M13-Kinetics_DiscPkt_KEY_1-23

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Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 1 of 22 PART A 1. List four factors which allow us to change the rate of a reaction or process, and give a brief explanation of how they do this. Concentrations : Molecules must collide to react. If concentration is increased, there are more reactants present to collide with one another within a smaller space. Physical state (phase, surface area): Reactants must come into contact for a successful reaction. The easier it is for reactants to collide, the faster the reaction will be. Increasing the ease of contact can be done by having a homogenous reaction mixture (either all gasses, all liquids, or by dissolving reactants into a solution) or by increasing the surface area of reactants with smaller particle sizes. Temperature: Molecules must collide with enough energy for a successful reaction to occur. Most reactions will proceed more quickly at higher temperatures because reactants will have a higher kinetic energy and therefore collide with more energy. Presence of a Catalyst : Catalysts increase the rate of reaction without being consumed by the process. More information about how catalysts work will come later in this module. 2. Exploring chemical kinetics using a sardine sandwich analogy! a. The procedure for making a sandwich is to combine 2 pieces of bread, 3 sardines, and 1 pickle to make one sandwich. Write a chemical equation to demonstrate this. 2 slices bread + 3 sardines + 1 pickle  1 sandwich b. What are two ways you might measure the rate of sandwich production? Any of the following ways will work:  Count number of sandwiches made  Count how many slices of bread are used and divide by 2  Count how many sardines are used and divide by three  Count number of pickles used c. Write a “rate expression” for the production of sardine sandwiches. ∆ 𝒔𝒂𝒏𝒅𝒘𝒊𝒄𝒉𝒆𝒔 ∆ 𝒕𝒊𝒎𝒆 = − 𝟏 𝟐 ൬ ∆ 𝒃𝒓𝒆𝒂𝒅 𝒔𝒍𝒊𝒄𝒆𝒔 ∆ 𝒕𝒊𝒎𝒆 ൰ = − 𝟏 𝟑 ൬ ∆ 𝒔𝒂𝒓𝒅𝒊𝒏𝒆𝒔 ∆ 𝒕𝒊𝒎𝒆 ൰ = −𝟏 ൬ ∆ 𝒑𝒊𝒄𝒌𝒍𝒆 ∆ 𝒕𝒊𝒎𝒆 ൰ d. If sardines are being used up at an average rate of 6 per minute, what is the rate of sandwiches being produced? What is the rate of bread being used? ൬ ∆ 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠 ∆ 𝑡𝑖𝑚𝑒 ൰ = − 1 3 ൬ ∆ 𝑠𝑎𝑟𝑑𝑖𝑛𝑒𝑠 ∆ 𝑡𝑖𝑚𝑒 ൰ = − 1 3 ൬ − 6 𝑠𝑎𝑟𝑑𝑖𝑛𝑒𝑠 𝑚𝑖𝑛𝑢𝑡𝑒 ൰ = 𝟐 𝒔𝒂𝒏𝒅𝒘𝒊𝒄𝒉𝒆𝒔 𝒎𝒊𝒏𝒖𝒕𝒆 ൬ ∆ 𝑏𝑟𝑒𝑎𝑑 𝑠𝑙𝑖𝑐𝑒𝑠 ∆ 𝑡𝑖𝑚𝑒 ൰ = 2 3 ൬ ∆ 𝑠𝑎𝑟𝑑𝑖𝑛𝑒𝑠 ∆ 𝑡𝑖𝑚𝑒 ൰ = 2 3 ൬ − 6 𝑠𝑎𝑟𝑑𝑖𝑛𝑒𝑠 𝑚𝑖𝑛𝑢𝑡𝑒 ൰ = −𝟒 𝒃𝒓𝒆𝒂𝒅 𝒔𝒍𝒊𝒄𝒆𝒔 𝒎𝒊𝒏𝒖𝒕𝒆 *Note that the 2/3 term comes from rearranging the relationship between bread slices and sardines. 3. Use the following axes and grid lines to plot the generic reaction: A  B

