First Order Rate Calculations

First Order Rate Calculations - First-Order Rate Laws:...

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Unformatted text preview: First-Order Rate Laws: Self-Study Assignment You will have a QUIZ on the attached pages on _____________________ . Your assignment is: READ the pages attached. WORK the examples in the lesson. Complete the pages as homework. To work the examples, • use a sheet of paper to cover below the * * * * * line, • try the problem on your paper, • then check your answer below the * * * * * line. Start early. This assignment will require 2-4 hours of work outside of class. . ©2009 ChemReview.net v. 2e viii Module 25 — Kinetics Lesson 25D: Logarithm Calculations Prerequisites: Lessons 1A to 1C only. Timing: This lesson must be done before first-order-integrated rate law calculations, but it may be helpful at any time where base 10 or base e calculations are encountered in science classes. Pretest: If you think you know this topic, try the last 4 calculations in Practice D at the end of the lesson. If you can do those calculations, skip the lesson. ***** Logarithms To solve first-order integrated rate law calculations, it will be necessary to • Take the natural log of numbers, and • Convert natural logs to numbers. To do so, you will need to calculate using the ln function and the number e. Let’s begin by reviewing the rules for powers and logarithms. As always, on the problems below, cover the answers below the * * * * * line and write your answer the questions above the line. I. Numbers, Bases, and Exponents Any positive number can be represented as another number to a power. Computer science often calculates in “base 2:” 24 = _________, 210 = 1,024 ***** 24 = 16. Try this one on your calculator: 3.52.7 = ________ (Use the calculator that you will use for quizzes and tests. Do not use a graphing calculator if it has too much storage to be allowed during tests in your course.) ***** • A standard TI-type calculator might use: 3.5 yx 2.7 = • On a graphing calculator (if allowed), try: 3.5 ^ 2.7 enter • On a reverse Polish (RPN) calculator, try: 3.5 enter 2.7 yx Find a key sequence that gives a result of 29.4431… In science, we most often calculate in base 10 and base e. Base 10 is more familiar, and the rules are parallel for all bases, so let’s consider base 10 calculations first. II. Powers of 10 Numbers can be written as numbers, or as numbers in exponential notation, or as 10 to a power. The power of ten can be either an integer or a number with decimals. For example: ©2009 ChemReview.net v. 2r Page 680 Module 25 — Kinetics You know that 102 = the number ________ and 103 = the number ___________. Estimate the value of this number: 102.5 = __________________. ***** 100 = 102 < 102.5 < 103 = 1,000 ; half-way between 100 and 1000 is 550…, but the answer is sure to be “somewhere between 100 and 1,000.” Now, use your calculator to get an exact answer. 102.5 = what number? __________. ***** • On a standard TI-type calculator, you might try: 2.5 10x and/or 10 yx 2.5 = and/or nd 2.5 2 or INV log . Try all three. • On a graphing calculator, you might try: 10 ^ 2.5 enter • On an RPN scientific calculator, try: 2.5 enter 10x Write down a sequence that works for you and gives this result: 316 Compare 316 to your estimate. They should be close: maybe off by 2 or 3 times, but not off by a whole decimal place or power. Most errors in your operation of a calculator will be caught if you use the rule: “first estimate, then calculate.” (On sf/rounding: when converting between numbers and exponentials, the statistical justification for significant figures breaks down. We will add a systematic rule when we study acid-base pH. Until then, round answers to 3 or 4 places.) ©2009 ChemReview.net v. 2r Page 681 Module 25 — Kinetics Q. Use the estimation logic above and your calculator key sequence to write either numbers or numbers in scientific notation for the problems below. Recall that in scientific notation, the decimal is written after the first digit in the significand (for review, see Lesson 1B. If you are unsure about an answer, check it below before doing the next part. a. 103.9 = (estimate): _____________________ (calculate): ________________ b. 1023.7798 = (estimate): __________________ (calculate): ______________ ***** Answers a. 1,000 = 103 << 103.9 < 104 = 10,000 ; less than 10,000, but close: ~9,000 ? 7,943 Compare your estimate to the calculator answer. b. 1023.7798 = (estimate): 1 x 1023 < 1023.7798 < 1024 = 10 x 1023 so the number will be between 1 x 1023 and 10.0 x 1023. On the calculator: 6.02 x 1023 ~8.0 x 1023 ?? Compare your estimate to the calculator answer. Here’s an estimating rule: compare the number written as 10 to a power to the number written in scientific notation. The exponents of the two 10’s must be within one of each other. Try that rule and your calculator on these. c. 10─3.7 = (in scientific notation): _______________________ (expos ± one?)___ d. 10─9.7 = (in scientific notation): _____________________ (expos ± one?) ____ e. 10─13.2 = (in scientific notation): ________________________ (expos ± 1 ?)