worksheet_05_104

1 CHEMISTRY 104 – WORKSHEET #5 – Module 13 Kinetics (Part II) Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) DIFFERENT METHODS TO DETERMINE RATE LAW I. Initial rates method (see Worksheet #4) II. Description (see Worksheet #4) III. Integrated Rate Law – Graphs (this Worksheet) IV. Half-life, t 1/2 : [A] versus time – Graphs (this Worksheet) V. Rate versus [A] – Graphs (this Worksheet) VI. Mechanisms (see Worksheet #6) III. Integrated Rate Laws : function of time (equations and graphs) 3 plots: [A] vs time ® linear ® 0 order; ln[A] vs time ® linear ® 1st order; vs time ® linear ® 2nd order; from the graph determine the order of A, k, and [A] o (see below); [A] t = amount that remains at time = t Order Rate Law Integrated Rate Law y-axis vs x-axis (y = mx + b) Slope y-intercept t 1/2 (half-life) Units of k 0 [A] t = -kt + [A] o [A] t vs time m = -k [A] o = Ms -1 1 ln[A] t = -kt + ln[A] o ln[A] t vs time m = -k ln[A] o = s -1 2 vs time m = k = M -1 s -1 1st order and % remaining : ; = % remaining (e.g., 22% consumed ® 78% remains ® = 0.78) Integrated Rate Law Graphs Zero Order 1st Order 2nd Order 1 [A] o rate = k[A] 0 = k t 1/2 = [A] o 2k M s rate = k[A] 1 t 1 / 2 = 0 . 693 k 1 s rate = k[A] 2 1 [A] = kt + 1 [A] o 1 [A] t 1 [A] o t 1 / 2 = 1 k[A] o 1 Ms ln [A] t [A] 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − kt [A] t [A] o [A] t [A] o ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = -0.005x + 0.5 k = -slope = 0.005 [A]o = y-intercept = 0.5M 0.00 0.10 0.20 0.30 0.40 0.50 0 20 40 60 80 100 [A](M) (me (s) [A] versus (me y = -0.005x - 0.6931 k = -slope = 0.005 ln[A]o = y-intercept = - 0.6931 [A]o = exp(-0.6931) = 0.5M -5.00 -4.00 -3.00 -2.00 -1.00 0.00 0 200 400 600 800 ln[A] 'me (s) ln[A] versus 'me y = 0.005x + 2 k = slope = 0.005 1/[A]o = y-intercept = 2 [A]o = 1/2 = 0.5M 0.0 1.0 2.0 3.0 4.0 5.0 6.0 0 200 400 600 800 1/[A] 'mes (s) 1/[A] versus 'me

2 IV. Half-Life, t 1/2 : [A] versus time graphs ; determine order of A from half-life: • if t 1/2 decreases (linear graph) ® A is zero order; • if t 1/2 is constant (curve plot) ® A is 1st order; • if t 1/2 increases (curve plot) ® A is 2nd order Zero Order 1st Order 2nd Order ; t 1/2 decreases ® Zero ; t 1/2 is constant ® 1st ; t 1/2 increases ® 2nd V. Rate versus [A] – Graphs • If rate vs [A] is a horizontal line (slope = 0) ® Zero order (Rate = k); y-intercept = k • If rate vs [A] is linear and not horizontal ® 1st order (Rate = k[A] 1 ); slope = k • If rate vs [A] is a curve plot ® 2nd order (Rate = k[A] 2 ) Zero Order 1st Order 2nd Order From Rate versus [A] plot: k = y-intercept [A] o : find largest value of [A] From Rate versus [A] plot: k = slope [A] o : find largest value of [A] From Rate versus [A] plot: Curve instead of line! FLOODING TECHNIQUE: “Pseudo-rate constant”, k obs ; need when two reactants are present ; A(aq) + B(aq) ® AB(aq) must use flooding! 1. Write rate law: Rate = k[A] n [B] m 2. Flood with Chemical B such that the concentration of B is large and approximately constant 3. Combine the rate constant, k, and [B] m which is approximately constant together: k obs = k[B] m 4. Re-write new rate law: Rate = k obs [A] n 5. Determine the order of A from graphs given. 6. Determine the order of B by inspection (e.g., [B] doubled and k obs doubled ® 1st order) or use: t 1/2 = [A] 0 2k t 1/2 = 0.693 k t 1/2 = 1 k[A] 0 k = y-intercept = 0.005 0 0.001 0.002 0.003 0.004 0.005 0.006 0.00 0.10 0.20 0.30 0.40 0.50 0.60 Rate(M/s) [A] (M) y = 0.005x k = slope = 0.005 [A] o = largest [A] value [A] o = 0.5M 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0 0.1 0.2 0.3 0.4 0.5 0.6 Rate(M/s) [A] (M) 0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0 0.1 0.2 0.3 0.4 0.5 0.6 Rate(M/s) [A] (M) m = ln k obs1 k obs2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln [B 2 ] 1 [B 2 ] 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