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 2 of 22 a. Plot [A] versus time in seconds. Let [A] 0 = 0.5 M. b. Plot [B] versus time in seconds. Let [B] 0 = 0 M. Use the same graph provided below for answering 3.a. and 3.b. Be sure to label axes and provide a title. 4. Hydrogen peroxide decomposes to form water vapor and oxygen gas. 2 H 2 O 2 (g)  2 H 2 O(g) + O 2 (g) A. Imagine there are four hydrogen peroxide molecules in a beaker before the reaction occurs. Use the space provided to draw the products after the reaction has gone to completion. Concentration vs Time of A à B Time (s) Concentration (M) A B

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 3 of 22 B. The graph at right shows the change in concentration as a function of time for the reaction. Which substance do each of the curves A, B, and C represent? A= H 2 O (product) B = O 2 (product) C = H 2 O 2 (reactant) Stoichiometry: 2 mol H 2 O 2 used up to generate 2 mol H 2 O and 1 mol O 2 5. Cyclobutane can decompose to form ethene: C 4 H 8 (g)  2 C 2 H 4 (g) The cyclobutane concentration can be measured as a function of time by mass spectrometry. a. Determine the initial rate of reaction. Initial Rate= – Slope = – ∆y/∆x (negative because we’re using up a reactant) Initial Rate= –(4.5 M – 3.5 M) / (0 – 2s) = 0.50 Ms –1 b. Determine the instantaneous rate of reaction at 20 s. Find this by determining the slope of the tangent line for the point at t = 20 s. (Line shown on plot.) Two points on this tangent line are: (10 s, 1.4 M) and (30 s, 0.6 M). Use these to determine slope of the line tangent to the point at 20 s. Instantaneous Rate = –(1.4 M – 0.6 M) / (10s – 30s) = 0.040 Ms –1 c. The average rate of reaction between 10 and 30 s is 0.048 Ms -1 . Suggest a reason why the average rate of reaction is different from the instantaneous rate calculated in part b, even though the average of 10 and 30 seconds is 20 seconds. How would a chemist change their calculations in order to get a more similar average and instantaneous rate? The average rate of reaction between 10 s and 30 s is the slope of the line connecting the points (10 s, 1.65 M) and (30 s, 0.7 M). This line has a different slope than the line that is tangent to the plot at t = 20 s. To calculate an average that is more similar to the instantaneous rate, the time range should be narrowed down. For example, the average rate of reaction between 18 s and 22 s could be used instead. 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.2 0.4 0.6 0.8 1 Concentration (M) Time (s) 10 20 30 40 50 Concentration vs. Time (C 4 H 8 à 2 C 2 H 4 ) Time (s) Concentration (M) A B C

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Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 4 of 22 These points of (18 s, 1.1 M) and (22 s, 0.93 M) give an average rate of 0.043 M, which is closer to the instantaneous rate of 0.040 M/s. 6. Consider the reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) a. Write a rate expression that defines the relationship between the rate of the reaction and the disappearance or appearance of each reactant and product. 𝑅𝑎𝑡𝑒 = − ∆[N ଶ ] ∆𝑡 = − 1 3 ∆[H ଶ ] ∆𝑡 = 1 2 ∆[NH ଷ ] ∆𝑡 After a period of time, hydrogen gas is being consumed at a rate of 0.080 M/s. b. At the same point, what is the rate of NH 3 being formed? − 1 3 ∆[H ଶ ] ∆𝑡 = 1 2 ∆[NH ଷ ] ∆𝑡 ∆[NH ଷ ] ∆𝑡 = − 2 3 ∆[H ଶ ] ∆𝑡 = − 2 3 ൬−0.080 𝑀 𝑠 ൰ = 𝟎. 𝟎𝟓𝟑 𝑴/𝒔 c. At the same point, what is the rate that nitrogen gas being consumed? − 1 3 ∆[H ଶ ] ∆𝑡 = − ∆[N ଶ ] ∆𝑡 ∆[N ଶ ] ∆𝑡 = 1 3 ∆[H ଶ ] ∆𝑡 = 1 3 ൬−0.080 𝑀 𝑠 ൰ = −0.027 𝑀/𝑠 The word “consumed” in the question implies a negative rate, so the answer is 0.027 M/s d. At the same point, what is the overall rate of reaction? 𝑅𝑎𝑡𝑒 = − 1 3 ∆[H ଶ ] ∆𝑡 = − 1 3 ൬−0.080 𝑀 𝑠 ൰ = 𝟎. 𝟎𝟐𝟕 𝑴/𝒔 e. If the overall rate of reaction is found to be 2.00 M/s, what is the rate of consumption of hydrogen gas? 𝑅𝑎𝑡𝑒 = − 1 3 ∆[H ଶ ] ∆𝑡 ∆[H ଶ ] ∆𝑡 = −3(𝑅𝑎𝑡𝑒) = −3 ൬2.00 𝑀 𝑠 ൰ = −6.00 𝑀/𝑠 The term “rate of consumption” in the question implies a negative rate, so the answer is 6.00 M/s

