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Chapter 3 Chemical Kinetics

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					Chapter 3 Chemical Kinetics
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Objectives: Describe, explain and calculate the dynamic characteristics of chemical reactions. Describe and explain the applications of rate laws to catalysis.

3.1. Reaction Rates
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THOUGHT TEASER: movies: * “Collage of reaction rates”. What is the difference between the reactions? Definition: Reaction rate = speed of a reaction = change of concentration of reactant or product per unit time General case: A  B Reaction rate = r = -D[A] / Dt for the consumption of reactants = = D [B] /Dt for the formation of products D[A] = Change in [A]; Dt = change in time. Example: 2NO2  2NO + O2

Chemical Kinetics (RQ217)
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Examine the graph on the previous slide, does the speed of the reaction increase or decrease or stay the same with time? Explain. a. It decreases. The concentration of N2O5 decreases by more than 0.2 mol/L from 2nd to 3rd hour, vs a decrease of less than 0.1 mol/L from 7th to 8th hour. b. It increases. The concentration of N2O5 increases by more than 0.2 mol/L from 2nd to 3rd hour, vs an increase of less than 0.1 mol/L from 7th to 8th hour. c. It decreases. The concentration of N2O5 increases by more than 0.1 mol/L from 2nd to 3rd hour, vs a decrease of less 0.2 mol/L than from 7th to 8th hour.

3.2. Reaction Rate Laws
Definition: Rate law = a relation between the change in reaction rate and the change in concentrations of reactants or time. General case: A  B  Differential rate law: shows how rate depends on concentrations of reactants r = k[A]n k = rate constant n = order of reaction. Must be determined experimentally.  Integrated rate law: shows how the concentrations change with time. r = -D[A] / Dt
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3.3. Determining the form of the rate law
Most common method: method of initial rates.  Uses instantaneous rates of reaction just after the reaction begins.  Helps determine the order of reaction. Uses differential rate law: r = k[A]n[B]m
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Determining the form of the differential rate law
Tool: r = k[A]n[B]m  Procedure: *1. Determine the orders (n, m, …) of the reaction relatively to each reactant. The way: by changing the concentration of one reactant and keeping the other reactant constant, checking effect on rate.
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The Form of the Differential Rate Law (Continued)
*2. Find Overall Order of reaction: sum of the individual orders (n + m + …).  *3. Determine the reaction rate constant (k) k = r / [A]n[B]m using known numeric values for r, [A], [B], n and m. *4. Write the equation of the rate law. Must have constant and variable parameters.  RQ2-18: What is the shape of the expected rate law?
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The Form of the Differential Rate Law (RQ2-18)
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Question: based on the procedure described in the previous two slides, what is the shape of the expected rate law? r = k[A]n[B]m k, n, and m are constants. r, [A] and [B] are variables r = k[A]n[B]m k, n, and m are variables. r, [A] and [B] are constants r = D[A]/Dt + D [B]/ Dt r, [A], [B] and t are variables

The form of the differential rate law (Illustration)
Example: Reaction of CO and NO2  Information provided: * Reaction: CO + NO2 -> CO2 + NO * Experimental data table: various concentrations of CO and NO2 and their corresponding initial reaction rates.  Information requested: find * Reaction rate constant * Reaction rate equation
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Differential rate law (Illustration)
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Exprmt [CO] mol/L 5.10E-4 5.10E-4 1.02E-3

[NO2] mo/L 3.50E-5 7.00E-5 3.50E-5

1 2 4

Initial Rate mol/L*h 3.4E-8 6.8E-8 6.8E-8

RQ2-18b
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How do you proceed to solve the problem using the KNU method? Are all needed parameter available? If not, which one should be determined next? k = r / ([CO]n x [NO2]m). [CO] and [NO2] are unknown and should be determined next. k = ([CO]n x [NO2]m) / r. n and m are unknown and should be determined next. k = r / ([CO]n x [NO2]m). n and m are unknown and should be determined next.

