Rate Laws Example: Determine the rate law for the following reaction given the data below. H 2 O 2 (aq) + 3 I - (aq) + 2H + (aq)  I 3 - (aq) + H 2 O (l) [H 2 O 2 ] [I - ] [H + ]Initial Expt # (M) (M) (M)Rate (M /s) x x x x 10 -6

Rate Laws For the reaction: 5Br - (aq) + BrO 3 - (aq) + 6 H + (aq)  3 Br 2 (aq) + 3 H 2 0 (l) the rate law was determined experimentally to be: Rate = k[Br - ] [BrO 3 - ] [H + ] 2 The reaction is first order with respect to Br -, first order with respect to BrO 3 -, and second order with respect to H +.

Rate Laws The previous reaction is fourth order overall. The overall reaction order is the sum of all the exponents in the rate law. Note: In most rate laws, the reaction orders are 0, 1, or 2. Reaction orders can be fractional or negative.

Rate Laws The value of the rate constant can be determined from the initial rate data that is used to determine the rate law. Select one set of conditions. Substitute the initial rate and the concentrations into the rate law. Solve the rate law for the rate constant, k.

Rate Laws The value of the rate constant, k, depends only on temperature. It does not depend on the concentration of reactants. Consequently, all sets of data should give the same rate constant (within experimental error).

Rate Laws Example: The following data was used to determine the rate law for the reaction: A + B  C Calculate the value of the rate constant if the rate law is: Expt #[A] (M)[B] (M)Initial rate (M /s) x x x ??? Rate = k [A] 2 [B]

Rate Laws Select a set of conditions: Substitute data into the rate law: Solve for k

Rate Laws Important: The units of the rate constant will depend on the overall order of the reaction. You must be able to report your calculated rate constant using the correct units.

Rate Laws Example: The following data was used to determine the rate law for the reaction: A + B  C Using the rate constant calculated in the previous example, calculate the initial reaction rate for experiment 4 if the rate law is: Expt #[A] (M)[B] (M)Initial rate (M /s) x x x ??? Rate = k [A] 2 [B]

First Order Reactions A first order reaction is one whose rate depends on the concentration of a single reactant raised to the first power: Rate = k[A] Calculus (integration) gives us the integrated form of the rate law which allows us to predict the concentration of a reactant after a given amount of time has elapsed: ln[A] t = -kt + ln[A] 0

First Order Reactions To predict the concentration of a reactant at a given time during a first order reaction: ln[A] t = -kt + ln[A] 0 where ln = natural logarithm (not log) t =time (units depend on k) [A] t = conc.or amount of A at time t [A] 0 = initial concentration or am’t of A k = rate constant

First Order Reactions Example: A certain pesticide decomposes in water via a first order reaction with a rate constant of 1.45 yr -1. What will the concentration of the pesticide be after 0.50 years for a solution whose initial concentration was 5.0 x g/mL?

First Order Reactions

The time required for the concentration of a reactant to drop to one half of its original value is called the half-life of the reaction. t 1/2 After one half life has elapsed, the concentration of the reactant will be: [A] t = ½ [A] 0 For a first order reaction: t 1/2 = k ½

First Order Reactions Given the half life of a first order reaction and the initial concentration of the reactant, you can calculate the concentration of the reactant at any time in the reaction OR the amount of time needed for the concentration to decrease to a given value. Use t 1/2 = 0.693/k to find the rate constant Substitute k and other given values into the integrated rate law.

First Order Reactions The integrated rate law can also be re- written in terms of the half life of the reaction: [A] t = [A] 0 x (0.5) This equation is easier to use when solving for A t or A 0. The original integrated rate law is easier to use if you need to find t elapsed. t elapsed t 1/2

First Order Reactions Example: A 25.0 mass % solution of a certain drug in water has a half life of 14 days. Calculate the concentration of the drug present in the solution after 30. days.