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 5 of 22 7. If a reaction has the experimental rate law, rate = k [A] 2 , explain what happens to the rate when a. The concentration of A is tripled. The new rate is 9 times faster. b. The concentration on A is halved. The new rate is ¼ its previous value. 8. Consider the following reaction and its rate expression. Answer the following questions about the effects of reactant concentrations on reaction rate. 2 NO (g) + O 2 (g) → 2 NO 2 (g) rate = k [NO] 2 [O 2 ] a. [O 2 ] is doubled The new rate is 2 times faster. b. [NO] is tripled The new rate is 9 times faster. c. [NO] is halved The new rate is ¼ its previous value. d. [O 2 ] is halved and [NO] is doubled The new rate is 2 times faster. e. [O 2 ] is doubled and [NO] is halved The new rate is ½ its previous value. 9. The hypothetical reaction below was studied by measuring the initial rate of appearance of C. The following data was obtained: 2 A + 2 B  C + 3 D [A] (mol/L) [B] mol/L Initial Rate (mol/(L∙s)) 6.0 × 10 -3 1.0 × 10 -3 0.012 6.0 × 10 -3 2.0 × 10 -3 0.024 2.0 × 10 -3 1.5 × 10 -3 0.0020 4.0 × 10 -3 1.5 × 10 -3 0.0080 a. What is the order of the reaction with respect to substance A? Second order. Keeping [B] constant, when [A] is doubled from 0.0020 M to 0.0040 M, the initial rate quadruples. b. What is the order with respect to B? First order. Keeping [A] constant, when [B] is doubled from 0.0010 M to 0.0020 M, the initial rate also doubles. c. What is the overall order? Third order. Second order in [A] + first order in [B] = third order overall: 2 + 1 = 3. d. What is the rate law? Rate = k[A] 2 [B] e. Calculate the rate constant (include units).

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 6 of 22 𝑘 = ோ௔௧௘ [஺] మ [஻] = ଴.଴ଵଶ ெ௦ షభ (.଴଴଺଴ ெ) మ (଴.଴଴ଵ଴ ெ) = 𝟑. 𝟑 × 𝟏𝟎 𝟓 𝑴 ି𝟐 𝒔 ି𝟏 f. Suppose A is disappearing at a rate of 0.034 mol/(L∙s). What is the rate of appearance of C? What is the rate of appearance of D? 1 mol C and 3 mol D are generated for every 2 mol A that is reacted (based on the balanced reaction). Therefore, if A is disappearing at a rate of 0.034 Ms -1 , then C is appearing at a rate of 0.017 Ms -1 (half as fast) and D is appearing at a rate of 0.051 Ms -1 (3/2 as fast). 10. The transfer of an oxygen atom from NO 2 to CO has been studied at 540 K: CO (g) + NO 2 (g)  CO 2 (g) + NO (g) This data set was collected: Initial Concentration (mol/L) Initial Rate (mol/(L∙h) [CO] [NO 2 ] 5.1 × 10 -4 3.5 × 10 -5 3.4 × 10 -8 5.1 × 10 -4 7.0 × 10 -5 1.7 × 10 -8 5.1 × 10 -4 1.8 × 10 -5 6.8 × 10 -8 1.0 × 10 -3 3.5 × 10 -5 6.8 × 10 -8 1.5 × 10 -3 3.5 × 10 -5 1.02 × 10 -7 Use the data in the table to: a. Determine the reaction order with respect to each reactant. First order in [CO]. Keeping [NO 2 ] constant at 6.8 X 10 -8 M, when [CO] is doubled from 1.8 X 10 -5 M to 3.5 X 10 -5 M, the initial rate also doubles. First order in [NO 2 ]. Keeping [CO] constant at 3.5 X 10 -5 M, when [NO 2 ] is doubled from 3.4 X 10 -8 M to 6.8 X 10 -8 M, the initial rate also doubles. b. Write the rate law. Rate = k[CO][NO 2 ] c. Calculate the rate constant (include units). 𝑘 = 𝑅𝑎𝑡𝑒 [𝐶𝑂][𝑁𝑂 ଶ ] = 0.0015 𝑀ℎ𝑟 ିଵ (3.5 × 10 ିହ 𝑀)(1.02 × 10 ି଻ 𝑀) = 𝟒. 𝟐 × 𝟏𝟎 𝟖 𝑴 ି𝟏 𝒉𝒓 ି𝟏 11. Cisplatin (Pt(NH 3 ) 2 Cl 2 ) is an inorganic compound used in the treatment of various cancers. It is helpful for doctors and cancer researchers to understand cisplatin ’ s kinetics in aqueous solutions. The reaction of Pt(NH 3 ) 2 Cl 2 with water and the corresponding rate law are shown below. Pt(NH 3 ) 2 Cl 2 (aq) + 4 H 2 O (l) ⇄ [Pt(H 2 O) 4 ] 2+ (aq) + 2 NH 3 (aq) + 2 Cl - (aq) Rate = k [Pt(NH 3 ) 2 Cl 2 ] with k = 0.090 s -1