___ ***** c. 10─3.7 = 0.00020 or 2.00 x 10─4 . (expos ± one?) ─3.7 and ─4 √ To enter a negative number, usually a +/- or (─) key must be used. On a standard TI-type calculator, try: 3.7 On an RPN calculator, try: enter d. 10─9.7 = 2.00 x 10─10 3.7 +/- +/- 10x 10x (expos ± 1?) √ Note how your calculator displayed the exponent. You will need to translate the calculator display into scientific notation when writing answers. e. 10─13.2 = 6.31 x 10─14 (expos ± 1?) √ ***** ©2009 ChemReview.net v. 2r Page 682 Module 25 — Kinetics Practice A: Use the calculator that you will use on tests. Answers are at the end of this lesson. 1. 10+16.5 = _______________________________________________ (expos ± 1 ?) ____ 2. 10─16.5 = ______________________________________________ (expos ± 1 ?) ____ 3. 102.2 = ________________________________________________ (expos ± 1 ?) ____ 4. 10─11.7 = ______________________________________________ (expos ± 1 ?) ____ 5. 10─0.7 = ________________________________________________ (expos ± 1 ?) ____ III. Logarithms a. A logarithm is simply an exponent. A logarithm answers the question: if a number is written as a base number to a power, what is the power? b. A logarithm can be a power of any base. Since 24 = 16, the base 2 log of 16 is written: log216 = 4, In science, base 10 and base e are used most often. The symbol for a base 10 log is simply log . If no base is specified, you should assume that log means a base 10 log. The symbol for a base e log (a natural log) is ln . IV. Base 10 Logs a. The log function on your calculator finds a base 10 log. The log function answers the question: if a number is written as 10 to a power, what is the power? Using that rule, do these without your calculator. Write the log of 1) 102 ***** 1) The log of 102 is 2. 2) 1000 2) log 1000 = log 103 = 3 3) 0.001 3) log 0.001 = log 10─3 = ─3 b. The equation defining log is log 10x = x . It must be memorized, but it may be easier to remember by repeatedly reciting this example: the log of 100 is 2. ©2009 ChemReview.net v. 2r Page 683 Module 25 — Kinetics c. For numbers greater than one, a log value will be a positive number. Positive numbers between 0 and 1 have a negative power of 10 when written in scientific notation, and a negative log, as in example 3) above. To check that you are doing calculator operations properly, do a simple calculation, first in your head or on paper, then using the calculator. Make sure that the two answers agree. Let’s try that method on some simple examples. Using your head, write the log of 1) 100 Using a calculator, write the log of 1) 100 ***** Using your head: 2) 10,000 2) 10,000 3) 0.01 3) 0.01 1) The log of 100 = the log of 102 = 2. 2) The log of 10,000 = log 104 = 4 3) The log of 0.01 = log 10─2 = ─2 For part 1) on a calculator, • A standard TI-type calculator might use: 100 • On an RPN calculator, try: 100 enter log • Some graphing calculators do not have a log button. You can learn a workaround (log x = ln x/2.303) or buy an inexpensive calculator with log and ln buttons. Practice with the calculator that will be allowed on tests. log Did the calculator agree with the answers done in your head? It must. d. For cases where you cannot do the log in your head, you can estimate the log. Try this in your head: if log 100 = _____, and log 1000 =_____, log 500 ≈ ______. ***** Log 100 = 2, and log 1000 = 3 , so log 500 is between 2 and 3, about 2.5? Now do log 500 on the calculator: _________ . ***** 2.70 When estimating in this manner, the digits before the decimal of the estimate and on the calculator should agree, but the numbers after the decimal point will often vary. Do this one in your head and then the calculator: Log 50 = (estimate): ______________________ (on the calculator): ________ ***** Log 10 = 1, and log 100 = 2 , so log 50 is between 1 and 2, about 1.5? 1.70 e. For easy checking: the log of a number, and the exponent of the number when it is written in scientific notation, will agree within ± one. See if that rule works to calculate, then check: 1) Log(7.4 x 106 ) = (on the calculator): ________________ ( expo ± 1 ? ___) ***** ©2009 ChemReview.net v. 2r Page 684 Module 25 — Kinetics 1) Log(7.4 x 106 ) = (on the calculator): 6.87 ( expo ± 1 ? √ ) Do these on your calculator: 2) Log(7.4 x 10─6 ) = ___________________________________ (expo ± 1 ? ___) 3) Log 2,000 = ____________________________________________ (expo ± 1 ? ___) ***** 2) Log(7.4 x 10─6 ) = ─ 5.13 (expo ± 1 ? √ ) Keys: 7.4 E or EE 6 +/- log 3) Log 2,000 = log(2 x 103) = or if RPN: 7.4 E or EE 6 +/- enter log 3.30 (expo ± 1 ? √ ) Summary: Log Rules To Memorize 1. A logarithm is simply an exponent: the power to which a base number is raised. 2. A logarithm answers the question: if a number is written as a base to a power, what is the power? 3. Calculator log buttons find the power for a number written as 10 to a power. 4. Checking log results: when a number is written in scientific notation, its power of 10 must agree with its base 10 logarithm within ± 1. 5. Equations: The definition of a log is log 10x = x ; the log of 100 is 2 . Practice B: Practice with the calculator you will use on tests. 1. 10─5.4 = (in scientific notation):________________________________ (expos ± 1 ?) ____ 2. 10─11.5 = _____________________________ (expos ± 1 ?) ____ 3. 