This document discusses reaction rates and kinetics concepts including: - Instantaneous reaction rates can be calculated from the slope of concentration-time graphs at specific points. - Reaction orders and rate laws can be determined experimentally using methods like the initial rate method or integrated rate law method. - First-order reactions follow the integrated rate law that the natural log of the concentration is linear with time. Second-order and zero-order reactions also have defining rate laws and kinetics equations.

1. General Chemistry II CHEM 152 Week 2

2. Week 2 Reading Assignment Chapter 13 – Sections 13.2 (rate), 13.3 (rate law), 13.4 (integrated rate law)

3. A more refined description The average reaction rate gives us information about the speed of the reaction over a certain period of time. But what if we are interested in a more detailed description. How do we calculate the instantaneous reaction rate at any particular time?

4. START END Different Rates During a Reaction Overall average rate Initial Rate Instantaneous Rate (slope of the line At a point)

5. . From your lab experience, how could we follow the course of this reaction? Measure rates? What’s changing? What could we measure? Measuring Rates Consider this reaction: Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time

6.  [Br 2 ]   Absorption Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time 393 nm light Detector 393 nm Br 2 ( aq )

7. Instantaneous rate = rate for specific tiny instance in time Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) Average rate =   [Br 2 ]  t = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent

8. rate  [Br 2 ] rate = k [Br 2 ] + 0.0 = rate constant = 3.50 x 10 -3 s -1 Y=mX+b Constant = Slope = k How do the RATES change with [Br 2 ] ? Rate = k [Br 2 ] 1 this is called a 1 st order reaction Rate Law The rate law for a reaction tells us how rate varies with concentration of the reactants. Br 2 (aq) + HCOOH(aq)  HBr(aq) + CO 2 (g) R A T E k = rate [Br 2 ]

16. Rate Laws In general, for a A + b B  x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction order with respect each reactant Overall reaction order= m + n + p These numbers can be 0, 1, 2 or fractions and must be determined by experiment ! They are NOT related to the stoichiometric coefficients a,b,x

17. Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) 1 2 3 4 5 O 2 NO 1.10x10 -2 1.30x10 -2 3.21x10 -3 1.10x10 -2 3.90x10 -2 28.8x10 -3 2.20x10 -2 1.10x10 -2 3.30x10 -2 1.30x10 -2 2.60x10 -2 1.30x10 -2 6.40x10 -3 12.8x10 -3 9.60x10 -3 Your Turn

18. Determining Reaction Orders Using Initial Rates Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. k [O 2 ] 2 m [NO] 2 n k [O 2 ] 1 m [NO] 1 n = = [O 2 ] 2 m [O 2 ] 1 m = 6.40x10 -3 mol/L*s 3.21x10 -3 mol/L*s = ; 2 = 2 m m = 1 Do a similar calculation for the other reactant(s). Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. O 2 ( g ) + 2NO( g ) 2NO 2 ( g ) rate = k [O 2 ] m [NO] n rate 2 rate 1 [O 2 ] 2 [O 2 ] 1 m 1.10x10 -2 mol/L 2.20x10 -2 mol/L m

19. Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) 1 2 3 4 5 O 2 NO 1.10x10 -2 1.30x10 -2 3.21x10 -3 1.10x10 -2 3.90x10 -2 28.8x10 -3 2.20x10 -2 1.10x10 -2 3.30x10 -2 1.30x10 -2 2.60x10 -2 1.30x10 -2 6.40x10 -3 12.8x10 -3 9.60x10 -3 Your Turn

20. Expected Solution : Another Example Consider this reaction inside a car engine: How can we use this data to determine the rate law and the rate orders with respect to each reactant? rate = k [NO 2 ] m [CO] n Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20

21. 16 = 4 m and m = 2 The reaction is 2 nd order in NO 2 . First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies . One Variable at a Time 0.080 0.0050 rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = 0.40 0.10 = m Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20

22. Now, choose two experiments in which [NO 2 ] remains constant and the [CO] varies. One Variable at a Time The reaction is zero order in CO . rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2 Overall order Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = 0.0050 0.0050 = 0.20 0.10 n ; 1 = 2 n and n = 0

