Chem27kinetics - Calculations In Chemistry Modules 19 and...

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Unformatted text preview: Calculations In Chemistry Modules 19 and above have been re-numbered. Module 25 on Kinetics is now Module 27 and is in this packet The former Module 27 on Acid-Base Fundamentals is now Module 29 If you are looking for Acid-Base pH topics, check Module 29 At www.ChemReview.Net ***** Module 27 – Kinetics: Rate Laws Module 27 – Kinetics: Rate Laws................................................................................748 Lesson 27A: Lesson 27B: Lesson 27C: Lesson 27D: Lesson 27E: Lesson 27F: Lesson 27G: Lesson 27H: Lesson 27I: Kinetics Fundamentals .....................................................................................748 Rate Laws ............................................................................................................753 Integrated Rate Law --Zero Order ...................................................................762 Base 10 Logarithms ...........................................................................................770 Natural Log Calculations .................................................................................779 Integrated Rate Law -- First Order ..................................................................787 Reciprocal Math..................................................................................................797 Integrated Rate Law -- Second Order..............................................................802 Half-Life...............................................................................................................809 For additional modules, visit www.ChemReview.Net ©2009 ChemReview.net v. 2e ii Module 27 — Kinetics: Rate Laws Module 27 — Kinetics: Rate Laws Prerequisites: It will be helpful if you have completed Module 20 – Graphing before this module. ***** Lesson 27A: Reaction Rates Reaction kinetics includes the study of • the speed (the rate) of chemical reactions; • the mechanism of chemical reactions: the “intermediate” particles that form in the transition from reactants to products; and • reaction energetics: how much energy is needed to form intermediates, and how reaction rates are affected by temperature. Why is kinetics important? We would like to speed up certain reactions that are important to society, such as the • burning of gasoline in internal combustion engines, so that incompletely burned combustion products do not escape into the environment; and • converting graphite (pencil lead) into “industrial diamonds” for drills and cutting tools. Other reactions, we would like to slow down, such as • the conversion of the iron in steel to iron oxide (rust), and • the decay of ozone (O3) in earth’s upper atmosphere, which protects life forms from the harmful elements of solar radiation. Kinetics helps in understanding, and potentially controlling, these processes. Definitions A rate is a change in a quantity per unit of time. A rate is a ratio that always has a time unit in the denominator. • Speed (or velocity) is a measure of how fast an object is moving: the rate of change of position per unit of time, in units such as miles per hour or meters per second. • An interest rate is the percentage of a loan amount you must pay, per month or per annum (per year), in addition to re-paying the principal. The rate of a chemical reaction measures how fast the reactants are used up, and/or how fast the products are formed. A reaction mechanism breaks down an overall chemical reaction into individual steps, identifying the temporary particles that form in the transition from reactants to products. The rate of a reaction is determined by the slowest step in the mechanism: the ratedetermining step. Rate-law equations measure the characteristics of the rate-determining step of a reaction. ©2011 ChemReview.net v. 2z Page 748 Module 27 — Kinetics: Rate Laws Catalysts are substances that cause a reaction to go faster, but are not used up in the reaction. • Metals such as platinum and palladium are used in a car’s catalytic converter to help ensure that incompletely burned gasoline reacts with oxygen. • Enzymes are biological catalysts: carbon-based molecules that regulate living processes. Average and Instantaneous Reaction Rates In chemical reactions, the rate at which one reactant is used up, or one product is formed, is defined as the rate of change of the concentration of the reactant or product. Rate = change in [A] change in time For most reactions, a graph of the concentration of a reactant or product ([A]) versus time will produce a smooth curve (an exception is the zero-order reaction, discussed below, for which the graph is a line). Two types of reaction rates are the average rate and the instantaneous rate. • The average reaction rate between two times t1 and t2 is defined as = Average Rate = change in [A] change in time Δ[A] = [A]2 ─ [A]1 Δt t2 ─ t1 This equation must be memorized. On a graph [A] versus time, the average rate will be the slope of the line between the two points (t1, [A]1) and (t2, [A]2). The instantaneous rate is, in the notation of calculus, d[A]/dt , the rate at a specific time t. At a given time, the instantaneous rate can be calculated by graphing measurements for [A] 0.12 versus time, drawing a line tangent to the 0.1 curve at the time, and DATA calculating the slope of Average Rate 0.08 the tangent line. Instantaneous For most reactions, the two slopes, representing the values for the average rate between two close points, and the instantaneous rate at a time half-way between the two points, will be close but not the same. Rate [X] • 0.06 0.04 0.02 0 0 10 20 30 40 50 Time (Seconds) ©2011 ChemReview.net v. 2z Page 749 60 Module 27 — Kinetics: Rate Laws Average Rate Calculations To calculate an average reaction rate from concentration versus time measurements: 1. Select the two measurements of concentration versus time to be averaged. 