10─0.5 = (number): _________( scientific notation): ______________(expos ± 1 ?) ____ 4. Log(6.8 x 1012 ) = ___________________________________ ( expo ± 1 ? __) 5. Log(6.8 x 10─12 ) = ______________________________ (expo ± 1 ? __) 6. Log 4.6 = ____________________________________ (expo ± 1 ? __) 7. Log 0.0020 = ____________________________________ (expo ± 1 ? __) ©2009 ChemReview.net v. 2r Page 685 Module 25 — Kinetics V. Converting From Logs to Numbers a. Knowing a log, we need to be able to write the number. This is called taking the antilog, but it is easier to remember what this means (and what buttons to press) if you remember what a log is. A log is __________. *** ** A log is an exponent. If the base 10 log is 2, what is the number? ___________ ***** Since a log is an exponent of 10, if the log is 2, the number is 102 = 100 . As an equation, the rule is: 10log x = x . Repeat to remember: “10 to the log x equals x.” As an easy example, remember: 10log 100 = 102 = 100 Try these without a calculator, then check your answer below. If these are the logs, what are the numbers? 1) 6 2) ─2 3) 0 ***** 1) If log is 6 , number is 106 3) If log = 0 , number = 100 = 1 2) If log is ─2 , number is 10─2 = 0.01 (anything to the zero power is one) b. Knowing the log, to find the number, take the antilog. On a calculator, • input the log, then take the antilog: press INV LOG or 2nd LOG • input the log, then press 10x . A log is simply an exponent of 10. • On some calculators, the steps are: input 10, x^y , input the log value, = . or The first key sequence is logical: to go from number to a log, take the log; to go from log to number, go backward (take the antilog). The second and third sequences are logical. To find the number, make the log what it is: a power of 10. Write a key sequence (or two) that works on your calculator for this calculation: • If the log is 2, what is the number? ______________________________ Test that sequence on these. 1) If the log is 1 , the number is (in your head): __________ (calculator): _______ 2) If the log is ─2 , the number is (in your head): __________ (calculator): _______ ***** ©2009 ChemReview.net v. 2r Page 686 Module 25 — Kinetics 1) If the log is 1 , the number is: 101 = 10 2) If log = ─2 , the number is: 10─2 = 0.01 . (Use: 2 +/- to change sign.) Using the same key sequences, try these on your calculator. 3) If log x = 8.7 , x = ___________________________________ (expo ± 1 ? ___) 4) Log A = ─10.7 , A = ___________________________________ (expo ± 1 ? ___) Note the same “is it reasonable?” quick check. The log and the exponent of the number in scientific notation should agree, ± 1. ***** 3) Log x = 8.7 , x = 501,200,000 or 5.012 x 108 4) Log x = ─10.7 , x = 2.0 x 10─11 (expo ± 1 ? √ ) (expo ± 1 ? √ ) In cases like 3) and 4), your calculator may convert the answer to exponential notation and display the exponent far to the right -- where you may miss it when looking for just the significant digits. That’s another reason to estimate to check your answers. Summary: Add these to your memorized log rule list. 6. Knowing the log, to find the number, take the antilog. A log is a power of 10. 7. 10log x = x . Recite and repeat to remember: “10 to the log x equals x.” As an easy example, remember: 10log 100 = 102 = 100 8. On a calculator, to convert a log value to a number, • input the log value, then press INV LOG ; or 2nd LOG ; or • Input the log, then press 10x . or input 10, x^y , input the log , = . Practice C 1. Log x = 12.4 , x = _____________________________ (expo ± 1 ? ___) 2. Log A = ─5.9 , A= ____________________________ (expo ± 1 ? ___) 3. Log D = ─0.25 , D = ____________________________ (expo ± 1 ? ___) 4. Log x = 1.1 , antilog = ____________________________ (expo ± 1 ? ___) 5. 10─3.3 = ____________________________________ (expos ± 1 ?) ____ 6. Log (2.0 x 10─9 ) = __________________________________ (expo ± 1 ? __) 7. Log 0.50 = ________________________________________ (expo ± 1 ? __) ©2009 ChemReview.net v. 2r Page 687 Module 25 — Kinetics VI. Base e Rules a. The Symbol e A lower-case e is the symbol for the natural exponential. The symbol e is an abbreviation for a useful number: 2.718… (which should be memorized). The number e has many interesting mathematical properties. The number e is important in science because it is found in many equations that predict natural phenomena. One example is first-order reaction rates. A process with a constant rate of growth will obey the equation [A]t = [A]0 · ekt where k is the rate constant and t is the time after t = 0 . This first-order rate law for decay can be written in these two equivalent ways: [A]t = [A]0 · e─kt or as ln[A]t = ─kt + ln[A]0 To calculate first-order rate laws from time data, we will use both e and ln. b. Calculating with natural exponentials We know that e1 = __________(what number?) ***** 2.718…. Using 1 and the ex button, write or circle the key sequence that produces that answer on your calculator. 1 ex . • A standard TI-type calculator might use: • On an RPN scientific calculator, try: 1 enter ex . Use the same key sequence to do these, and then check your answers below. 1) e2 = 2) e2.5 = 3) e─1 = 4) e─2.5 = ***** 1) e2 = 7.389 2) e2.5 = 12.18 Recall that to enter a negative number, you usually need a +/- key. 3) e─1 ( = 1/e = 1/2.718.. ) = 0.3679 4) e─2.5 ( = 1/e2.5 ) = 0.08208 c. Calculating Natural Logs The ln function (the natural log) answers this question: if a number is written as e to a power, what is the power? Just as log 10x = x , the natural log definition is ln ex = x Use the natural log definition to do these without a calculator. 