4 3. A reaction: A(aq) + B(aq) ® AB(aq) was studied. In one experiment, [A] o = 0.0050M and [B] o = 0.50M In another experiment, [A] o = 0.0050M and [B] o = 1.0M. Plots of ln[A] t versus time are shown. I. What is the rate law for this reaction? a. rate = k[A] b. rate = k[B] c. rate = k[A][B] d. rate = k[A] 2 [B] e. rate = k[A][B] 2 II. What is the value of the rate constant k; include units? a. 0.00225M -2 s -1 b. 0.00900M -1 s -1 c. 0.00225M -1 s -1 d. 0.00900M -2 s -1 e. 0.0045M -2 s -1 4. A reaction, D(aq) ® C(aq), was studied and a plot of concentration versus time was prepared. What is the rate law for the reaction, and what is the value of k (include units)? a. rate = k[D] 1 ; k = 13s -1 b. rate = k[D] 1 ; k = 0.077s -1 c. rate = k[D] 1 ; k = 7.7s -1 d. rate = k[D] 2 ; k = 0.077M -1 s -1 e. rate = k[D] 2 ; k = 7.7M -1 s -1 y = -0.00225x - 5.29832 [B] = 0.50M y = -0.00900x - 5.29832 [B] = 1.0M -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 ln[A] time (s) ln[A] versus time 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 [D](M) time [D] versus Time

5 5. If 18.5% of a sample of tritium, 3 H, used in nuclear warheads decays in 3.64years, what is the half-life of this isotope in years? (Reminder: Nuclear decays are always first order!) a. 0.0562yr b. 3.64yr c. 10.5yr d. 12.3yr e. 19.7yr 6. The reaction: F(aq) + G(aq) ® FG(aq) was analyzed. In Experiment 1, [F] o = 0.0025M and [G] o = 2.00M (T = 298K), and the linear data was plotted below. In Experiment 2, [F] o = 0.0050M and [G] o = 4.00M (T = 298K), and again the linear data was plotted. Experiment 1 Experiment 2 I. What is the order with respect to F? a. 0 b. 1 c. 2 d. 1 / 2 II. What is the order with respect to G? a. 0 b. 1 c. 2 d. 1 / 2 III. What is the value of the rate constant k (no units have been included in the answers)? a. 0.45 b. 0.90 c. 0.68 d. 0.225 e. 0.0025 y = 0.45x + 400 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 0 500 1000 1500 2000 2500 3000 1/[F]t time (s) Run#1: 1/[F] versus time(s) at 298K [F] o = 0.0025M; [G] o = 2.00M y = 0.90x + 400 0 500 1,000 1,500 2,000 2,500 3,000 0 500 1000 1500 2000 2500 3000 1/[F]t time (s) Run#2: 1/[F] versus time(s) at 298K [F] o = 0.0025M; [G] o = 4.00M