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Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 7 of 22 a. Calculate the initial rate of reaction when the initial concentration of Pt(NH 3 ) 2 Cl 2 is: i. 0.010 M 𝑅𝑎𝑡𝑒 = (0.090 𝑠 ିଵ )(0.010𝑀) = 𝟗. 𝟎 × 𝟏𝟎 ି𝟒 𝑴𝒔 ି𝟏 ii. 0.050 M 𝑅𝑎𝑡𝑒 = (0.090 𝑠 ିଵ )(0.050𝑀) = 𝟒. 𝟓 × 𝟏𝟎 ି𝟑 𝑴𝒔 ି𝟏 iii. 0.0050 M 𝑅𝑎𝑡𝑒 = (0.090 𝑠 ିଵ )(0.0050𝑀) = 𝟒. 𝟓 × 𝟏𝟎 ି𝟒 𝑴𝒔 ି𝟏 b. Using the observed trends form your answers above, fill in the table below: [Pt(NH 3 ) 2 Cl 2 ], initial Initial Rate of Reaction Rate of disappearance, Pt(NH 3 ) 2 Cl 2 Rate of appearance, Pt(H 2 O) 4 Rate of appearance, Cl - 0.010 M 9.0 X 10 -4 Ms -1 9.0 X 10 -4 Ms -1 9.0 X 10 -4 Ms -1 1.8 X 10 -3 Ms -1 0.020 M 1.8 X 10 -3 Ms -1 1.8 X 10 -3 Ms -1 1.8 X 10 -3 Ms -1 3.6 X 10 -3 Ms -1 0.040 M 3.6 X 10 -3 Ms -1 3.6 X 10 -3 Ms -1 3.6 X 10 -3 Ms -1 7.2 X 10 -3 Ms -1 c. Write an expression defining the rate of reaction in terms of the rate of disappearance of Pt(NH 3 ) 2 Cl 2 . Do the same thing in terms of the rate of appearance of Cl - . 𝑹𝒂𝒕𝒆 = −∆[𝐏𝐭(𝐍𝐇 𝟑 ) 𝟐 𝐂𝐥 𝟐 ] ∆𝒕 = 𝟏 𝟐 ∆[𝐂𝐥 ି ] ∆𝒕 12. The rate constants k of three different reactions are below.  Reaction A: k = 2.3 M -1 s -1  Reaction B: k = 1.8 Ms -1  Reaction C: k = 0.75 s -1 Which reaction represents a zero-order reaction? Why? Reaction B is the zero-order reaction. The units of the rate constant are in Ms -1 , indicating that no terms with units of M are multiplied by the rate constant to arrive at the desired rate units of Ms -1 . (Note that M/time – in this case, M/s – are the units for rates of reaction.) By the same logic, Reaction A is second-order and Reaction C is first-order.