1) ln e0 = ______ 2) ln e1 = ______ 3. ln e─4 = ________ ***** ©2009 ChemReview.net v. 2r Page 688 Module 25 — Kinetics 1) ln e0 = 0 3) ln e─4 = ─4 2.) ln e1 = 1 By definition, ln e = _____ . ***** ln e = ln e1 = 1. Try this one in your head: ln(2.718) should equal about ____________ . ***** ln(2.718) ≈ ln e = ln e1 = 1. Now on your calculator, do the same calculation: ln(2.718) = ___________ ***** Is the calculator answer close to the mental arithmetic answer? Write down the key sequence that works: ln(2.718) = ____________________ Your calculator will take the natural log of any positive number. Try these. 1) ln 314 = ________________ ***** 1) ln 314 = 5.75 To check an answer, after writing it down, use the ex key and see if you return to the number you were taking the ln of. Try that as a check on these: 2) ln 0.0050 = _________________ (after writing answer, use ex . Check? ___ 3) ln (6.02 x 1023) = _________________________ 4) ln (19.29 x 10─15) = _________________________ Check? ___ Check? ___ ***** 2) ln 0.0050 = ─ 5.30 3) ln(6.02 x 1023) = 54.75 4) ln(19.29 x 10─15) = ─ 31.58 Note in part 4), a calculator does not require the input of scientific notation. However, if you use the ex key to check your answer, it will likely return the original number converted to scientific notation. When units are attached to a number, the unit may be included in the ln value. For example, the answer to ln(2.718 M) can be written 1.00 ln M Using that convention, try these on the calculator: 5) ln(10 M) = __________________ ***** 5) ln(10 M) = 2.303 ln M 6) ln(0.050 M) = _______________ 6) ln(0.050 M) = ─3.00 ln M Often, however, when the log or ln of a number and unit is taken, in many textbooks the unit in the answer is simply omitted. We will discuss how to handle these non-homogeneous cases later in this lesson. ©2009 ChemReview.net v. 2r Page 689 Module 25 — Kinetics d. Converting ln Values to Numbers A base 10 definition: 10log x = x A base e definition: eln x = x Note the similarities. Using the bottom equation, for some calculations involving ln and e you will not need a calculator. Try this: eln(─11) = ________ ***** eln(─11) = ─11 The equation eln x = x also means that if you know the ln value, to find the corresponding number, make the ln value a power of e. If the ln value = 1, the number (in your head) is ___________ * * * ** e1 = 2.718… Knowing that answer, do the same ln to number conversion on your calculator by taking the antilog. If the ln value = 1, the number obtained using the calculator is ___________ ***** Input the ln, then press INV or 2nd ln or press ex . Write down or circle the key sequence that converted ln = 1 to the number 2.718… Use your key sequence to convert the following ln values to numbers. Write first write the number in terms of e, then the number, then the number in scientific notation. 1) If ln = 6 , number = e = (number): ________ = (sci. notation):_____________ ***** 1) If ln = 6 , number = e6 = (nbr): 403 = (sci. notation): 4.03 x 102 In base e calculations, unlike base 10, there is no obvious correlation between the scientific notation exponent and the base e logarithm that helps in checking your answer. However, you can check by taking the ln of the number answer and see if it returns to the original ln value. Try these. 2) If ln = ─4.5 , number = e = (nbr): _________ = (sci. notation):_____________ 3) ln = 57.2 , number =________________________ 4) ln [A] = 0.03 ln M , [A] = e = (nbr. and unit): _________________ ***** ©2009 ChemReview.net v. 2r Page 690 Module 25 — Kinetics 2) If ln = ─4.5 , number = e─4.5 = 0.0111 = 1.11 x 10─2 3) ln = 57.2 , number = e 57.2 = 6.94 x 1024 4) If ln [A] = 0.03 ln M , [A] = e0.03 ln M = 1.03 M Note that the unit in 3) is the unit expected for a concentration WANTED. If a concentration is WANTED, but a log or ln value does not include a unit, add the unit M to the answer. Apply that rule to the following problems. Write the answer as a number or in scientific notation. When in doubt, check answers as you go. 5) ln [Z] = ─12.5 , [Z] =________________________ ***** 5) [Z] = e ─12.5 = 3.73 x 10─6 M (add the unit of concentration: mol/L) 6) ln [R] = ─ 0.17 , [R] =________________________ 7) [D] = e ─1.39 , [D] =________________________ 8) ln(0.250 M) = _______________________ 9) ln [A] = ─ 2.63 , [A] = ______________ ***** 6) [R] = e ─0.17 = 0.844 M 8) ln(0.250 M) = ─ 1.386 ln M 7) [D] = 0.249 M 9) [A] = e─ 2.63 = 0.0721 M ***** e. Notes on Notation with e and ln • Some calculators use an E at the right side of the answer screen to show the power of 10 for numbers in scientific notation. This is not the same as the symbol e for the natural exponential. • Be careful to distinguish “taking the ln” from “the ln value.” Ln(7.389) = _____________ . Try it. You should get close to 2. But if ln = 7.389 , the number with that ln is __________ Try it. ***** If ln = 7.389, the number is e7.389 = 1,618 If you get lost on a natural log calculation, a good strategy is to do a similar and simple base-10 mental and calculator computation, and then apply the same logic to the natural log case. Simple base 10 calculations can often be figured out in your head, and the formulas and steps for base 10 and base e calculations are parallel. ©2009 ChemReview.net v. 2r Page 691 Module 25 — Kinetics f. Converting between base 10 and natural logs A general rule for logarithms of any base is logb(x) = ln(x)/ln(b) , where b is the base. For base 10 logs, this equation becomes Log10(x) = ln(x)/ln(10) = Log10(x) = ln(x)/2.303 This relationship is generally memorized as 2.303 log(x) = ln(x) The value of a natural log is always 2.303 times higher than the base 10 log. Summary: Add these rules to your memorized log-rule list. 9. The symbol e is an abbreviation for a number with special properties: e = 2.718... 