Unit 4 Chemical Kinetics and Chemical Equilibrium Reaction Rates Rate Laws First and Second Order Reactions Chemical Equilibrium Equilibrium Constants LeChatelier’s Principle

Reaction Rates Questions to consider: What makes “superglue” bond instantly while Elmer’s glue does not? What factors determine how quickly food spoils? Why do “glow sticks” last longer when stored in the freezer? How do catalytic converters remove various pollutants from car exhaust?

Reaction Rates These types of questions can be answered using chemical kinetics. The study of the speed or rate at which chemical reactions occur

Reaction Rates The rate of a chemical reaction is affected by many factors, including: concentration of reactants as concentration of reactants increases the rate of reaction generally increases

Reaction Rates The rate of a chemical reaction is affected by many factors (cont): reaction temperature food spoils more quickly at room temperature than in a refrigerator bacteria grow faster at RT than at lower temperatures

Reaction Rates The rate of a chemical reaction is affected by many factors (cont): presence of a catalyst a substance that increases the rate of a reaction without being consumed in the reaction Enzymes biological catalysts proteins that increase the rate of biochemical reactions

Reaction Rates The rate of a chemical reaction is affected by many factors (cont): surface area of solid or liquid reactants or catalysts as surface area increases the rate of reaction generally increases

Reaction Rates The speed of an object or event is the change that occurs in a given time interval. Speed of a car = change in distance time interval =  d  t Remember, the term change always refers to final value minus initial value.

Reaction Rates Similarly, the rate (or speed) of a reaction can be determined: Rate = change in concentration (or moles) of product time interval Rate =  (conc. or moles)  t

Reaction Rates Consider the chemical reaction: A B Time = mol A t = 20. min 5.0 mol A 5.0 mol B t = 40. min 2.0 mol A 8.0 mol B

Reaction Rates If the number of moles of A and B are measured and plotted, a graph such as this one can be obtained This data can be used to find the reaction rate.

Reaction Rates The reaction rate for a chemical reaction can be expressed as either: the increase in concentration (or number of moles) of a product as a function of time. the decrease in concentration (or number of moles) of a reactant as a function of time OR

Reaction Rates In this reaction: Average rate of appearance of B = change in # of moles of B change in time =  (mol B)  t  We can calculate the average rate for any time interval involved in the reaction.

Reaction Rates If we consider the rate of appearance of B over the first 20 minutes of reaction: Average rate of appearance of B =  (mol B)  t = 5.0 mol B – 0.0 mol B 20. min – 0. min = 0.25 mol/min

Reaction Rates The average rate of appearance of B during the second 20 minutes of the reaction: Avg. rate = 8.0 mol B – 5.0 mol B 40. min – 20. min =0.15 mol/min Notice that the average rate of reaction decreases over the course of the reaction.

Reaction Rates The rate of a reaction can also be expressed as the disappearance of A as a function of time. For this particular reaction, when 1 mole of B is formed, 1 mole of A must disappear. A B

Reaction Rates Time  B  A Interval  t  t 0 – 20.0 min0.25 mol min 20.0 – 40.0 min0.15 mol min Notice:  B/  t = -  A/  t

Reaction Rates We don’t want to report two different values for the rate of the reaction. For reactions with 1:1 stoichiometry: Avg. rate =  (moles product)  t =-  (moles reactant)  t

Reaction Rates For most reactions, the reaction rate is expressed as a change in concentration of a particular reactant or product Average Rate =  [Product] = -  [Reactant]  t where [Product] = concentration of product [Reactant] = concentration of reactant units: M / sec or M / min M = molarity = moles/liter

Reaction Rates On the exam, you will be expected to find the average rate of reaction for a specific time interval when given the concentration or number of moles of either reactants or products as a function of time.

Reaction Rates Example: Given the following data, what is the average rate of the following reaction over the time interval from 54.0 min to min? CH 3 OH (aq) + HCl (aq) CH 3 Cl (aq) + H 2 O (l) Time (min)[HCl] (M)

Reaction Rates Given: [HCl] 54 min = 1.58 M [HCl] 215 min = 1.02 M Find: avg. rate of disappearance of HCl Avg. rate = -  [HCl]  t = - (1.02 M M) 215 min - 54 min = M / min

Reaction Rates Example: Calculate the average reaction rate for the reaction A  B during the first 60.0 minutes using the following data: Time[A] 0.0 min1.50 M 20.0 min1.00 M 40.0 min0.80 M 60.0 min0.75 M 80.0 min0.70 M

Reaction Rates So far, all reactions have had a one-to-one stoichiometry. What happens when the coefficients are not all 1? 2 A 3B

Reaction Rates Consider the following reaction: 2 HI (g) H 2 (g) + I 2 (g) Timemolmolmol (min) HI H 2 I

Reaction Rates Calculate the change in HI and H 2 as a function of time for the first 20.0 minutes of reaction: 2 HI (g) H 2 (g) + I 2 (g)  HI  H 2 Time molmolmol  t  t (min) HI H 2 I 2 (mol/min)

Reaction Rates The average reaction rate must be numerically the same, regardless of whether you express it as the rate of appearance of product or the rate of disappearance of reactant. HI disappears twice as fast as H 2 appears. To make the rates equal: Rate = - 1  [HI] =  [H 2 ] 2  t  t

Reaction Rates In general, for a reaction: a A + b B c C + d D the rate of the reaction can be found by: Rate = - 1  [A] = - 1  [B] = 1  [C] = 1  [D] a  tb  tc  td  t

Reaction Rates Rate = - 1  [A] = - 1  [B] = 1  [C] = 1  [D] a  tb  tc  td  t This equation can be used to establish the relationship between rate of change of one reactant or product to another reactant or product. You have to be able to do this on the test, too!