23. Your Turn rate = k [NO 2 ] 2 What is the value of the constant, k, in this reaction? Exp. Initial Rate (mol/L*s) Initial [NO 2 ] Initial [CO] 1 2 3 0.0050 0.080 0.0050 0.10 0.10 0.40 0.10 0.10 0.20

25. Overall Reaction Order Units of k (t in seconds) 0 mol/(L*s) (or mol L -1 s -1 )  M/s 2 L/(mol*s) (or L mol -1 s-1)  1/Ms 3 L 2 /(mol 2 *s) (or L 2 mol -2 s -1 )  1/M 2 s Units: Rate Constant k 1 1/s (or s -1 )

26. Concentration-Time Relations Consider a FIRST ORDER REACTION: A  B The rate law is Rate= k[A] RATE= -d[A]/dt = k[A] How can we predict the concentration of reactants at any moment in time? How do you find [A] = f(t) ?

27. First-Order Rate Law A  B k

28. Integrating - (d[A]/d t) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law Integrated Rate Laws

33. First-Order Reactions The half-life , t ½ , is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 ) ln [A] 0 [A] 0 /2 k = t ½ ln2 k = 0.693 k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t ½ ln2 k = 0.693 5.7 x 10 -4 s -1 =

34. A plot of [N 2 O 5 ] vs. time for three half-lives

35. Second-Order Rate Law 2A  B k

37. Second-Order Half Life

38. Zero-Order Reactions rate = k [A] 0 = k k = = M /s [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t =0 t ½ = t when [A] = [A] 0 /2 [A] = [A] 0 - kt A product rate = - D[A] D t rate [A] 0 D[A] D t = k - t ½ = [A] 0 2 k

39. Consider the following data for the reaction: N 2 O 5 -> 2NO 2 + ½ O 2 Graphical Analysis

40. Summary ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0

41. At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1 , to two molecules of ethylene (C 2 H 4 ). If the initial C 4 H 8 concentration is 2.00M, what is the concentration after 0.010 s? _____________ M

42. At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1 , to two molecules of ethylene (C 2 H 4 ). (b) What fraction of C 4 H 8 has decomposed in this time?

43. SOLUTION: [C 4 H 8 ] = 0.84 mol/L = 0.58 ln 2.00 [C 4 H 8 ] = -(87s -1 ) (0.010s) ln [C 4 H 8 ] 0 [C 4 H 8 ] t = - k t (a) (b) [C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M - 0.84M 2.00M

44. PLAN: Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 0 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000 0 C via the following first-order reaction: The rate constant is 9.2s -1 , (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? One-quarter of the initial value means two half-lives have passed. Use t 1/2 = ln2/k to find the half-life

45. SOLUTION: t 1/2 = 0.693/9.2s -1 = 0.075s (a) 2 t 1/2 = 2(0.075s) = 0.15 s (b)

46. 0 th Order 1 st Order 2 nd Order Plot for straight line Slope, y-intercept Half-life Rate law rate = k rate = k [A] rate = k [A] 2 Units for k mol/(L*s) 1/s L/(mol*s) Int. rate law (straight-line form) [A] t = -k t + [A] 0 ln[A] t = - k t + ln[A] 0 1/[A] t = k t + 1/[A] 0 [A] t vs. t ln[A] t vs. t 1/[A] t vs t -k, [A] 0 -k, ln[A] 0 k, 1/[A] 0 [A] 0 /(2 k) (ln 2)/ k 1/( k [A] 0 ) Overview

47. Summary Activity: N 2 O 5 (g)  NO 3 (g) + NO 2 (g) Consider the following graphs and reaction data to predict the concentration of N 2 O 5 at 275 sec. Time (s) [N 2 O 5 ] 0 1.000 25 0.822 50 0.677 75 0.557 100 0.458 125 0.377 150 0.310 175 0.255 200 0.210

48. What is the concentration of N 2 O 5 after 275 sec? ______________ mol/L