10. The ln (natural log) answers the question: if a number is written as e to a power, what is the power? 11. Note the patterns: A useful memory device is log 10x = x and 10log x = x . “The log of 10 to the x is x; 10 to the log x is x.” ln ex = x and eln x = x . Write the base 10 rules, then substitute e and ln. Note the logic: A log is an exponent. 12. Knowing the ln, to find the number, take the antilog. On a calculator, • • 13. input the ln value, then press INV ln ; or Input the ln value, then press ex . An ln is simply an exponent of e. 2.303 log(x) = ln(x) Practice D: Do the odd-numbered problems, and the evens for later review or additional practice. 1. e5.2 = 4. ln 1066 = 2. e─1.7 = 5. ln 0.0050 = 7. ln(19.29 x 10─6) = 8. ln e6.2 = 3. e─20.75 = 6. ln(3 x 108) = 9. eln(─42) = 10. If ln = ─6.8 , number = e 11. If ln D = 7.4822 , D = 12. If ln = ─12.5 , antilog = 13. If log [A] = ─9 , [A] = 14. If log x = 13.7 , x = 15. Log A = ─13.7 , A = 16. 10─11.7 = ©2009 ChemReview.net v. 2r = (number in sci. notation):___________________ Page 692 Module 25 — Kinetics 17. 19. ln(0.050 M) = 21. If log(x) = 5.0 , ln(x) = 23. Given that 18. e ─11.7 = ln [B] = ─13.7 , [B] = 20. e ─0.693 ln M = 22. if ln(x) = 34.5 , log(x) = ln[A]t = ( ─ 0.0173 s─1) ( t ) + 0.693 a. If t = 20. s, [A] = ? 24. Given that b. If [A] = 0.710 M, t = ? ln[A]t = ( ─ 0.0241 ln M•yrs.─1 ) ( t ) ─ 4.61 ln M a. If [A] = 0.0025 M, t = ? b. If t = 28.8 years, [A] = ? ANSWERS Practice A 1. 10+16.5 = 3.16 x 1016 2. 10─16.5 = 3.16 x 10─17 3. 102.2 = 1.58 x 102 4. 10─11.7 = 2.00 x 10─12 (expos ± 1 ?) √ 5. 10─0.7 = (expos ± 1 ?) √ 2.00 x 10─1 (expos ± 1 ?) √ Practice B 1. 10─5.4 = 3.98 x 10─6 (expos ± 1 ?) √ 2. 10─11.5 = 3.16 x 10─12 (expos ± 1 ?) √ 3. 10─0.5 = (number): 0.316 (std. notation): 3.16 x 10─1 (expos ± 1 ?) √ 4. Log (6.8 x 1012 ) = 12.83 5. Log (6.8 x 10─12 ) = ─11.17 6. Log 4.6 = 0.663 7. Log 0.0020 = ─2.70 Practice C 1. Log x = 12.4 , x = 2.51 x 1012 5. 10─3.3 = 5.01 x 10─4 2. Log x = ─5.9 , x = 1.26 x 10─6 6. Log (2.0 x 10─9 ) = ─8.70 3. Log x = ─0.25 , x = 0.562 = 5.62 x 10─1 7. Log 0.50 = ─0.30 4. Log x = 1.1 , antilog = 12.6 = 1.26 x 101 Practice D 1. e5.2 = 181 2. e─1.7 = 0.183 4. ln 1066 = 6.97 5. ln 0.0050 = 7. ln (19.29 x 10─6) = ─10.86 ─5.30 8. ln e6.2 = 6.2 3. e─20.75 = 9.74 x 10─10 6. ln (3 x 108) = 19.52 9. eln(─42) = ─42 10. If ln = ─6.8 , number = e─6.8 = (number in std. notation): 1.11 x 10─3 11. If ln D = 7.4822 , D = 1776 12. If ln = ─12.5 , antilog = 3.73 x 10─6 ©2009 ChemReview.net v. 2r Page 693 Module 25 — Kinetics 13. If log [A] = ─9 , [A] = 10─9 M 14. If log x = 13.7 , x = 5.01 x 1013 16. 10─11.7 = 2.00 x 10─12 15. Log A = ─13.7 , A= 2.00 x 10─14 18. e ─11.7 = 8.29 x 10─6 17. ln [B] = ─13.7 , [B] = 1.12 x 10─6 M 20. e ─0.693 ln M = 0.500 M 19. ln(0.050 M) = ─3.00 ln M 21. If log(x) = 5.00 , ln(x) = ? 22. if ln(x) = 34.5 , log(x) = ? 23. Given that 2.303 log(x) = ln(x) . ; 2.303 log(x) = ln(x) . ; ln(x) = (2.303)(5.00) = 11.5 log(x) = 34.5/2.303 = 15.0 ln[A]t = ( ─ 0.0173 s─1) ( t ) + 0.693 a. Strategy: to find [A]t , first solve for ln[A]t ? = ln[A]t = ( ─ 0.0173 s─1 ) ( 20.0 s ) + 0.693 = ( ─ 0.346 + 0.693 = 0.347 WANTED is [A] at 20 s. Known is: ln[A]20 s = 0.347 ***** [A] = eln[A] = e0.347 = 1.41 M Solve for [A]20 s. (If a concentration is wanted, add M as unit) b. Strategy: Solve for t in symbols first. ***** t = ln[A] ─ 0.693 = ln[0.710 M] ─ 0.693 = ─ 0.342 ─ 0.693 ─ 0.0173 s─1 ─ 0.0173 s─1 ─ 0.0173 s─1 = 24. Given that = ─ 1.035 ─1 ─ 0.0173 s 59.9 s = t = ( 1/ s─1 = s --- see lesson 25F ) ln[A]t = ( ─ 0.0241 ln M/yr. ) ( t ) ─ 4.61 ln M a. Strategy: Solve for t in symbols first. ***** t = ln[A] + 4.61 ln M = ln[0.0025 M] + 4.61 ln M = ─ 5.99 ln M + 4.61 ln M = ─ 0.0241 ln M/yr. ─ 0.0241 ln M/yr. ─ 0.0241 ln M/yr. = = ─ 1.38 ln M ─ 0.0241 ln M/yr. 57.3 years = t b. Strategy: to find [A]t , first solve for ln[A]t ? = ln[A]t = ( ─ 0.0241 ln M/yr. ) (28.8 yr. ) ─ 4.61 ln M = ─ 0.694 ln M ─ 4.61 ln M = ─ 5.30 ln M WANTED is [A]. Known is: ln[A] = ─ 5.30 ln M ***** [A] = eln[A] = e─ 5.30 ln M = 0.0050 M Solve for [A]. ***** ©2009 ChemReview.net v. 2r Page 694 Module 25 — Kinetics Lesson 25E: Integrated Rate Law – First Order Prerequisites: Do Lesson 25D before this lesson. ***** For first-order reactants, the integrated rate law is ln[A]t = ─kt + ln[A]0 The most commonly encountered reaction that is first order is the nuclear process of radioactive decay, which is always first-order. However, the steps of standard chemical reactions can also be first order as well. The easiest way to learn to use this rate law is by example. First-Order Calculations In some rate law calculations, the order of a reactant is supplied, and the problem can be solved by simply writing and solving the appropriate rate law. For the following problem, begin by listing and assigning symbols to the WANTED and DATA. Then decide what equation relates those symbols. Solve the equation in symbols before