Reaction Rates Example: How is the rate of disappearance of N 2 O 5 related to the rate of appearance of NO 2 in the following reaction? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)

Reaction Rates Example: If the rate of decomposition of N 2 O 5 in the previous example at a particular instant is 4.2 x M /s, what is the rate of appearance of NO 2 ? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Given: -  [N 2 O 5 ] = 4.2 x M /s  t

Reaction Rates 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Rate = - 1  [N 2 O 5 ] = 1  [NO 2 ] 2  t4  t So:  [NO 2 ] = - 4  [N 2 O 5 ]  t 2  t = 2 x 4.2 x M /s = 8.4 x M/s

Reaction Rates Recall that the average reaction rate changes during the course of the reaction. Until now, we have calculated average reaction rates. The reaction rate at a particular time (not time interval) is called the instantaneous reaction rate.

Reaction Rate The instantaneous reaction rate is found by determining the slope of a line tangent to the curve at the particular time of interest. Fortunately (for you), you won’t have to do this on the exam or HW!

Rate Laws Consider the data presented earlier for the disappearance of HCl as a function of time for the following reaction. CH 3 OH (aq) + HCl (aq) CH 3 Cl (aq) + H 2 O (l) Time (min)[HCl] (M)

Rate Laws The average reaction rate decreases with time. The reaction slows down as the concentration of reactants decreases. CH 3 OH (aq) + HCl (aq) CH 3 Cl (aq) + H 2 O (l) Time (min)[HCl] (M)Avg. Rate (M /min)

Rate Laws In general, the rate of any reaction depends on the concentration of reactants. The way in which the reaction rate varies with the concentration of the reactants can be expressed mathematically using a rate law. An equation that shows how the reaction rate depends on the concentration of the reactants

Rate Laws For a generalized chemical reaction: w A + x B y C + z D the general form of the rate law is: Rate = k[A] m [B] n where k = rate constant m, n = reaction order

Rate Laws Rate Constant (k) a proportionality constant that relates the concentration of reactants to the reaction rate Reaction Order the power to which the concentration of a reactant is raised in a rate law Overall reaction order The sum of all individual reaction orders

Rate Laws Rate laws must be determined experimentally. Measure the instantaneous reaction rate at the start of the reaction (i.e. at t = 0) for various concentrations of reactants. You CANNOT determine the rate law by looking at the coefficients in the balanced chemical equation!

Rate Laws First Order Reaction Overall reaction order = 1 Rate = k[A] Expt[A] (M)Rate (M/s)

Rate Laws Second Order Reaction Overall reaction order = 2 Rate = k[A] 2 Expt[A] (M)Rate (M/s)

Rate Laws Third Order Reaction Overall reaction order = 3 Rate = k[A] 3 Expt[A] (M)Rate (M/s)

Rate Laws Zero Order Reaction Overall reaction order = 0 Rate = k[A] 0 = k Expt[A] (M)Rate (M/s)

Rate Laws REMEMBER Rate laws must be determined experimentally. Determine the instantaneous reaction rate at the start of the reaction (i.e. at t = 0) for various concentrations of reactants. You CANNOT determine the rate law by looking at the coefficients in the balanced chemical equation!

Rate Laws To determine the rate law from experimental data, identify two experiments in which the concentration of one reactant has been changed while the concentration of the other reactant(s) has been held constant determine how the reaction rate changed in response to the change in the concentration of that reactant.

Rate Laws To determine the rate law from experimental data (cont) Repeat this process using another set of data in which the concentration of the first reactant is held constant while the concentration of the other one is changed.

Rate Laws Example: The initial reaction rate of the reaction A + B C was measured for several different starting concentration of A and B. The following results were obtained. Determine the rate law for the reaction. Expt #[A] (M)[B] (M)Initial rate (M /s) x x x 10 -5

Rate Laws Rate = k [A] m [B] n Compare experiments 1 and 2 to find n: [A] = constant [B] = doubles Expt #[A] (M)[B] (M)Initial rate (M /s) x x x Rate doubles: 8.0 x = x x 2 [2] n = 2.0 n = 1 Rate = k[A] m [B] 1

Rate Laws Rate = k [A] m [B] n Compare experiments 1 and 3 to find m: [A] = doubles [B] = constant Expt #[A] (M)[B] (M)Initial rate (M /s) x x x Rate quadruples: 16.0 x = x x 2 x 4 [2] m = 4.0 n = 2 Rate = k[A] 2 [B]

Rate Laws You can also solve this using algebra: Rate = k [A] m [B] n Compare experiments 1 and 3 to find m: Rate 2 = k [0.200 M] m [0.100 M] n = 16.0 x =4.0 Rate 1 k [0.100 M] m [0.100 M] n 4.0 x [0.200 M] m = 4.0 [0.100 M] m 2 m = 4.0 only if m = 2 [2.00] m = 2.0 Rate = k[A] 2 [B] n

Rate Laws You can also solve this using algebra: Rate = k [A] m [B] n Compare experiments 1 and 2 to find n: Rate 2 = k [0.100 M] m [0.200 M] n = 8.0 x = 2.0 Rate 1 k [0.100 M] m [0.100 M] n 4.0 x [0.200 M] n = 2.0 [0.100 M] n 2 n = 2.0 only if n = 1 [2.00] n = 2.0 Rate = k[A] 2 [B]