2 + Br2 2 NOBr (slow) Try This The first step of a mechanism involving the reaction of bromine is Br2 2 Br (fast, equilibrium) What is the expression relating the concentration of Br to that of Br2? 14.7 Catalysts
Speed up a reaction without being used up in the reaction. are biological catalysts. Catalysts are in the same phase as the reactants. Catalysts are in a different phase as the reactants. How Catalysts Work Catalysts allow reactions to proceed by a - a new pathway. New pathway has activation energy. More molecules will have this activation energy. Heterogenous Catalysts Catalysts will speed up a reaction but only to a certain point. Past a certain point adding more reactants won t change the rate. Zero Order Catalysts and rate. Rate increases until the. Then rate is.