2. Assign the lower time to be t1, and 3. Apply the average rate definition: = Average Rate = change in [A] change in time Δ[A] = [A]2 ─ [A]1 Δt t 2 ─ t1 Use those steps on the following example. Q1. For the data at the right, calculate the average rate of the reaction for the time between three and five minutes. Assign t1 to be the lower time: 3.0 min. [A]1 is then 0.275 M. 0 0.500M 0.395 M 2.0 min. 0.325 M 3.0 min. 0.275 M 4.0 min. ***** [A] 1.0 min. (When you come to a * * * * * line, cover below the line and answer the questions above the line in your notebook.) Time 0.240 M 5.0 min. 0.215 M Using the equation for average rate: Average Rate = Δ[A] = [A]2 ─ [A]1 = 0.215 M ─ 0.275 M = Δt t2 ─ t1 5.0 min. ─ 3.0 min. ─ 0.030 mol L • min. Note about this result: 1. The negative sign for the rate reflects that [A] is decreasing with time. 2. The units can be written in several equivalent ways. M = M • min.─1 = min. mol • 1 = L min. mol = mol • L─1 • min.─1 L • min. Follow the math from step to step between these equivalent forms. 3. Note the similarity between the equation for average rate between two times and the formula for the slope of a line on a graph: m = slope = rise = run change in y change in x = Δy Δx = y2 ─ y1 x2 ─ x1 The equation for the average rate between two times matches the form of a slope formula in which the x values are always time values. The average rate of a reaction is the slope of a line between two points on a graph that represent the concentrations plotted at y1 and y2 at times t1 and t2 plotted on the x-axis. ©2011 ChemReview.net v. 2z Page 750 Module 27 — Kinetics: Rate Laws Practice A 1. For the reaction A + 2B 2C, the average rate of reaction of A is ─1.50 x 10─3 M • s─1 during the first 2.00 minutes of the reaction. If [A]= 0.240 M after 2.00 minutes, what was the initial [A]? Rates of Disappearance versus Appearance If the balanced equation for a reaction is known, and the rate at which any one of the reactants is used up or products form is known, all of the rates of disappearance of reactants and appearance of products can easily be calculated. The coefficients of a balanced equation supply the simple whole-number ratios needed for these calculations. Try applying that rule to the following example. Q. The Haber process uses nitrogen gas (isolated from air) and hydrogen gas (which can be produced from water) plus catalysts and energy to produce ammonia. Ammonia can be converted to fertilizers that improve food production. The Haber process reaction is: N2(g) + 3 H2(g) 2 NH3(g) At a time when the nitrogen gas is reacting at a rate of 0.060 M/s, a. What is the rate of reaction of the hydrogen gas? b. What is the rate of ammonia formation? Solve, then check your answer below. ***** a. For every one N2 molecule used up, three H2 must be used up, so the rate of change in hydrogen gas concentration is: (─ 0.060 N2 M/s) x (3 H2/N2) = ─ 0.180 H2 M/s. Note that though a rate of disappearance can be labeled as positive, what is measured rate equations is a rate of change in concentration, so that in calculations, a rate at which substances are used up must be assigned a negative sign. Try Part B. ***** b. For every one nitrogen molecule used up, two ammonia molecules must be formed. The rate of ammonia formation must be 0.060 M/s x 2 = + 0.120 M/second . Practice B: Time [N2O5] 0 0.200 M 1.0 min. 0.141 M 2.0 min. 0.100 M Measurements of [N2O5] are recorded at the right. 3.0 min. 0.0707 M a. Calculate the average rate of reaction of N2O5 during the first three minutes of the reaction. 4.0 min. 0.0500 M Check your answer after each part. 1. The gas dintrogen pentoxide decomposes to form nitrogen dioxide gas and oxygen gas. The balanced equation is 2 N2 O 5 4 NO2 + 1 O2 ©2011 ChemReview.net v. 2z Page 751 Module 27 — Kinetics: Rate Laws b. Calculate the average rate of oxygen gas formation during the first three minutes of the reaction. c. Calculate the average rate of NO2 formation between 2.0 and 4.0 minutes. ANSWERS Practice A 1. WANTED: Initial [A] , which in symbols is [A]0 , the concentration of A at time = 0. The problem provides the average rate, and the average rate equation is the only one we know so far. . Average Rate = Δ[A] = [A]2 ─ [A]1 = ─ 1.50 x 10─3 M • s─1 Δt t 2 ─ t1 DATA: [A]2 = 0.240 M [A]1 = [A]0 = ? t2 = 2.00 min. = 120. seconds t1 = 0 s The rate constant uses seconds, but the data includes minutes. In equations, the units must be consistent. The easier conversion is to seconds. If needed, adjust your work and complete the