Differential rate law (Illustration Solution)
Way to answer part 1: k = r/([CO]n*[NO2]m)  Step 1: Goal : figure out n, order of the reaction for CO * comparison of rates 1 & 4. * Result: r4 / r1= ? In term of speeds =? In terms of concentrations * Solution: n = ?  Conclusion: order of reaction for reactant A (CO) = ?
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Differential rate law (Example Solution 2)
Step2: Goal: find m, reaction order for NO2 The way: comparison of rates 1 & 3. Result: r3/r1 = ? Solution: m = ?  Conclusion: order of reaction for reactant B (NO2) = ?  Step 3. Overall order of reaction: n + m  Step 4. Find k, the reaction rate constant  Step 4. Find expression of r  Extra exercise: #41, pg 609
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The form of the differential rate law (Continued 3)
NOTE: if ry / rx = an(or m) = z = not 2 Use logarithm relations to figure out n, m.  To find n : log (ry / rx ) = n * log(a) n = log (ry / rx ) / log(a)  Example: r2/r1 = 6.25 / 5.00 = 1.25 = 2n ln(1.25) = n*ln2 => n = ln(1.25)/ln2 = 0.322  Most cases simplify to n, m = 1, 2, …
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3.4 Integrated Rate Laws
Used when experimental reaction data are concentrations of reactants vs reaction times.  General case: A  B + C r = -D[A] / Dt = k[A]n  D[A] = - Dt x k  1st objective: find order of the reaction.  Way to reach the objective: process the data to obtain a straight line graph.
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a. First Order Law
General case: A -> B + C Rate: r = -D[A] / Dt = k[A]n = k[A] , because n = 1.  Integrate -D[A] / Dt = k[A]  Result: ln [A] = -kt + ln[A]0.  [A]0 = initial concentration of A.  Plotting ln[A] vs t gives a straight line, graph slope = -k; ln[A]0 = y intercept
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First Order Law (Continued)
Half-life = the time necessary to reduce the original concentration of A in half. At t1/2: [A] = ½ x [A]0 The 1st order reaction equation becomes: ln (½ x [A]0) = -kt1/2 + ln[A]0 ln(2) / k = t1/2 = 0.693 / k
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First Order Law (Illustration)
2N2O -> 2N2 + O2 Time(min) [N2O] (M) 15 0.0835 30 0.068 80 0.035 120 0.022 1/[N2O] ln[N2O] 11.97605 -2.48291 14.70588 -2.68825 28.57143 -3.35241 45.45455 -3.81671

First Order Law (Illustration Continued)
Info provided: * Reaction: 2N2O  2N2 + O2 (N2O Decomposition) * reaction data: [N2O] vs time  Info requested: * rate constant * rxn rate when [N2O] = 0.035 mol/L  Note [N2O]0 cannot be found directly on curve. Must be derived from ln relation.
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Integrated Rate Order of Rxn (RQ2-19)
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Based on the graphs shown for the reaction in the previous slide, what is the order of that reaction? a. Second order, because the graph of 1/[A] vs time shows a increase in products b. First order, because the graph shows of ln[A] vs time c. First order, because the graph of ln[A] vs time shows a straight line

First Order Law (Illustration Solution)
Plot [N2O], 1/[N2O] and ln[N2O] vs time to determine the order of the reaction.  Find the slope: divide the difference between two ln[N2O] data by their corresponding time difference (-2.6883 – (-2.4829))/(30 – 15) = ? k = - slope
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First Order Law (Illustration Solution, Continued)
At [N2O] = 0.035 mol/L r = previously calculated k x 0.035 mol/L =?  t1/2 = 0.693 / previously calculated k  Extra example: #47, pg 609
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Chemical Kinetics (llustration)
O2 + 2NO -> 2 NO2 Which reaction mixture reacts faster?
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b. Second Order Reaction Rate Law
General reaction case: A  B Rate law = r = -D[A]/Dt = k[A]n = k[A]2, because n = 2 (how come?) Integrate -D[A]/Dt = k[A]2 Result: 1/[A] = kt + 1/[A]0  RQ2-20: What is the relation between the variables of the above expression?
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Second Order Reaction Rate Law (RQ2-20)
RQ2-20: What is the relation between the variables of the above expression? Justify your answer. a. 1/[A]0 and k have a linear relation , similar to y = ax + b. Plotting them on a graph produces a straight line. a. 1/[A] and t have a linear relation, similar to y = ax + b. Plotting them on a graph produces a straight line. a. 1/[A] and t have a parabolic relation , similar to 1/y = ax + b. Plotting them on a graph produces a parabole.
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Second Order Reaction Rate Law (Continued)
Plotting 1/[A] vs t gives a straight line Slope = k; y intercept = 1/[A]0  Half-life = time when [A] = (1/2)*[A]0; Replace [A] by (1/2)*[A]0 in the general equation, which becomes 2/[A]o = kt1/2 + 1/[A]0 Result: t1/2 = 1/k[A]0
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Second Order Rate Law (Illustration)
2NH3 -> N2 + 3H2 Time (s) 0 25 50 75 [NH3] (mol/l) 8.00E-07 6.75E-07 5.84E-07 5.15E-07 ln[NH3] 1/[NH3] -14.03865 1250000 -14.20855 1481481 -14.35336 1712329 -14.4791 1941748

Second Order Rate Law (Illustration Continued)
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Info available: 2NH3  N2 + 3H2 Table showing [NH3] change over time. Info requested rxn order value of k Differential rate law Integrated rate law 1/[A] = kt + 1/[A]o half-life of rxn Value of the rate when [NH3] = 7.00E-7 M

