How fast reactants turn into products. Usually measured in Molarity per second units. Kinetics

How fast reactants turn into products. Usually measured in Molarity per second units. Kinetics

Reaction rated are fractions of a second for fireworks to explode. Reaction Rates takes years for a metal to corrode. The erosion of rocks takes thousands of years.

Section 14.1 The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can t tell how fast. Graphite will spontaneously turn to Diamond eventually. Reaction mechanism- the steps by which a reaction takes place.

Factors that Affect Rx Rates 1. The physical states of reactants. 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

Factors that Affect Rx Rates 2. The concentrations of the reactants. As concentration increases, the frequency with which the reactant molecules collide increases, leading to increased rates.

Factors that Affect Rx Rates 3. The temperature at which the reaction occurs. Rates increase as temp increases. Increases KE of molecules, move more rapidly, collide more frequently and also with higher energy Why we refrigerate foods

Factors that Affect Rx Rates 4. The presence of a catalyst. 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!

14.2 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 = - D [A ] = D [B] Dt Dt Rate = Conc. of B at t 2 -Conc. of B at t 1 t 2 - t 1 For this reaction N 2 + 3H 2 2NH 3

As the reaction progresses the concentration H 2 goes down C o n c e n t r a t i o n [H 2 ] Time

C o n c e n t r a t i o n As the reaction progresses the concentration N 2 goes down 1/3 as fast [H 2 ] [N 2 ] Time

As the reaction progresses the concentration NH 3 goes up. C o n c e n t r a t i o n [H 2 ] [NH 3 ] [N 2 ] Time

Calculating Rates Average 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. Instantaneous rates 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 C o n c e n t r a t i o n D[H 2 ] Dt Time

Instantaneous slope method. C o n c e n t r a t i o n D[H 2 ] D t Time

Reaction Rates and Stoichiometry In general, for the reaction a A + b B c C + d D The rate is given by Rate = - 1 D[A] = -1 D[B] = 1 D[C] = 1 D[D] a Dt b Dt c Dt d Dt In our example: N 2 + 3H 2 2NH 3 -D[N 2 ] = -1 D[H 2 ] = 1 D[NH 3 ] Dt 3 Dt 2 Dt

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 reversible. 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 Initial rate method.

The method of Initial Rates This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction.

Determining Rate Laws Rate Law shows how the rate depends on the concentrations of reactants. Rate Expression vs Rate Law 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 NO2 breaks down immed. into NO + O2

2 N 2 O 5 4 NO 2 + O 2 You will find that the rate will only depend on the concentration of the reactants. Rate = k[n 2 O 5 ] n This is called a rate law k is called the rate constant. (depends on temperature and the solvent used) n is the order of the reactant -usually a positive integer.

Practice:Graph this data L2 L1 [N 2 O 5 ] (mol/l) Time (s) 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000

2000 1800 1600 1400 1200 1000 800 600 400 200 0 1.2 1 0.8 0.6 0.4 0.2 0 To find rate we have to find the slope at two points Solve for instantaneous rates at.8m and.4 M.

2000 1800 1600 1400 1200 1000 800 600 400 200 0 At.80 M the rate =- 5.926x 10-4 M/s 1.2 1 0.8 0.6 0.4 0.2 0

2000 1800 1600 1400 1200 1000 800 600 400 200 0 At.40 M the rate is =- 2.963 x 10-4 M/s 1.2 1 0.8 0.6 0.4 0.2 0

Since the rate at twice the concentration is twice as fast the rate law must be.. Rate = -D[N 2 O 5 ] = k[n 2 O 5 ] 1 = k[n 2 O 5 ] Dt We say this reaction is first order 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 = k [A] m [B] n We use the letters m, n, p, q, etc. Letter o not used because it may be confused as a zero. NOT THE STOICHIOMETRY!

Reaction Orders The exponents m and n (etc.) are called reaction orders. If the exponent is one, the rate is first order with respect to that reactant. If the exponent is two, the rate is second order with respect to that reactant. The overall reaction order is the sum 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. First order the rate doubles when concentration doubles, triples when conc. triples, and so forth. Second order 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) Zero Order concentration has no effect on the rate

You Try It The experimentally determined rate law for the reaction 2NO + 2H2 N2 + 2H2O 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 H2?

Determine the order H2O2 + 3I -1 + 2 H + I3-1 + 2 H2O Given the following data, determine the rate law: (no data given for Hydrogen ion means assume order is zero) 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

Rate Constant, k Solve for the rate constant, k, and include its units. K units = 1/(time units)* 1-(total order K units = time -1 *[ ] -1

Another Example 2A + 2B C Initial Initial Initial Find the rate law. [A] Find k, with units. [B] rate (M/s).10.20 300.30.40 3600.30.80 14460

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 to determine the values of n,m,and p Solve for K, including units

Initial concentrations (M) Rate (M/s) BrO - 3 Br - H + 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 the rate of a reaction 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 time during the course of a reaction. (Calculus)

Types of Rate Laws Differential Rate law - describes how rate depends on concentration. Integrated Rate Law - Describes how concentration depends on time. 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 =- D[A] = k[a] m Dt An equation which shows how the rate depends on the concentrations of reactants. The concentration of the products does not appear in the rate law because this is an initial 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.