plugging in numbers. If you get stuck, read the answer until unstuck, then try again. Q. Radon-222 is a noble gas nuclide that decays to form other elements. The rate for this radioactive decay is first order, with a rate constant of 0.181 ln M/day. If the original [Rn-222] in a sample is 2.5 x 10─4 M, what will be the [Rn-222] after two weeks? **** * WANTED: [Rn]after 2 weeks = [Rn]t 0.181 ln M/day = k DATA: 2 weeks = t = 14 days (convert to units consistent with the rate constant) Initial concentration = 2.5 x 10─4 M = [Rn] 0 The rate is first order. We know two equations for first-order rates: Differential law: Rate = k [A] and Integrated law: ln[A]t = ─kt + ln[A]0 Which equation is the best match with the symbols in the data? ***** When the rate data includes time, the integrated law will usually be needed to solve. Write the specific equation: ln [Rn]t = ─kt + ln [Rn]0 See if you can solve for the WANTED amount. ***** Strategy: To find [Rn]t, first solve the rate equation for ln[Rn]t, then use [Rn] = eln [Rn] ***** ©2009 ChemReview.net v. 2r Page 695 Module 25 — Kinetics ? = ln [Rn] = ─kt + ln [Rn]0 = ─ (0.181 ln M/day)(14 days) + ln(2.5 x 10─4 M) = ─ 2.53 ln M ─ 8.29 ln M = ─ 10.82 ln M = ln [Rn]14 days (Math help? See Lesson 25D, section VI.) If needed, finish solving for the WANTED unit. ***** WANT: [Rn]after 14 days = eln [Rn] = e(─ 10.82 ln M) = 2.0 x 10─5 M Rn ***** Units of Concentration In rate-law calculations, the units for concentration may be either moles/liter, or units proportional to moles per liter. The units must also be consistent: the same for all measures of concentration in a given problem. Some of the “proportional to moles per liter” units found in rate problems are • grams per unit of volume for a substance; such as g/cm3; • atoms of a substance per unit of volume: e.g. atoms/liter; • counts of radioactive decay/time • atoms per unit proportional to volume for a given substance: such as “atoms/gram dried cotton;” • pressure of a gas in any pressure units: kPa, torr, or other pressure units; • partial pressure of a gas in any pressure units. Why is gas pressure proportional to concentration? Since, for an ideal gas, PV=nRT, rearranging terms we can write n/V = moles/liter = P/RT = P x (the constant 1/RT at constant temperature). This equation can be re-written as: [ideal gas] = (a constant) x P which is one of the forms for a direct proportion (Lesson 18A). At a constant temperature (which is required for a rate constant to be constant), molar concentration is therefore directly proportional to pressure for a gas with ideal behavior. Units of the First-Order Rate Constant The units for first-order rate constants (k) can be expressed as either “ln(concentration units)/time units,” or as “1/time” units, such as “sec─1.” In most textbooks, the 1/time format for first-order k units is used. Mathematically, either practice can be justified. From this point forward, these lessons will generally use the practice in most textbooks: to drop the unit when an ln is taken of a quantity. However, if units are dropped from ln values, solving for concentration may not produce a unit as part of the answer. When 1/time units are used for k, you must add units to any concentration calculated in the problem, and those units must be consistent with the concentration unit used elsewhere in the problem. If no unit for concentration is supplied, assume the unit is moles/liter. ©2009 ChemReview.net v. 2r Page 696 Module 25 — Kinetics Practice A 1. The earth’s atmosphere has a small amount of carbon dioxide that contains 14C, called carbon-14, a radioactive isotope of carbon. While a plant is alive, C-14 is stored in its cells during photosynthesis. Living plants have a relatively constant and predictable concentration of radioactive carbon. After the plant is harvested and/or dies, the radioactive carbon is no longer replenished, and the concentration of C-14 in the nonliving plant material falls as C-14 undergoes radioactive decay at a first-order rate. By measuring the amount of C-14 in plant remains, how long ago the plant was harvested can be determined. The rate constant for the decay of C-14 is 1.21 x 10─4 year─1. If the [C-14] in freshly harvested cotton fibers is 2.00 x 1010 atoms/gram, and the [C-14] in a cotton garment found in a burial tomb is found to be 1.12 x 1010 atoms/g, how many years ago was the cotton harvested? 2. Using the same k value for C-14 and the same [C-14] at harvest supplied in Problem One, calculate the atoms per gram that would be found in a sample after 10,000. years. 3. Using the same k value for C-14 and the same [C-14] at harvest supplied in Problem One, calculate the half-life of C-14: how many years are required for the [C-14] to be reduced to 1/2 of the concentration at harvest. 4. If in an experiment, a first-order rate equation for a decay reaction is found to be ln [B] = ─ (0.200 year─1)(t) ─ 3.22 , what was the original [B]? First-Order Reactions and Graphical Analysis Some rate law calculations require that the order of a reactant be determined from experimental data before applying a rate law to solve. When the order of a reactant is not known, it can be determined by graphical analysis. Graphical analysis is also used to find the specific rate law equation that explains the data. For the following question, cover below the * * * * * line, then answer the questions above the line. Q. For the reaction: A B , time and [A] are measured as a reaction proceeds. The data is recorded at the right. 