calculation. ***** SOLVING in symbols first: ? = [A]0 = [A]1 = ─ { ( ─1.50 x 10─3 M • s─1) ( t2 ─ t1 ) ─ [A]2 } = = ─ { ( ─ 1.50 x 10─3 M • s─1) (120 s ) ─ 0.240 M } = ─ ( ─ 0.180 M ─ 0.240 M ) = 0.420 M = [A]0 Practice B 1a. 1b. Average Rate = Δ[N2O5] = [N2O5]2 ─ [N2O5]1 = 0.0707 M ─ 0.200 M = ─ 0.043 mol Δt t2 ─ t1 3.0 min. ─ 0 min. L • min. For every two N2O5 used up, one is O2 is formed. Average Rate of O2 formation = ─ 0.043 M /min. x 1/2 = + 0.022 M/min. 1c. The only data available during that time period is for N2O5 being used up, but if we know that average rate, we can use coefficients to calculate rates for the other reactants and products. Average Rate = Δ[N2O5] = [N2O5]2 ─ [N2O5]1 = 0.0500 M ─ 0.100 M = ─ 0.025 mol N2O5 Δt t 2 ─ t1 4.0 min. ─ 2.0 min. L • min. In the balanced equation, NO2 is being produced at double that rate. Substances being used up should be given a negative, and substances being formed a positive, rate of change in concentration. (─ 0.025 M/min N2O5 ) x 2 = + 0.050 M/min. NO2 ***** ©2011 ChemReview.net v. 2z Page 752 Module 27 — Kinetics: Rate Laws Lesson 27B: Rate Laws A rate law (also called a differential rate law) expresses how the rate of a reaction depends on the concentration of reacting particles. When studying the impact of a single reactant particle A on a reaction rate, the rate law equation is Rate = Δ[A] = k [A]n Δt where k is the rate constant and n is an exponent that is termed the order of the reactant. The rate constant k is a number with units that will always be the same when a given reaction is carried out at the same temperature. For most reactions, the rate constant will change significantly even with a small temperature change. Typically, an increase in temperature of 10˚C will roughly double a reaction rate. In a chemical reaction: A + B + C predicted by the rate law: D + E , the rate of the forward reaction will be Rate = k [A]n[B]o[C]p where the exponents n, o, and p are the order of each reactant. If each exponent and the rate constant k can be calculated, a complete rate law for the reaction can be written. To understand the mechanism of a chemical reaction, the first step is to “determine the form of the rate law” by determining the order (the exponent) for each [reactant] in the rate-law equation. Zero Order Reactants If the exponent in a rate equation is zero (n = 0) for a reactant, the reaction is termed zero order for that reactant, and the rate equation is: rate = k[A]0 = k (anything to the zero power equals one). If a reaction has only one reactant, and that reactant is zero order, the reaction proceeds at a constant rate: the rate does not depend on the concentration of the reactant. Reactions where the rate is determined by a catalyst are often zero order. First Order Reactants For a first-order reactant, n = 1, and rate = k[A]1 = k[A] . The rate of the reaction is directly proportional to the concentration of A: if the reactant concentration is doubled, the reaction rate is doubled. The decay of radioactive isotopes is one type of process with first-order kinetics. Second Order Reactants For a second-order reactant, n = 2, and rate = k[A]2 . If the [reactant] is doubled, the reaction rate will be four times as fast. ©2011 ChemReview.net v. 2z Page 753 Module 27 — Kinetics: Rate Laws Summary: Commit this chart to memory. Zero First Second Rate Law rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 When [A] doubles: rate stays same rate doubles rate quadruples Order In a rate law, the value of the exponents is not determined by the coefficients of the balanced equation. The order (the exponent) for each reactant must be determined by experiment. Three methods that can determine the order for a reactant are: • Varying the reactant concentration, and analyzing the initial rate of the reaction; • Arithmetic analysis of how the reactant concentration changes with time; and • Graphical analysis of how reactant concentration changes with time. Let’s consider these one at a time. Finding the Order: Concentration versus Initial Rate When determining the order of a reactant by varying its concentration, the initial rate of the reaction is measured: the rate just after the reaction begins, when very few of the products have been formed. Why measure the initial rate? In theory, all chemical reactions are reversible: a reaction that goes forward can go backward. In practice, some reactions go forward and backward easily; the reversible color change of an acid-base indicator is one example. Other reactions strongly tend to go only one direction: most “burning” goes strongly to completion. Analysis of reaction-rate data can be difficult if a reaction is going both forward and backward, but for a reaction to go backward, the products must exist. After reactants are first mixed, when not much of the products have been formed, the reaction is going forward far more than it is going backward. This simplifies kinetics calculations. Use the characteristics of zero-, first-, and second-order reactants to solve this question. Q. For the reaction: A data at the right, B , based on the a. determine the order of A. b. Write the rate law. Expt. [A] Initial Rate (mol/L•s) 1 0.20 M 1.3 x 10─4 2 0.40 M 2.6 x 10─4 ***** 3 0.60 M 3.9 x 10─4 Answer a. The