Two Types of Rate Laws Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law Integrated- Data table contains TIME AND CONCENTRATION DATA. Uses graphical methods to determine the order of the given reactant. K=slope of best fit line found through linear regressions

Integrated Rate Law Can be used when we want to know how long a reaction has to proceed to reach a predetermined concentration of of some reagent

Graphing Integrated Rate Law Time is always on x axis Plot concentration on y axis of 1st graph Plot ln [A] on the y axis of the second graph Plot 1/[A] on the y axis of third graph Your are in search of a linear graph

Results of linear graph Zero order: time vs concentration= line y= mx+ b [A]= -kt + [A0 ] A- reactant A, A0 - initial concentration of A at t=0 l slope l= k, since k cannot be negative, and k will have a negative slope Rate law will be rate=k[A]0

Results of linear graph First order: time vs ln [ ]= line y= mx+ b ln [A]= -kt + ln [A0 ] A- reactant A, A0 - initial concentration of A at t=0 l slope l= k, since k cannot be negative, and k will have a negative slope Rate law will be rate=k[A]1

Results of linear graph second order: time vs 1/ [ ]= line y= mx+ b 1/[A]= kt + 1/ [A0 ] A- reactant A, A0 - initial concentration of A at t=0 k=slope Rate law will be rate=k[A]1

First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN © 2012 Pearson Education, Inc.

First-Order Processes CH3NC CH3CN This data were collected for this reaction at 198.9 C. © 2012 Pearson Education, Inc.

First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative of the slope: 5.1  105 s1. © 2012 Pearson Education, Inc.

Second-Order Processes The decomposition of NO2 at 300 °C is described by the equation NO2(g) NO(g) + O2(g) 1 2 and yields data comparable to this table: Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 © 2012 Pearson Education, Inc.

Using the graphing calculator L1=time L2=concentration-- if straight line, zero order L3=ln concentration-- if straight line- 1st order L4= 1/concentration– if straight line—2nd order Perform 3 linear regressions

Second-Order Processes The plot is a straight line, so the process is second-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 0.01000 4.610 50.0 0.00787 4.845 100.0 0.00649 5.038 200.0 0.00481 5.337 300.0 0.00380 5.573 © 2012 Pearson Education, Inc.

Second-Order Processes 1 [NO2] Graphing vs. t, however, gives this plot Fig. 14.9(b). Because this is a straight line, the process is second-order in [A]. Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263 Add problem after this slide. © 2012 Pearson Education, Inc.

Determine the rate law and calculate k What is the concentration of N2 O5 at 600s? At what time is the concentration equal to 0.00150 M?

The decomposition of N2 O5 was studied at constant temp 2 N2 O5 (g)  4 NO2(g) + O2(g) Time (s) 0.1000 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400

© 2012 Pearson Education, Inc. Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0. © 2012 Pearson Education, Inc.

Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln = kt1/2 ln 0.5 = kt1/2 0.693 = kt1/2 = t1/2 0.693 k Note: For a first-order process, then, the half-life DOES NOT DEPEND ON CONCENTRATION!!!!!!!!

Half-Life First order decay is what is seen in radioactive decay 0.693 k This is the equation used to calculate the half-life of a radioactive isotope

Problem Exercise A certain first-order reaction has a half-life of 20.0 minutes. a. Calculate the rate constant for this reaction. b. How much time is required for this reaction to be 75% complete? 3.47 × 10−2 min−1; 40 minutes

Half-Life For a second-order process, 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0 Add problem after this slide. 2  1 [A]0 = kt1/2 1 = = t1/2 1 k[A]0 © 2012 Pearson Education, Ic.

Half life zero order

Half life first order

Half life second order

Summary table order zero First second Rate Law Rate=k rate-=k[ A] Integrated rate law in form of y=mx+b [A]t = -kt + [Ao ] ln[A]t = -kt + ln[Ao ] Rate law in data packet on AP exam Does not appear ln[A]t -ln[Ao ] = -kt 1/[A]t -1/[Ao ] = kt slope Slope = -k Slope= -k slope=k Half-life [A0 ]/2k 0.693/k 1/kA0

© 2012 Pearson Education, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Reaction Mechanisms Chemical reactions proceed via a sequence of distinct stages. The sequence is known as the mechanism and each part of the mechanism is known as a “step”. The rate of the reaction is only dependant upon the slowest step, also known as the rate determining step or RDS. only reactants that appear in the rate determining step appear in the rate equation and vice-versa © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. © 2012 Pearson Education, Inc.

Example #1 The reaction below W + Y  Z Has the following mechanism W  R slow R + Y Q fast Q  Z fast Here the rate only depends on the concentration of _______________ and therefore the rate equation only contains this reactant The rate equation is, ______________ in the rate determining step is 1. Note: R and Q are not reactants or products, but are rather they are called ________________________, produced in one step, but then are used up in a subsequent step. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Note: 1. In all valid mechanisms the sum of the individual fast and slow steps must be the same as the overall chemical equation. 2.The stoichiometric coeffiecient of a substance that appears in the slow step is the power that the concentration of that substance is raised to in the rate equation. 3. If a substance is present at the beginning of a reaction AND present in the same form at the end of the reaction, it can be identified as a catalyst © 2012 Pearson Education, Inc.

Slow Initial Step A proposed mechanism for this reaction is Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) The NO3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The reaction below A + B C + D Has the following mechanism A  Q fast equilibrium Q + B  C + D slow Here the slow step contains Q and B, and _____is an intermediate Can intermediates be featured the rate equation? _____________ Since the formation of Q is dependent on A, Q can be replaced by A in the rate equation. Therefore the rate equation is given as_________________ The orders w.r.t A and B are________________________________ since the stoichiometric coefficient of B in the rate determining step is 1, and the stoichiometric coefficient of A (which Q depends upon) is also 1. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature-dependent. © 2012 Pearson Education, Inc.

Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore, E). The high point on the diagram is the transition state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation-energy barrier. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules has higher energy. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier. As a result, the reaction rate increases. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. −Ea/RT f = e © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: k = Ae where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. −Ea/RT © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k =  ( ) + ln A Ea R 1 T Add problem(s) after this slide. y = mx + b Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. . 1 T © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. © 2012 Pearson Education, Inc.