7 CHEMISTRY 104 – WORKSHEET #5 – ANSWERS – Module 13 Kinetics (Part II) Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) ANSWERS 1. b { ; ; t = 76190s } 2. I. b {if ln[A 2 B] vs time yields a line ® first order} II. b {find slope from ; slope = -k; ; k = 0.01; 1st order k units = s -1 } III. e {ln[A 2 B] 0 = -3.0; ; [A 2 B] 0 = e -3.0 = 0.0498M } IV. d {ln[A 2 B] t = -kt + ln[A 2 B] 0 = -(1 x 10 -2 )(500) + ln(0.05) = -7.996; ; [A 2 B] 500 = e -7.996 = 0.0003369M} V. c {first order half-life: ; } 3. I. e { order for A = 1 : since a plot of ln[A] t versus time yields a line ® 1st order; order for B = 2 : this comes from k obs1 = 0.00900 where k obs1 is the slope of the line given in the equation of the line shown on the plot and k obs2 = 0.00225 where k obs2 is the slope of the line given in the equation of the line shown on the plot; note that when [B] doubled going from 0.50M to 1.0M, k obs quadrupled (0.00900/0.00225 = 4); since k obs = k[B] y and k is constant, then if [B] doubled and k obs quadrupled ® order = 2; can also be determined mathematically using: ; ; } II. d {use k obs = k[B] y ; can use either experiment: 0.00900 = k(1.0) 2 ; k = 0.0090M -2 s -1 ; for the other experiment: 0.00225 = k(0.5) 2 ; k = 0.0090M -2 s -1 ; units of k = M -2 s -1 because it is a thir4d order reaction} 1 [Cl 2 ] t = kt + 1 [Cl 2 ] 0 1 0.00050 = (1.75 x 10 − 2 )t + 1 0.0015 Δ y Δ x slope = [ − 5 − ( − 3)] (200 − 0) = − 0.01 = − k e ln[A 2 B] 0 = e − 3 e ln[A 2 B] 500 = e − 7.996 t 1 / 2 = 0 . 693 k t 1 / 2 = 0 . 693 0 . 01 s − 1 = 69 . 3 s y = ln k obs1 k obs2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln [B 2 ] 1 [B 2 ] 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = ln 0.00900 0.00225 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln 1.00 0.50 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = ln(4) ln(2) = 2

8 4. b {determine the first half-life, t 1/2 ; time it takes for [D] = 0.010M to drop to 0.005M is ~9sec (from 9 – 0 = ~9); determine the second half-life, t 1/2 ; time it takes for [D] = 0.005M to drop to 0.0025M is ~9sec (from 18 – 9 = ~9); if desired (not needed) determine the third half-life, t 1/2 ; time it takes for [D] = 0.0025M to drop to 0.00125M is ~9sec (from 27 – 18 = ~9); since the t 1/2 is constant for this plot ® first order with respect to D; for first order: ; ; } 5. d {use ; = amount that remains = 100%-18.5% = 81.5%; and t = 3.64; solve for k: ln(0.815) = -k(3.64); k = 0.05620yrs -1 ; use k to find t 1/2 ; ; } 6. I. c {rate = k[F] x [G] y ; letting k obs = k[G] y this yields: rate = k obs [F] x ; since a plot of 1/[F] versus time yielded a line ® 2nd order with respect to F; x = 2} II. b {rate = k[F] x [G] y ; k obs1 = slope from Experiment 1; k obs1 = 0.45; k obs2 = slope from Experiment 2; k obs2 = 0.90; when [G] doubled going from 2.00M to 4.00M, k obs doubled going from 0.45 to 0.90. Since k obs = k[G] y and k is constant, then if [G] doubles and k obs doubles ® y = 1; can also be determined mathematically using: ; ; ; } III. d {k obs1 = slope from Experiment 1; k obs1 = 0.45; use k obs1 = k([G] 1 ) 1 ; 0.45 = k(2.0) 1 ; k = 0.225; could also have used k obs2 ; k obs2 = slope from Experiment 2; k obs2 = 0.90; using k obs2 = k([G] 2 ) 1 ; 0.90 = k(4.0) 1 ; k = 0.225; since this is a third order reaction, units of k are M -2 s -1 } t 1 / 2 = 0 . 693 k k = 0 . 693 t 1 / 2 k = 0 . 693 9 = 0 . 077 s − 1 ln [A] t [A] 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − kt [A] t [A] 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ [A] t [A] 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.815 t 1 / 2 = 0 . 693 k t 1 / 2 = 0 . 693 0 . 05620 yrs − 1 = 12 . 33 yrs y = ln k obs1 k obs2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln [G] 1 [G] 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = ln 0.45 0.90 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ln 1.00 2.00 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ y = ln(0.5) ln(0.5) y = 1