data shows that in Experiment 2, [A] is doubled compared to Experiment 1, and the reaction rate doubled as well. Comparing Experiment 3 to Experiment 1, when [A] was tripled, the reaction rate tripled. This reaction fits the definition of “first order in A.” b. The rate law is: Rate = k[A], where the exponent of [A] is one. ©2011 ChemReview.net v. 2z Page 754 Module 27 — Kinetics: Rate Laws Try one more example. Q. For the reaction: C D , the data at the left is collected. a. Determine the order of C. b. Write the rate law. c. Using the results of Expt. 2 and the rate law, calculate a value for the rate constant (k). ***** Expt. [C] Initial Rate (mol / L • s) 1 0.10 M 1.0 x 10─4 2 0.20 M 4.0 x 10─4 3 0.40 M 16.0 x 10─4 Answer a. In Experiment 2, the [C] is doubled compared to Expt. 1, and the reaction rate quadrupled. Comparing Expt. 3 to Expt. 1, when [C] is four times higher, the reaction rate is 16 times higher. This reaction is “second order in C.” b. This behavior fits the rate law: rate = k[C]2 . If needed, adjust your work and then try Part C. ***** c. Since k is a constant in the rate law equation, data from any one of the three experiments, substituted into the rate law, should give the same value for k. Solve Rate = k[C]2 for k in symbols first, then substitute data from Expt. 2 in the chart above. ***** k = Rate = 4.0 x 10─4 (mol/L • s) = 4.0 x 10─4 mol ● L2 = [C]2 [0.20 mol/L]2 0.040 L · s mol2 1.0 x 10─2 L mol · s Recall that in simplifying the units, when a fractional unit is in the denominator, it helps to separate the fractional unit into a reciprocal, then invert the reciprocal (see Lesson 17C). The units of the answer above can be written in several equivalent ways, including = 1 = M─1 •s─1 = L •mol─1 •s─1 L mol · s M·s Concentration versus Initial Rate for More Than One Reactant In an overall chemical reaction: A + B + C will be predicted by the rate law equation: D + E , the rate of the forward reaction Rate = k [A]n[B]o[C]p The overall reaction order for the reaction is the sum of the reactant exponents. For a reaction with more than one reactant, it is relatively easy to determine the order of each reactant if, in a series of experiments, the concentration of each reactant is varied while the other reactant concentrations are held constant. ©2011 ChemReview.net v. 2z Page 755 Module 27 — Kinetics: Rate Laws This is a key principle of experimental design: when multiple variables affect an outcome, vary one and measure the impact on the second while holding the others constant. Try this problem. Q. For the reaction A + B at the right is collected. C, the data Expt. [A] [B] Initial Rate (mol / L • s) b. Write the overall rate law. 1 0.10 M 0.10 M 2.1 x 10─5 c. Find the overall reaction order. 2 0.20 M 0.10 M 8.0 x 10─5 3 0.20 M 0.20 M 16.1 x 10─5 a. Determine the order of A and B. d. Using the rate law and Expt. 1 data, find a numeric value and units for k. ***** Answers a. Comparing Experiment 2 to Experiment 1, [B] is held constant, while [A] doubles, and the reaction rate approximately quadruples. Allowing for experimental error, this indicates that the power of [A] is 2, and A is therefore a second-order reactant. Comparing Experiment 3 to Experiment 2, the [A] is held constant, while the [B] doubles, and the reaction rate approximately doubles. This fits the profile for B being a first-order reactant. b. What is the overall rate equation for this reaction? Rate = k[A]2[B]1 = k[A]2[B] . c. The “overall reaction order” for the reaction is the sum of the exponents: 2 + 1 = 3. The reaction is said to be “third-order overall.” d. Since k is constant in all of the experiments, you can use data from any one of the experiments to calculate k. Since Rate = k[A]2[B] , if we use Expt. 1 data: k= Rate = [A]2[B] 2.1 x 10─5 mol/L • s = 2.1 x 10─5 mol/L • s = 10─3 mol3/L3 (0.10 mol/L)2 (0.10 mol/L) = 2.1 x 10─2 mol ● L3 = 3 L•s mol 2.1 x 10─2 L2 mol2 • s ***** Relationships to Memorize 1. Average Rate = change in [A] = Δ[A] = [A]2 ─ [A]1 = slope between points change in time Δt t2 ─ t1 (t1, [A]1) and (t2, [A]2). 2. Zero First Second Rate Law: rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 If [A] doubles, the rate: Stays the same Doubles Quadruples Order: 3.. For reaction A + B + C D + E , the rate-law equation is: Rate = k [A]n[B]o[C]p 4. Overall reaction order = sum of the exponents in the rate law. ©2011 ChemReview.net v. 2z Page 756 Module 27 — Kinetics: Rate Laws 5. Knowing the rate of disappearance or appearance for any one of the reactants or products, coefficients will calculate the rates for the other reactants and products. Practice A: Finding Order From Concentration versus Initial Rate Answers are at the end of this lesson. 