Integrated Rate Law **Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Changes Rate = D[A] n Dt 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[n 2 O 5 ] 1 If concentration doubles rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law

First Order Integrated Rate Law ln[n 2 O 5 ] t = - kt + ln[n 2 O 5 ] 0 ln is the natural log [N 2 O 5 ] 0 is the initial concentration.

First Order Differential form Rate = D[A] / Dt = k[a] Integrated form ln[a] = - kt + ln[a] 0 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 straight line.

First Order By getting the straight line you can prove it is first order Often expressed in a ratio [ ] A ln A [ ] 0 = kt

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 reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 ln ln(2) = kt 1/2 [ ] A [ ] A 0 0 2 = kt 1 2

Half Life t 1/2 = 0.693/k The time to reach half the original concentration does not depend on the starting concentration. An easy way to find k

First Order Half-Life Does not depend on the starting concentration. Half-life remains constant throughout the reaction The concentration of the reactant decreases by ½ in each of a series of regularly spaced time intervals, namely, t1/2 Radioactive decay 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 = -D[A] / Dt = k[a] 2 Integrated Rate Law 1/[A]t = kt + 1/[A] 0 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 first order. If the 1/[A]t plot is linear, the reaction is second 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 In contrast to the behavior of first-order reactions, the half-life for second order and other reactions depends on reactant concentrations and therefore changes as the reaction progresses. t 1 = 1 2 k[a] 0 The lower the initial concentration, the greater the half-life.

Zero Order Rate Law Rate = k[a] 0 = k Rate does not change with concentration. Integrated [A] = -kt + [A] 0 When [A] = [A] 0 /2 t = t 1/2 t 1/2 = [A] 0 /2k

Zero Order Rate Law Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry.

14.5 Temperature and Rate The rates of most chemical reactions increase as the temperature rises. Why? Collision Model 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. Figure 14.3 Page 594

Review Molecules must collide to react. Concentration 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. Activation energy - 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. How does figure 14.4 shows Ea. Pg 595

P o t e n t i a l E n e r g y Reactants Reaction Coordinate Products

P o t e n t i a l E n e r g y Reactants Reaction Coordinate Activation Energy E a Products

P o t e n t i a l E n e r g y Reactants Activated complex Reaction Coordinate Products

P o t e n t i a l E n e r g y Reactants Reaction Coordinate }DE Products

P o t e n t i a l E n e r g y 2BrNO Br---NO Br---NO Reaction Coordinate Transition State 2NO + Br 2

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 exponentially.

Arrhenius 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 Euler s number (opposite of ln) E a = activation energy R = ideal gas constant T is temperature in Kelvin

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 reactionrate data 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 rate constant Ea is activation energy R is gas constant T is absolute temperature A = frequency factor, is constant as temperature is varied. As the magnitude of Ea increases, k decreases 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 ln k = - Ea + ln A RT Straight line equation predicts that a graph of ln k versus 1/T will be a line with a slope equal to Ea/R, y intercept will be ln A.

Non-graphical Way to Calculate Ea We need to know the rate constant of a reaction at two or more temperatures. ln k1 = Ea ( 1 1 ) k2 R (T2 T1)

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 reaction mechanism. 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.

This reaction takes place in three steps

E a First step is fast Low activation energy

E a Second step is slow High activation energy

E a Third step is fast Low activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

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.

Reaction Mechanisms 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 intermediate It is formed then consumed in the reaction

Reaction Mechanisms Each of the two reactions is called an elementary step. The rate for a reaction can be written from its molecularity. Molecularity is the number of pieces that must come together.

Unimolecular step involves one molecule - Rate is first order. Bimolecular step - requires two molecules - Rate is second order Termolecular 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: If we know that a reaction is an elementary reaction, then we know its rate law. 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 rate constant and activation energy. 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 = RATE DETERMINING STEP.

Rate Determining 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: NO2 + NO2 NO3 + NO (slow) Step 2: NO3 + CO NO2 + CO2 (fast) Overall:? Step 2 is much faster, k2 >>k1 Rate = k1 [NO2] 2 k1 k2

Practice The decomposition of nitrous oxide, N2O, is believed to occur by a twostep 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 k1 Step 1: NO + Br2 NOBr2 (fast) k -1 k2 Step 2: NOBr2 + NO 2 NOBr (slow) 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. k1 [NO][Br2] = k-1 [NOBr2] Rate of forward reaction = rate of reverse reaction

Solve for [NOBr2] [NOBr2] = k1/k-1 [NO][Br2] Substituting this relationship into the rate law for the rate-determining step: Rate = k2 (k1/k-1) [NO][Br2][NO] Rate = k [NO] 2 [Br2] 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 an 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. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.

How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy. More molecules will have this activation energy. Do not change DE

Heterogenous Catalysts H H Hydrogen bonds to surface of metal. Break H-H bonds H H H H H H Pt surface

Heterogenous Catalysts H H H C C H H H H H Pt surface

Heterogenous Catalysts The double bond breaks and bonds to the catalyst. H H H C C H H H H H Pt surface

Heterogenous Catalysts The hydrogen atoms bond with the carbon H H H C C H H H H H Pt surface

Heterogenous Catalysts H H H C C H H H H H Pt surface

Homogenous Catalysts Chlorofluorocarbons catalyze the decomposition of ozone. Enzymes regulating the body processes. (Protein catalysts)

Catalysts and rate 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. R a t e Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration Concentration of reactants