First Order Reactions Example: A certain pesticide has a half life of yr. How many years will it take for the concentration of a 2.30 x M solution of the pesticide to decrease to 2.3 x M.

Identifying Zero, First & Second Order Reactions Experimentally Notice that the integrated rate law for first order reactions follows the general formula for a straight line: ln[A] t = -kt + ln[A] 0 y = mx + b Therefore, if a reaction is first-order, a plot of ln [A] vs. t will give a straight line, and the slope of the line will be -k.

First Order Reactions Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

First Order Reactions This data was collected for the reaction at 198.9°C. CH 3 NCCH 3 CN Remember that the partial pressure of the gas will be directly proportional to the number of moles of that gas at constant T and V.

First Order Reactions When ln P is plotted as a function of time, a straight line is obtained. Therefore, the process is first-order.

First Order Reactions Slope = 4.0 – ,000 – 5,000 s Slope = -5 x s -1 k = -slope k = 5 x s -1 20,000 – 5,000 s 4.0 – 4.75 To find the rate constant for the reaction, calculate/determine the slope:

Second Order Processes A reaction that is second order with respect to A has the following rate law: Rate = k[A] 2 The integrated form of the second order rate law follows the general formula for a straight line: 1 [A] t = kt + 1 [A] 0

Second Order Processes For a reaction that is second order with respect to A, a plot of 1/[A] versus time will give a straight line. k = slope of line

Zero Order Reactions A zero order reaction is one whose rate is independent of the reactant concentration. Rate = k The integrated form of this rate law follows the general form for a straight line: [A] t = -kt + [A] o Plotting [A] vs. time gives a straight line for a zero order reaction. k = - slope of line

Distinguishing Zero, First, & Second Order Reactions Graphically On your exam, you must be able to use experimental data to graphically determine if a reaction is zero, first, or second order with respect to a given reactant and determine the rate constant of the reaction. Zero Order: A plot of [A] vs. time is a straight line. First Order: A plot of ln[A] vs. time is a straight line. Second Order: A plot of 1/[A] vs. time is a straight line.

Distinguishing Zero, First, & Second Order Reactions Graphically Example: The concentration of A in the reaction: A  2 B was monitored as a function of time. Graphically determine if the reaction is zero, first, or second order with respect to A. Determine the rate constant for the reaction. Time (min)[A] M The data obtained:

Distinguishing Zero, First, & Second Order Reactions Graphically A plot of [A] vs. time is a straight line so the reaction is zero order with respect to A. Slope = 0.50 M – 1.50 M 40. min – 0. min Slope = M/min k = - slope k = M/min

Distinguishing Zero, First, & Second Order Reactions Graphically Notice that the plots of ln[A] vs. time and 1/[A] vs. time are both curves instead of straight lines! This means that the reaction is not first or second order with respect to [A].

Distinguishing Zero, First, & Second Order Reactions Graphically Example: The decomposition of NO 2 at 300°C is described by the equation: NO 2 (g)  NO (g) + 1/2 O 2 (g) and yields the data below. Determine if the reaction is zero, first, or second order with respect to NO 2. Determine the value of k [NO 2 ], MTime (s)

Distinguishing Zero, First, & Second Order Reactions Graphically A plot of [NO 2 ] vs. time is a curve and not a straight line. The reaction is not zero order with respect to [NO 2 ]

Distinguishing Zero, First, & Second Order Reactions Graphically A graph of ln [NO 2 ] vs. t is a curve and not a straight line. Reaction is not first order in [NO 2 ] [NO 2 ], M −5.573 −5.337 −5.038 −4.845 −4.610 ln [NO 2 ] Time (s)

Distinguishing Zero, First, & Second Order Reactions Graphically The graph of 1/[NO 2 ] vs. time is a straight line so the process is second- order in [NO 2 ]. Slope = 263 M -1 – 100. M s – 0.0 s Slope = M -1 s -1 k = slope = M -1 s [NO 2 ], M /[NO 2 ] Time (s).