1. For the reaction X + Y Z , the data at the right is collected. a. Determine the order of X and Y. b. Write the overall rate equation. c. Find the overall order for the reaction. Expt. [X] [Y] Initial Rate (mol / L • s) 1 0.25 M 0.10 M 1.0 x 10─5 2 0.25 M 0.20 M 4.0 x 10─5 3 0.50 M 0.10 M 4.0 x 10─5 Expt. [A] [B] Initial Rate (mol / L • s) 1 0.20 M 0.50 M 3.5 x 10─5 2 0.40 M 1.0 M 1.4 x 10─4 3 0.40 M 0.50 M 1.4 x 10─4 d. Calculate a value for k. 2. For the reaction A + B C, the data at the right is collected. a. Determine the order of A and B. b. Write the overall rate equation. c. Find the overall order for the reaction. Finding the Order Using Rate, Time, and Half-Life The half-life of a reactant measures how much time is required for half of the reactant to be used up. In problems where the goal is simply to determine the order of a reactant from concentration versus time data, the order can often be determined by half-life analysis. Half Lives and Order If sufficient data is available, the order of a reactant can be determined by estimating the percentage of the original concentration remaining at a time equal to double the first half-life. • For zero-order reactants: after the first half-life, half of the reactant is used up. At a time equal to double the first half-life, all of the reactant is used up. A zero-order reactant is used up at a constant rate. • First-order reactants have a constant half-life: the time for the second half-life is the same as the first. For first-order reactants: after the first half-life, half of the reactant is used up, and half remains. At a time equal to double the first half-life, half of that remaining 1/2 is used up, and 1/4 of the original amount remains. After triple the first half-life, 1/2 of the 1/4, = 1/8th of the original amount remains. ©2011 ChemReview.net v. 2z Page 757 Module 27 — Kinetics: Rate Laws • For second-order reactants: after the first half-life, half of the reactant remains. At a time equal to double the first half-life, 1/3 of the reactant remains. At a time equal to triple the first half-life, 1/4 of the original reactant remains. For second-order reactants, the time to reduce 1/2 to 1/4 is twice as long as to reduce the original amount by half; the second half-life is twice as long as the first. Summary: memorize these relationships. 1 Order Zero First Second 2 Rate Law rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 3 If [A] doubles, the rate: Stays the same Doubles Quadruples 4 At double the first halflife, [A] remaining is: None 25% 33% To Use the “Double the First Half-Life” Method to Determine Reaction Order If sufficient data is available: 1. Estimate the first half-life: the time required for one-half of the original [reactant] to be used up. 2. Estimate what percentage of the original concentration is remaining at a time that is double the first half-life. 3. Use the “double the first half-life” rules to determine the order of the reactant. Let’s apply this method to a problem. Time [C] 0 1.00 M 10. s 0.71 M 20. s 0.50 M 30. s 0.35 M ***** 40. s 0.25 M Answer 50. s 0.17 M Q1. For the reaction: C D , time and [C] are measured as a reaction proceeds. Based on the rules above, a. estimate the first half-life of C in the reaction. b. Determine the order of reactant C in the rate law, The original concentration is cut in half after 20. seconds, so the first half-life of C is 20 s. Double the first half-life is 40 s. The [C] at 40 s is 1/4 (25%) of the original concentration. This behavior fits: rate = k[C]1 = k[C] . This reaction is first order in C. Try another example. Time [R] 0 0.200 M a. Estimate the first half-life of R in the reaction. 5.0 s 0.125 M b. Estimate the [R] after double the first half-life. 9.0 s 0.095 M c. Determine the order of reactant R, 14 s 0.074 M 18 s 0.063M 22 s 0.054 M Q2. For the reaction: R ***** ©2011 ChemReview.net v. 2z S , [R] and time are measured. Page 758 Module 27 — Kinetics: Rate Laws Answer a. The original concentration is cut in half, to 0.100 M, after about 8 s. b. Double the first half-life is about 16 s. The [R] at about 16 s is about 0.069 M. Is this zero, first, or second order? ***** c. Since 0.069 M/0.200 M = 0.345; about 34% of the original concentration remains at double the first half-life, which is close to 33%. This approximately fits the profile for a second-order reactant: rate = k[R]2 . In many problems, not enough data is provided to allow analysis at “double the first halflife,” and this method cannot be used. In those cases, however, the reaction order can often be determined by the graphical analysis methods that we will learn in the next lesson. Practice B: Order From Concentration, Time, and Half-life 1. For the reaction: X reaction proceeds. Y , [X] and time are measured as a Time [X] a. Estimate the first half-life of X in the reaction. 0 0.100 M 25 s 0.079 M b. Determine the order in X. 50. s 0.065 M c. Write the rate equation. 100. s 0.048 M d. Calculate the average rate of the reaction during its first 100 seconds. 150. s 0.038 M 200. s 0.032 M e. Which of these can be written as an answer unit for part d? After each, write YES or NO. 1) M s 2) M •s─1 3) mol •L─1 •s─1 2. For the reaction: D E , [D] and time are measured at the right. Write the rate-law expression. 