Second Order Rate Law (Illustration Solution)
Plot [NH3], 1/[NH3] and ln[NH3] vs time. * Which plot shows a straight line? * What is the subsequent rxn order  Next step: RQ2-21: how do you find the rate constant of the reaction?
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Second Order Rate Law (RQ2-21)
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RQ2-21: how do you find the rate constant of the reaction? Average out the difference between two values of 1/[NH3] by their corresponding time differences Divide the sum between two values of 1/[NH3] by their corresponding time sums Divide the difference between two values of 1/[NH3] by their corresponding time differences

Second Order Rate Law (Solution, Continued)
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r = previously calculated k x [NH3] = ? t1/2 = 1/ previously calculated k x [NH3]0 1/[NH3] = previously calculated k x t + 1/[NH3]0 At [NH3] = 7.00E-7 M, r = previously calculated k x 1/7.00E-7 M Extra exercise: #45 pg 609

c. Zero Order Rate Law
Read pg  r = D[A] / Dt = k[A]n where n = 0, so: r=k  For a 0 order reaction, the rate is constant, does not change with concentration  Integrated rate law: [A] = -kt + [A]0  Plotting [A] vs t shows straight line, slope = -k; y intercept = [A]0  Used in cases of catalysis.
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3.5. Kinetic Model of a Chemical Reaction
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Thought teaser: show collisions from “Reaction of NO and O3” (CD2) What conditions must be fulfilled in order for a reaction to occur?

Collisions between reactants
Reaction: result of collisions between reactants  Conditions for a successful collision: *1. Orientation requirement: reactants must be oriented in a way for the collisions to result into a reaction *2. Energy Requirement Reaction must have enough energy to occur
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Energy of Activation (Ea)
Definition: Minimum amount of energy required for a collision to result into a reaction. Show Ea curve  Relation between collision energy & orientation; k = A*e-Ea/RT (Arrhenius) k = rate constant A = (collision) frequency factor lnk = lnA – Ea/RT x ln(e) = – Ea/RT + lnA
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Energy of Activation (RQ2-22)
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In the relation lnk = – Ea/RT + lnA, what variables should be used to obtain a straight line graph? Explain a. ln(k) and 1/T. They have a linear relation, with –Ea and ln(A) as the slope and the y intercept respectively b. ln(k) and T. They have a linear relation, with – Ea and ln(A) as the slope and the y intercept respectively c. ln(k) and Ea. They have a linear relation, with –T and ln(A) as the slope and the y intercept respectively

Energy of Activation (Continued)
Plotting lnk vs 1/T  straight line. Slope = - Ea/R  Ea can be determined 2 ways: * finding the slope of the curve * using the difference between 2 values of k at 2 different temperatures ln(k2) – ln(k1) = ln(k2/k1) = (Ea/R)(1/T1- 1/T2). Therefore Ea = ln(k2/k1) x R / (1/T1- 1/T2)
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Energy of Activation (Illustration)
Case: finding the Ea of a reaction  Info provided: * Initial temp: 3.00E2 K * Final temp: 3.10E2 K * k triples in the given temp range  Info requested: energy of activation.  Extra exercises : #61, Pg 611
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Energy of Activation (Illustration Solution)
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Use the relation ln(k2/k1) = Ea/R)(1/T1 – 1/T2) to figure out the expression of Ea. Based on info provided, if k triples within the temp range, then k2/k1 = ? => ln(k2/k1) = ? Plug the values of ln(k2/k1), R, and (1/T1 – 1/T2) in the expression of Ea. Ea = ?

Energy of Activation (Illustration 2)
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Experimental data of a chemical reaction
k (1/Mxs) 0.000000623 0.00000242 1/T 0.0018018 0.00173913 ln(k) -14.288719 -12.931743

T (K) 555 575

645
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0.000144
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0.00155039
0.00142857

-8.8456973
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Energy of Activation (Illustration 2)
Info provided: experimental data of a chemical reaction  Info requested: Ea  RQ2-23: How can Ea be calculated a. Divide the sum between 2 values of ln(k) by the sum between 2 corresponding values of 1/T b. Divide the difference between 2 values of ln(k) by the difference between 2 corresponding values of 1/T c. Divide the difference between 2 values of ln(k) by the difference between 2 corresponding values of T
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Energy of Activation (Illustration 2 Solution)
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Ea = R x (ln(k2) –ln(k1)) / (1/T11/T2) = 8.3145 J/(mol*K) * (-12.93 + 14.29) / (1/555 – 1/575) K^-1 = ?

Chemical Kinetics (RQ2-22b)
The half-life of the following chemical reaction, A -> B, is found to depend on the initial
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concentration of reactant. What plot of the reaction experimental data gives a straight line? Justify your answer a. [A] vs t, because it is a first order reaction and t1/2 = k[A]o /(0.693). b. 1/[A] vs t, because it is a second order reaction and t1/2 = 1/(k[A]o). c. ln[A] vs t, because it is a zero order reaction and t1/2 = 1/(k[A]o).


				
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