0 20.0 1.41 40.0 1.00 60.0 0.710 0.500 0.353 120. ln[A] in ln(M) 2.00 100. ©2009 ChemReview.net v. 2r [A] in M 80.0 a. Use the “first half-life” method and [A] to determine the order of the reaction, and then check your answer below. ***** seconds 0.250 Page 697 Module 25 — Kinetics The first half-life is 40. s. Double the first half-life is 80 s. At t = 80 sec., [A] = 25% of the original concentration. This fits the behavior of first order in A. b. Using a calculator, add values for ln[A] to the last column of the table above, then check your answers below. ***** Sample: If [A] = 2.00 M, seconds [A] in M ln[A] ln[A] = ln(2.00 M) = 0.693 ln M = 0.693 0 2.00 0.693 We will use the shortcut: if you take the ln of a unit, omit the unit in the answer. 20.0 1.41 0.344 40.0 1.00 0 Your values should match those at the right. 60.0 0.710 ─ 0.342 80.0 0.500 ─ 0.693 100. 0.353 ─ 1.04 120. 0.250 ─ 1.39 c. On the two grids below, graph the data: first [A] vs. time, then ln[A] vs. time. [A] versus time 2 [A ] in m ol/L 1.5 1 0.5 0 0 40 80 120 seconds ln[A] versus time d. For which graph does the data fit closer to linear behavior? f. Based on the two graphs, does the data better fit the behavior of a reaction that is zero order in A, or first order in A? 0.5 ln[A] e. For the graph that is linear, calculate the slope of the line. 1 0 0 40 80 120 -0.5 -1 -1.5 seconds Check your answers below. ***** ©2009 ChemReview.net v. 2r Page 698 Module 25 — Kinetics d. Your graphs should be similar to these. ln [A] versus time [A] versus time 1 2 0.5 ln [A ] [A ] in m o l/L 1.5 1 0 0 40 80 120 -0.5 0.5 -1 0 0 40 80 120 -1.5 seconds seconds The linear graph is the plot of ln[A] vs. time. e. For the slope calculation, using the lowest and highest x values on the graph: At t1 = lowest time = 0 s, ln[A] = 0.693 ; at t2 = 120 s, ln[A] = ─ 1.39 m = rise = Δln[A] = ln[A]2 ─ ln[A]1 = (─1.39 ─ 0.693) = ─ 0.0174 s─1 run Δt t 2 ─ t1 (120 ─ 0) s ***** f. For this data, • The slope of [A] vs. time, which was constant for zero-order reactants, is not constant for this first-order reactant data. • The slope of ln[A] vs. time is constant; Let’s analyze the first-order rate law to see why this is the case. 1. Write the two forms of the rate law for a reaction A B that is first-order in A. ***** Differential rate law: Integrated rate law: Rate = k [A] ln[A]t = ─kt + ln[A]0 2. Compare the first-order integrated rate law: to the equation for a line on a graph: ln[A]t = ─kt + ln[A]0 y = mx + b Write the symbols in the first-order integrated rate law next to the corresponding symbols in the equation for a line. y= m= x= b= ©2009 ChemReview.net v. 2r Page 699 Module 25 — Kinetics Then, for the data in the problem above, after the symbols for the two constants, add the values for the two constants. Include numbers and units. ***** y = ln[A]t = a variable amount. m = ─ k = the constant slope = ─ 0.0174 s─1 from Part (e) above. x= t = time, a variable. b = ln [A]0 = ln ( [A] initially, at t = 0 ) = ln(2.00 M) = + 0.693 3. Write the value for k, with its units. ***** Since m = ─ k , k = ─ m = ─ (the slope) = + 0.0174 s─1 4. Write the first-order integrated rate law, a. as memorized; then b. re-write the law, keeping the same variable symbols, but substituting for the two constants the values and units of the constants. This result is the calculated integrated rate law. ***** a. The rate law using symbols: b. Substituting the constants: ln[A]t = ─kt + ln[A]0 ln[A]t = ─ ( + 0.0174 s─1 ) ( t ) + 0.693 5. Test your rate law: choose a time in the original data table that was not used to calculate the slope. Enter that time into the calculated integrated rate law and calculate [A]. Then compare that calculated [A] to the actual [A] in the data table. See if the law, with your calculated constants, predicts the [A] in the data at that time. (Use t = 60.0 s to match the answer below.) ***** Equation: DATA: ln[A]t = ─ ( + 0.0174 s─1 ) ( t ) + 0.693 (list the symbols for the variables in the equation, but don’t re-write the known constants.) ln[A]t = ? t = 60.0 s WANTED: Strategy: [A]t at t = 60.0 s To find [A]t, use the equation to find ln[A]t, then use [A] = eln[A] If you needed that hint, adjust your work and finish from there. ***** ©2009 ChemReview.net v. 2r Page 700 Module 25 — Kinetics ? = ln[A]t = ─ ( + 0.0174 s─1 ) ( t ) + 0.693 = ─ ( + 0.0174 s─1) (60.0 s) + 0.693 { ( s─1 )( s1 ) = s0 = 1 } = ─ ( 1.044) + 0.693 = ─ 0.351 ln[A]60 s = ─ 0.351 WANTED is [A] at 60.0 s. Known is: Solve for [A]60 s. ***** [A] = eln[A] = e ─ 0.351 [A] = 0.704 M (add the unit for concentration used in the problem: mol/L) In the original data table at t = 60. s, [A] = 0.710 M Allowing for rounding and experimental error, this calculated answer and the data in the original table agree. In predicting the experimental data, the first-order integrated rate law, with its calculated constants, worked. ***** 6. Test your equation again: use the calculated rate law to find the time at which [A] will equal 0.353 M. ***** Rate Law: ln[A] = ─ ( + 0.0174 s─1 ) ( t ) + 0.693 DATA: [A] = 0.353 M ln[A] = ln (0.353 M) = ─ 1.041 t=? Solve the equation in symbols for the wanted symbol. ***** t = ln[A] ─ 0.693 = ─ 1.041 ─ 0.693 ─ 0.0174 s─1 ─ 0.0174 s─1 t= = ─ 1.734 ─ 0.0174 s─1 99.7 s Compare this calculated time to the time in the original table at [A] = 0.353 M. The first-order integrated rate law, with its calculated constants added, explains and predicts, within experimental error, the results of the experiment. Practice B. Answers are at the end of this lesson. 