3. For the reaction: A B, [A] and time are measured. Write the rate equation. [A] 0 0.100 M 1.0 hour 0.095 M 2.0 hours 0.090 M 3.0 hours mol L •s Time [D] 0 0.050 M 25 s 0.044 M 50. s 0.038 M 100. s 0.025 M 200. s 0M 0.085 M 4.0 hours ©2011 ChemReview.net v. 2z Time 4) 0.081 M Page 759 Module 27 — Kinetics: Rate Laws ANSWERS Practice A: Order From Concentration versus Rate 1 a. The rate of the forward reaction will be predicted by the rate equation: Initial Rate Expt. 0.25 M 0.10 M 1.0 x 10─5 2 0.25 M 0.20 M 4.0 x 10─5 3 To find the order for X, choose two experiments where the other variable [Y] is held constant while [X] is changed. In experiments 1 and 3, [Y] is held constant. [Y] 1 Rate = k [X]n[Y]p [X] 0.50 M 0.10 M 4.0 x 10─5 (mol / L • s) Comparing experiments 1 and 3, [X] is doubled, and the rate goes up four times. This indicates that the reaction is “second order in X;” the exponent of [X] in the rate equation is two. To find the order for Y, choose two experiments where the other variable [X] is held constant while [Y] is changed. In experiments 1 and 2, [X] is constant. Comparing experiments 1 and 2, [Y] is doubled, and the rate goes up four times. This indicates that the exponent for [Y] is 2, and the exponent of [Y] in the rate equation is two. b. The overall rate equation is: Rate = k [X]2[Y]2 c. The overall order for the reaction is the sum of the exponents: 2 + 2 = 4. d. Since k is constant, you can use the rate law with data from any one of the experiments to calculate k. Solving Rate = k [X]2[Y]2 for k, if we use data from Experiment 3: ?=k= Rate [X]2[Y]2 = 4.0 x 10─5 mol / L • s (5 x 10─1 mol / L)2 (1 x 10─1 mol / L)2 = 0.16 x 10─1 mol ● L4 L•s mol4 2 = 1.6 x 10─2 a. The rate of the forward reaction will be predicted by the rate equation: Rate = k [A]n[B]p To find the order for A, choose two experiments where the other variable (B) is held constant: experiments 1 and 3. = 4.0 x 10─5 mol / L • s = 25 x 10─4 mol4 / L4 L3 mol3 • s Initial Rate Expt. [A] [B] 1 0.20 M 0.50 M 3.5 x 10─5 2 0.40 M 1.0 M 1.4 x 10─4 (mol / L• s) To compare the rates more easily, rewrite 3 0.40 M 0.50 M 1.4 x 10─4 the values of the initial rates so that they all have the same exponential term. It is easiest if you give all of the values the same exponential term as the smallest exponential term in the series, and change the others (see Lesson 1B) to match the exponential of the smallest. If you needed that hint, try the problem again. ***** ©2011 ChemReview.net v. 2z Page 760 Module 27 — Kinetics: Rate Laws For experiments 2 and 3, the rate becomes 14 x 10─5. Comparing experiments 1 and 3, [A] is doubled, and the rate goes up 14/3.5 = 4 ; it quadruples. This indicates that the exponent of [A] in the rate equation is two (when the concentration goes up by a factor of 2, the rate goes up by a factor of 2 squared). The reaction is said to be “second order in A.” To find the order for B, choose two experiments where the other variable (A) is held constant: experiments 2 and 3. From experiment 3 to 2, [B] is doubled, but the rate stays the same. This indicates that the exponent for [B] is 0. b. The overall rate equation is: Rate = k [A]2[B]0 = k [A]2 c. The overall order for the reaction is the sum of the exponents: 2 + 0 = 2 . Practice B: Order From Concentration, Time, and Half-life 1. a. The initial concentration is reduced to one-half after about 95 seconds. b. Double the first half-life would be about 190 s. At that point, about 34% of the initial concentration remains (0.034/0.100). After double the first half-life, 0% remains if zero order, 25% if first order, and 33% if second order. The data fit second order in X. c. Rate = k [X]2 d. Average rate = change in [X] per unit of time. Set t1 as the lower t value (t = 0). Time [X] 0 0.100 M 25 s 0.079 M 50. s 0.065 M 100. s 0.048 M 150. s 0.038 M 200. s 0.032 M Average rate = Δ [X] = [X]2 ─ [X]1 = 0.048 M ─ 0.100 M = ─ 0.052 M = ─ 5.2 x 10─4 M Δt t2 ─ t1 100 s ─ 0 s 100 s s e. All are YES. All four forms are equivalent 2. a. The initial concentration has been cut in half after 100 seconds. Double the first half-life is 200 s. At 200 s , no D remains. This fits behavior that is “zero order in D.” Rate = k [D]0 = k Zero order in D would also mean a constant rate of change for [D]. Check: do the [D] numbers drop by about the same amount every 50 seconds? Time [D] 0 0.050 M 25 s 0.044 M 50. s 0.038 M 100. s 0.025 M 200. s 0M 3. Some problems will not offer enough data to get close to “double the first halflife.” This is an example. Though [A] seems to be falling at a constant rate, which would indicate zero order, zero and first order have similar rates initially. This problem does not provide enough data to use the “double the first half-life” method, and at this point the rate equation cannot be determined. ***** ©2011 ChemReview.net v. 2z Page 761 Module 27 — Kinetics: Rate Laws Lesson 27C: The Integrated Rate Law – Zero Order Prerequisites: Lessons 27A and 27B. In addition, you will need to know the rules for graphing in Graphing Module 20. ***** Two Forms of the Rate Law In a chemical reaction: A + B + C predicted by the rate equation: D + E , the rate of the forward reaction will be Rate = k [A]n[B]o[C]p In the previous lesson, the rate constant and the order of the reactants was determined by measurements of the concentration of each reactant versus • the initial rate of the reaction, and • the concentration at a time that is double the first half-life. A third method of determining the rate law is to graphically analyze the measurements of concentration versus time. The rate law above uses concentration and rate as variables. To calculate a rate law graphically, the rate laws are stated using concentration and time as variables. To begin, commit the following