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 8 of 22 13. The hypothetical reaction B  C + D occurs. Determine the rate law and the rate constant, k. Rate Law: Rate = k[B] because ln[B] vs. t is linear. Rate Constant: k = – (slope of linear plot) = – ቀ ିଽ.଻ହ ି (ି଺.ଽ) (ଶ଴଴ ௦ ି ଴ ௦) ቁ = 0.014 s -1 (or similar answer) 14. Again, the hypothetical reaction B  C + D occurs. Determine the rate law and rate constant, k. Rate Law: Rate = k[B] 2 because 1/[B] vs. t is linear. Rate Constant: k = (slope of linear plot) = ቀ (ଵ.ଷாସ ெ షభ ) ି (ଵ.଴ாଷ ெ షభ ) (ଶଶହ ௦ ି ଴ ௦) ቁ = 53 M -1 s -1 (or similar answer)

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 9 of 22 15. Now consider a hypothetical reaction, A + B  C. In the plot below, the curves for the two different concentrations of A vs. time overlap perfectly. What is the rate law? Justify the order of each reactant with a brief explanation. Rate law: Rate = k[A] 0 [B] 1 = k[B] The reaction is zero order with respect to A because changing the concentration of A has no effect on the rate. The reaction is first order with respect to B because each half-life requires the same amount of time. 16. Again, consider the reaction, A + B  C. A plot of ln[B] vs. time for two different [A] is shown. What are the rate law and rate constant? Justify the order of each reactant with a brief explanation. Rate law: Rate = k[A] 1 [B] 1 = k[A][B] We can tell that the reaction is first order with respect to B because the plot of ln[B] vs. time is linear. The reaction is first order with respect to A because doubling the concentration of A causes the rate slope to double. Rate constant: −𝑠𝑙𝑜𝑝𝑒 = 𝑘 ௢௕௦ = 𝑘[𝐴] ଵ 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 1 𝑀 𝑙𝑖𝑛𝑒 = (−10.25) − (−6.90) (240. 𝑠) − (0 𝑠) = −0.01395 𝑠 ିଵ = −0.0140 𝑠 ିଵ 𝑘 = − 𝑠𝑙𝑜𝑝𝑒 [𝐴] ଵ = − −0.0140 𝑠 ିଵ (1.0 𝑀) ଵ 𝒌 = 𝟎. 𝟎𝟏𝟒𝟎 𝑴 ି𝟐 𝒔 ି𝟏 17. Still considering the reaction, A + B  C, use the following plot of the instantaneous rate as a function of [B] to determine the rate law. Justify the order of each reactant with a brief explanation.

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Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 10 of 22 Rate law: Rate = k[A] 2 [B] 1 = k[A] 2 [B] We can tell that the reaction is first order with respect to B because the plot of instantaneous rate vs. [B] is linear (see below). The reaction is second order with respect to A because doubling the concentration of A causes the rate slope to quadruple . Think about the following plots (taken from a whole class slide) and how they relate to this problem: 18. The decomposition of dinitrogen pentoxide is first order with a rate constant of 5.00 × 10 -4 s −1 at 490 °C. 2 N 2 O 5 (g) → 4 NO 2 (g) + O 2 (g) a. If [N 2 O 5 (g)] is 0.000800 M initially, what will be its concentration 6.0 minutes later? In a first order reaction, ln[N 2 O 5 ] vs. time will be linear. 6 minutes = 360 s 𝑙𝑛(𝐴 ௧ ) = −𝑘𝑡 + 𝑙𝑛(𝐴 ଴ ) 𝐴 ௧ = 𝑒 ି௞௧ା௟௡(஺ బ ) [𝑁 ଶ 𝑂 ହ ] = 𝑒 ି൫ହ.଴଴ × ଵ଴ షర ௦ షభ ൯(ଷ଺଴. ௦)ା௟௡ (଼.଴଴ ×ଵ଴ షర ) = 𝟔. 𝟔𝟖 × 𝟏𝟎 ି𝟒 𝑴 b. What is the half-life of N 2 O 5 ? 𝑙𝑛(𝐴 ௧ ) = −𝑘𝑡 + 𝑙𝑛(𝐴 ଴ ) 𝑙𝑛 ൬ 𝐴 ௧ 𝐴 ଴ ൰ = −𝑘𝑡 𝑡 ଵ/ଶ = − 𝑙𝑛(0.5) 𝑘 = − 𝑙𝑛(0.5) 5.00 × 10 ିସ 𝑠 ିଵ = 𝟏𝟑𝟗𝟎 𝒔