1. Write the integrated rate law for a a. zero order reactant: _________________________________________ b. First-order reactant: _________________________________________ ©2009 ChemReview.net v. 2r Page 701 Module 25 — Kinetics 2. To get a straight line graph using [A] and time data, a. For a zero-order reactant, plot _________ on the y-axis and _____ on the x-axis. b. For a first-order reactant, plot __________ on the y-axis and _____ on the x-axis. 3. What ratio that uses concentration and time must be constant for a a. zero-order reactant: ___________________________ b. First-order reactant: ___________________________ 4. What will be the term for the y-intercept in an integrated rate law that is a. Zero order: ___________________________ b. First-order: ___________________________ 5. If reactant A is first order, and [A] versus time data is collected, a. Will a plot of [A] versus time have points on a line? ______________ b. What plot will produce points on a line? _______________________ ANSWERS Practice A 1. The WANTED and DATA include terms for both time and a first-order rate constant. What equation includes those terms? The first-order integrated rate law: ln [A] = ─kt + ln [A]0 That equation will work for any units that are proportional to concentration. Assume that atoms/g is proportional to molar concentration for cotton. DATA: ln[A] = ln[C-14]after decay = ln(1.12 x 1010 atoms/g) = + 23.14 k = 1.21 x 10─4 year─1 t=? ln [A]0 = ln [C-14]0 = ln(2.00 x 1010 atoms/g) = + 23.72 Solving for the WANTED symbol in symbols first: t = ln [A] ─ ln [A]0 = ─k ©2009 ChemReview.net v. 2r (+ 23.14 ─ 23.72) = ─ 4 y r─ 1 ─ 1.21 x 10 ─0.58 yr. = 4,800 yr. ─4 1.21 x 10 Page 702 Module 25 — Kinetics 2. For first-order decay, use the first-order integrated rate law: ln [A] = ─kt + ln [A]0 WANTED: [C-14]after 10,000. years, in atoms/gram DATA: Use the equation symbols to make the data table. ln[A] = ln[C-14]after 10,000.years = ? k = 1.21 x 10─4 year─1 t = 10,000. years ln [A]0 = ln [C-14]0 = ln(2.00 x 1010 atoms/g) = + 23.72 Strategy: To find the [C-14]after decay, use the rate law to find ln[C-14]after decay first. Then use: ? = [C-14] = eln [C-14] If you needed that hint, adjust your work and finish. ***** ln[C-14]after decay = ─ (1.21 x 10─4 year─1 )(10,000 yrs ) + 23.72 = ─ 1.21 + 23.72 = + 22.51 WANTED is [C-14 ]10,000 yrs. Known is: ln[C-14]10,000 yrs = + 22.51 Solve for [C-14]. ***** [C-14]10,000 yrs = eln[C-14] = e+22.51 = 5.97 x 109 atoms/g When solving for concentration, add the concentration unit used elsewhere in the DATA. 3. For first-order decay, use the first-order integrated rate law: WANTED: Strategy: ln [A] = ─kt + ln [A]0 Half-life of C-14 Half-life is time it takes for initial concentration to be cut in half. Since [C-14]original = 2.00 x 1010 atoms/g [C-14]at half-life = 1.00 x 1010 atoms/g (half as much) If you needed that hint, adjust and try the problem again. ***** DATA: ln [C-14]at half-life = ln(1.00 x 1010 atoms/g) = 23.03 k = 1.21 x 10─4 year─1 t = ? years ln [A]0 = ln [C-14]0 = ln(2.00 x 1010 atoms/g) = + 23.72 Solve the rate equation for the WANTED symbol, in symbols first. ©2009 ChemReview.net v. 2r Page 703 Module 25 — Kinetics t = ln [A] ─ ln [A]0 = ─k (+ 23.03 ─ 23.72) = 0.69 yr. = 5,700 yr. ─4 yr ─1 ─4 ─ 1.21 x 10 1.21 x 10 Does this answer make sense? In problem one, a little less than half of the initial C-14 had decayed in 4,800 years, so 5,700 years for exactly half is in close agreement. In problem 2, the 10,000 year time of decay was a little less than two 5,700 year half lives. After 2 half lives, 25% of the original amount should be remaining for first-order decay. In problem 2, 10,000 years is a little less than 2 half lives and 0.597 x 1010/2.00 x 1010 = 30%, or a bit more than 25% of the original, remains. This is about what would be expected by estimation. These three answers are consistent. 4. WANTED: [B]0 (Write the symbol for the WANTED initial concentration of B) This is a first-order reaction. The first-order equation that includes [B]0 is DATA: ln [B] = ─ kt + ln [B]0 . Compare that to the form of the given equation: ln [B] = ─ (0.200 year─1)(t) ─ 3.22 ln [B]0 = ─ 3.22 (Finish from here) ***** [B]0 = eln[B]0 = e─ 3.22 = 0 0400 M B If a concentration is wanted, add M as unit unless other concentration units are used in the problem. Practice B 1. a. Zero order: [A] = ─kt + [A]0 b. First-order: ln [A] = ─kt + ln [A]0 2. a. To get a straight line for a zero-order reactant, plot [A] on the y-axis and t on the x-axis. b. For a first-order reactant, plot ln [A] on the y-axis and t on the x-axis. 3. a. Constant ratio for a zero-order reactant: Δ[A] /Δt b. First-order reactant: Δ ln[A] /Δt 4. a. The y-intercept in an integrated rate law that is zero order: [A]0 5. a. Will a plot of first order [A] vs. time have points on a line? No b. First order: ln [A]0 b. What plot will? ln [A] vs. t ***** ©2009 ChemReview.net v. 2r Page 704 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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