chart to memory. Rate Law Summary Zero First Second 2 Differential Rate Law rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 3 If [A] doubles, the rate: Stays the same Doubles Quadruples 4 At double the first halflife, [A] remaining is: None 25% 33% [A]t = ─kt + [A]0 ln[A]t = ─kt + ln[A]0 1 = +kt + 1 [A]t [A]0 1 Order 5 Integrated Rate Law In the chart: • The Row 2 equations are usually called the rate laws, but the more proper term, to distinguish them from Row 5, is the differential rate laws. In calculus, the instantaneous rate is defined in terms of the differential d[A]/dt . • Row 5 is termed the integrated rate laws because those laws can be obtained by integrating (using calculus) the form of the rate law in Row 2. • In the differential rate laws, concentration and rate are variables. In the integrated rate laws concentration and time are variables. • The ln in the first-order integrated rate law represents the natural logarithm; and • [A]0 represents the initial concentration of a reactant, the concentration at time = 0. ©2011 ChemReview.net v. 2z Page 762 Module 27 — Kinetics: Rate Laws The two rate equations in each column are mathematically equivalent: the differential and integrated rate laws for each order are the same relationship stated in two different ways. Since the Row 2 equations are more simple, why bother with Row 5? Row 5 has one advantage: all three of the Row 5 equations are in the form y = mx + b , which is the general equation for a line. (The zero- and first-order differential-rate laws can be matched to y = mx + b, but the second order rate law is in the form y = mx2, which is algebraically quite different. The form y = mx + b fits all three of the integrated rate laws.) Let’s briefly review the rules for the equation for a line. Analyzing Linear Data When values of x and y are graphed in Cartesian (x,y) coordinates, if the points fall on a straight line, the data will fit the equation y = mx + b . In this equation, • y and x are variables, and m and b are constants. • m is the slope of the line. A line has a constant slope; m will have the same value between all points on the line. If m is positive, the slope of the line is up (/); if m is negative the slope is down (\). • To calculate m, memorize: • b is the y-intercept: the value of y when x = 0, which is the value of y when the line crosses the y-axis. b will be positive if a line crosses the y-axis above the origin of the graph and negative if the line crosses below the origin. • Once the value of the slope is calculated, b can be calculated by substituting into y = mx +b for any data point on the line. • While m and b have constant values in the equation for a line, y and x can change. If a value is chosen for either y or x, and the values for the constants m and b are known, the equation y = mx + b can be solved for the other variable. • If the values for any three of the terms y, m, x, and b are known, the equation can be solved for an unknown fourth term. m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δ x x2 ─ x1 Practice A: Answers are at the end of this lesson. If you are unsure about an answer, check after each part. If you need additional help with this topic, see Module 20. 1. If graphed data fits the equation: y = mx + b a. What ratio will be constant between any two points? b. What formula is used to calculate slope? ©2011 ChemReview.net v. 2z Page 763 Module 27 — Kinetics: Rate Laws 2. Using the data at the right, with ˚C values on the x-axis, ˚F ˚C ─40. ─40. b. Calculate the slope between any other two points. 32 0 c. Are the slope results consistent with the equation y = mx + b ? Why or why not? 68 20. 122 50. 212 100. a. Calculate the slope between any two points. Include units. d. Substituting the variable symbols in the data, write the equation for the line. e. What is the numeric value and unit of m ? f. Using the equation, the m value, and any x and y value in the data table, calculate the y-intercept. g. Write the equation for the line, substituting both the symbols for the y- and x-axis variables and the numeric values and units for the two constants. h. Test the equation. Pick a ˚C value from the table that you have not used in your calculations. Plug that ˚C value into your Step g equation and calculate a ˚F. Compare the calculated ˚F to the actual ˚F in the table at that ˚C. i. Starting from the data table, what were you able to accomplish by the steps above? The Zero-Order Integrated Rate Law Compare the zero-order integrated rate law: to the equation for a line: [A]t = ─k t + [A]0 y= mx + b The two equations have the same form: • Like y and x, [A]t and t are variables: quantities that are varied to study the rate. • [A]0 is constant: the [A] at the start of the experiment (when t = 0). • ─k is a constant. Its value is minus the rate constant for the reaction. If data for [A] and t obey the zero-order rate law above, then • A graph of [A]t versus t must be a straight line; and • A value for k can be calculated from the slope of the graph of [A]t versus t . • The slope of the line (a constant) must be equal to minus the rate constant k; Once k is known at the temperature conditions of the reaction, a complete rate law, including values for the constants, can be written. Whenever