Rate Laws Example: Determine the rate law for the following reaction given the data below. H2O2 (aq) + 3 I- (aq) + 2H+ (aq) I3- (aq) + H2O (l) [H2O2] [I-] [H+] Initial Expt # (M) (M) (M) Rate (M /s) 1 0.010 0.010 0.00050 1.15 x 10-6 2 0.020 0.010 0.00050 2.30 x 10-6 3 0.010 0.020 0.00050 2.30 x 10-6 4 0.010 0.010 0.00100 1.15 x 10-6

Rate = k[Br-] [BrO3-] [H+]2 Rate Laws For the reaction: 5Br- (aq) + BrO3- (aq) + 6 H+ (aq) 3 Br2 (aq) + 3 H20 (l) the rate law was determined experimentally to be: Rate = k[Br-] [BrO3-] [H+]2 The reaction is first order with respect to Br-, first order with respect to BrO3-, 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: Rate = k [A]2[B] Expt # [A] (M) [B] (M) Initial rate (M /s) 1 0.100 0.100 4.0 x 10-5 2 0.100 0.200 8.0 x 10-5 3 0.200 0.100 16.0 x 10-5

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.

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] Using calculus, it is possible to derive an equation that can be used to predict the concentration of reactant A after a given amount of time has elapsed.

First Order Reactions To predict the concentration of a reactant at a given time during the 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 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 yield a straight line, and the slope of the line will be -k.

First Order Reactions CH3NC CH3CN Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN

First Order Reactions CH3NC CH3CN This data was collected for this reaction at 198.9°C. Remember that the partial pressure of the gas will be directly proportional to the number of moles of that gas

First Order Reactions When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative slope: 5.1  10-5 s−1.

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 10-4 g/mL?

First Order Reactions

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. t1/2 After one half life has elapsed, the concentration of the reactant will be: [A]t = ½ [A]0 For a first order reaction: t1/2 = 0.693 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. Use t1/2 = 0.693/k to find the rate constant Substitute in the value for k and the initial concentration to find the value for [A]t: ln[A]t = -kt + ln[A]0

First Order Reactions Example: A certain pesticide has a half life of 0.500 yr. If the initial concentration of a solution of the pesticide is 2.5 x 10-4 g/mL, what will its concentration be after 1.5 years?

First Order Reactions Find the rate constant, k: Substitute data into equation:

First Order Reactions Solve for [A]1. 5 yr using the inverse natural logarithm:

First Order Reactions There are two other (simpler!!) ways to solve this problem: Draw a picture: 1.5 yr = 3 half lives 2.5 x 10-4 g/mL 0.5 yr 1.25 x 10-4 g/mL 0.5 yr 0.625 x 10-4 g/mL 0.5 yr 0.31 x 10-4 g/mL

First Order Reactions For a first order reaction, you can also determine the concentration (or mass or moles) of a material left at a given amount of time by: [A]t = 1 x [A]0 2 Elapsed time Half life

First Order Reactions 2 [A]t = 1 x[A]0 For the previous example: [A]1.5 yr = 1 (1.5 yr/0.5 yr) x (2.5 x 10-4 g/mL) [A]1.5 yr = 3.1 x 10-5 g/mL Elapsed time Half life

First Order Reactions Example: Cobalt-57 has a half life of 270. days. How much of an 80.0 g sample will remain after 4.44 years if it decomposes via a first order reaction?

First Order Reactions Calculate the rate constant, k. Substitute data into lnAt = -kt + lnA0

First Order Reactions Solve for A4.44 yr using the inverse natural logarithm:

First Order Reactions Since 4.44 years is exactly 6 half lives, we can also draw a picture to solve this problem. 4.44 yr = 6.00 0.740 yr When one half life (270 days or 0.740 yr) has elapsed, half of the original 80.0 g will have decomposed.

First Order Reactions 80.0 g 270 days 40.0 g 270 days After 6 half-lives have elapsed, 1.25 g of Co-57 are left. 20.0 g 270 days 10.0 g 270 days 5.00 g 270 days 2.50 g 270 days 1.25 g

First Order Reactions You can also solve this problem using the third approach: [A]4.44 yr = 1 (4.44 yr/0.74 yr) x (80.0 g) 2 [A]4.44yr = 1.25 g

Second Order Processes If the rate law for a reaction that is second order with respect to reactant A is integrated, the integrated rate law becomes: 1 [A]t = kt + [A]0 y = mx + b

Second Order Processes If a process is second order with respect to A, a plot of 1/[A] vs. time will give a straight line slope of line = k To determine if a reaction is first order or second order with respect to a reactant A, plot both ln[A] vs time and 1/[A] vs. time. Whichever plot gives a straight line indicates if the reaction is 1st order or second order.

Second Order Processes Example: The decomposition of NO2 at 300°C is described by the equation: NO2 (g) NO (g) + 1/2 O2 (g) and yields the data below. Is this reaction first order or second order with respect to NO2? 0.00380 300.0 0.00481 200.0 0.00649 100.0 0.00787 50.0 0.01000 0.0 [NO2], M Time (s)

Second Order Processes Graphing ln [NO2] vs. t yields: The plot is not a straight line, so the process is not first-order in [A]. 0.00380 0.00481 0.00649 0.00787 0.01000 [NO2], M −5.573 −5.337 −5.038 −4.845 −4.610 ln [NO2] 300.0 200.0 100.0 50.0 0.0 Time (s)

Second Order Processes Graphing ln 1/[NO2] vs. t, however, gives this plot. Because this is a straight line, the process is second-order in [A]. 0.00380 0.00481 0.00649 0.00787 0.01000 [NO2], M 263 208 154 127 100 1/[NO2] 300.0 200.0 100.0 50.0 0.0 Time (s)