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 11 of 22 19. The rate constant for the radioactive decay of radium-223 is 0.0606 day -1 . What is the half-life of radium-223? 𝑡 ଵ/ଶ = − 𝑙𝑛(0.5) 𝑘 = − 𝑙𝑛(0.5) 0.0606 𝑑𝑎𝑦 ିଵ = 𝟏𝟏. 𝟒 𝒅𝒂𝒚𝒔 20. For the reaction 2 NO 2 (g) → 2 NO (g) + O 2 (g), the rate constant is 0.775 M −1 s −1 at 330 °C. a. Write the rate law for this reaction. Rate = k[NO 2 ] 2 b. Draw an NO 2 concentration vs time plot for this reaction. c. What is an appropriate way to find k graphically? Since this is a second order reaction with respect to NO 2 , we can plot 1/[NO 2 ] vs. time. The slope of this linear plot will be the rate constant, k. d. If the initial concentration of NO 2 is 0.0050 M, what is the final concentration of NO 2 after 30.0 min? 30.0 𝑚𝑖𝑛 × 60 𝑠 1 𝑚𝑖𝑛 = 1800 𝑠 1 [𝐴 ௧ ] = 𝑘𝑡 + 1 [𝐴 ଴ ] 1 [𝐴 ௧ ] = (0.775 𝑀 ିଵ 𝑠 ିଵ )(1800 𝑠) + 1 0.0050 𝑀 = 1595 𝑀 ିଵ [𝑨 𝒕 ] = 𝟔. 𝟑 × 𝟏𝟎 ି𝟒 𝑴

Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 12 of 22 21. Looking at the graph below, what can you determine about the reaction order for the nuclear decay of carbon-14? What is the half-life? Calculate the rate constant. The nuclear decay of carbon-14 is first order since the number of 14 C nuclei drops by ½ at consistent time intervals. The half-life for nuclear decay of carbon-14 is 5,730 years. 𝑘 = − 𝑙𝑛(0.5) 𝑡 ଵ ଶ = − 𝑙𝑛(0.5) 5,730 𝑦𝑟 = 𝟏. 𝟐𝟏 × 𝟏𝟎 ି𝟒 𝒚𝒓 ି𝟏 22. Below are three concentration vs. time graphs. Assign each of them as 1 st , 2 nd , or 0 th order. What is a general rule relating reaction order and half-lives? This plot is for a first order reaction since the concentration drops by ½ at consistent time intervals. This plot is for a second order reaction. It is not first order since the concentration drops by ½ at inconsistent time intervals. It is not zeroth order since the plot of concentration vs. time is not linear. This plot is for a zeroth order reaction since the plot of concentration vs. time is linear . In general, half-life is related to reaction order in the following ways:  If half-life stays constant over time, the reaction is first order.  If half-life increases over time, the reaction is second order (or higher).  If half-life decreases over time, the reaction is zeroth order.

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Chemistry 104 Discussion Activity Packet ANSWER KEY Chemical Kinetics Kinetics, Module 13 Date of last revision: 7/17/23 Page 13 of 22 23. The reaction A + B  C is shown at various concentrations of A. Use the method of half-lives to determine the order of reaction with respect to reactant B. What is the half-life of B when [A] = 0.5 M? Order with respect to B: First order in [B] since the concentration drops by ½ at consistent time intervals. Only one plot is needed to determine this, but both plots verify this. Half-life of B when [A] = 0.5 M: Half-life is about 70 seconds. The drop from 6.00E-04 M to 3.00E-04 M takes about 70 s (120 s – 50 s). The drop from 3.00E-04 M to 1.50E-04 M also takes about 70 s (190 s – 120 s). [A] = 0.5 M

M13-Kinetics_DiscPkt_KEY_1-23

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