the reaction is run at that temperature, that rate law can then be used to predict [A] at any stated t, or to predict the time when [A] will reach a certain value. This fulfills a central purpose of science: to develop equations that accurately predict results of a process under a variety of conditions. ©2011 ChemReview.net v. 2z Page 764 Module 27 — Kinetics: Rate Laws Calculating the Slope to Find the Rate Constant To practice using rate laws, let’s go back to data from the previous lesson that we thought might be zero order and apply graphical analysis. Q. For the reaction: D E , measurements of [D] and time are recorded at the right. Time [D] 0 0.050 M 25 s 0.044 M 50. s 0.038 M 100. s 0.025 M 150. s 0.013 M Do the following steps. a. Assuming the reaction is zero order in D, write the differential and integrated forms of the rate law for this reactant. ***** Differential Rate Law: rate = k[D]0 = k Integrated Rate Law: [D]t = ─kt + [D]0 b. Match the symbols in the zero-order integrated rate law to the symbols in the equation for a line. y= m= x= b= ***** The zero-order integrated rate law: The equation for a line: [D]t = ─k t + [D]0 y = mx+ b y = [D]t = the variable concentration of D at a variable time t m = ─k = a constant = minus the rate constant x= t = time, a variable b = [D]0 = 0.050 M = the concentration of D initially, at t = 0 . c. What data plotted on y and x will result in a straight line with a slope = ─k ? ***** When y = mx+b , a graph of y vs. x will be a straight line with a slope m. For this data, plotting [D] on y and t on x should result in a straight line with a slope = m = ─k d. Graph the data in the table above that should result in a straight line. Either use a graphing program or use your own graph paper and the “graphing by hand” rules in Module 20. ***** ©2011 ChemReview.net v. 2z Page 765 Module 27 — Kinetics: Rate Laws Your graph should look similar to the graph at the right. [D] versus time 0.05 e. Calculate the slope between two widely separated points on the line. 0.04 [D] in mol/L The straight line means that the data fits the behavior expected for a zero-order reaction. 0.03 0.02 0.01 0 0 50 100 150 seconds ***** The calculation below uses the two points at the left and right sides of the plot. However, the value for the slope should be the same no matter which two points on the line are used to calculate the slope. On a straight line, the slope between any two points is the same: a constant. Setting x1 = t1 = 0 s (the lower time), y1 will then = 0.050 M. m = slope = rise = Δ y = y2 ─ y1 = 0.013 M ─ 0.050 M = ─0.037 M = run Δ x x2 ─ x1 150 s ─ 0 s 150 s = ─ 2.5 x 10―4 M/s f. The graph is consistent with [D]t = ─kt + [D]0 ; the integrated rate law equation for a zero order reactant. If the values of the two constants ─k and are [D]0 are known, the integrated rate law can be used to make predictions about [D] and t. Find the numeric value and units of the rate constant in the integrated rate law that fits this data. ***** The integrated rate law is: [D]t = ─kt + [D]0 and the equation for a line is: m = ─k , so k = ─m y = mx + b = + 2.5 x 10―4 mol • L─1 • s─1 g. Write the value of [D]0 that can be substituted into the integrated rate law for the data above. ***** [D]0 is the [D] in the data at time = 0 seconds, so [D]0 = 0.050 M h. Write the integrated rate law, substituting the numeric values of the two constants found above in place of the symbols for the two constants. ***** ©2011 ChemReview.net v. 2z Page 766 Module 27 — Kinetics: Rate Laws [D]t = ─kt + [D]0 ; [D]t = ─ ( + 2.5 x 10―4 mol/L • s)( t ) + 0.050 M g. Test the equation: plug in data for a time in the data table that was not used to calculate the slope. See if the equation predicts the [D] in the data at that time. To match the answer below, use t = 50 s. ***** Use [D]t = ─ ( 2.5 x 10―4 mol/L • s)( t ) + 0.050 M . At t = 50 s, = ─ ( 2.5 x 10―4 mol/L • s)(50 s) + 0.050 M = ─ (12.5 x 10―3 mol/L) + 0.050 M = ─ ( 0.0125 mol/L) + 0.050 M = 0.0375 M = 0.038 M which agrees with the data in the table at t = 50. s. The integrated rate law, with its constants added, correctly predicts the experimental result. ***** Summary: Use these steps to calculate the constants for a zero-order-rate equation from experimental data. 1. Write the zero-order integrated rate law: [A]t = ─k t + [A]0 and, under it, the equation for a line: y= mx + b 2. Graph the values of y and x in the data table for the experiment. If the data points fall close to a straight line, the data fit the prediction for a reactant that is zero order in A. 3. Calculate the slope for two widely separated points on the graph. m = Δy/Δx = Δ[A] / Δt 3. Write the equation that explains and predicts the data. a. Use the slope (m) to write a value for k. b. Write the zero-order integrated rate law substituting numeric values in place of the symbols for the two constants: k and [A]0. c. Test the rate law: pick an unused time from the data table, plug that time into the rate law, and see if the equation predicts the [A] that is in the data at that time t. d. If yes, the data is consistent with a zero-order reactant. If no, and if the graph was a line, check your values for k and [A]0. ***** ©2011 ChemReview.net v. 2z Page 767 Module 27 — Kinetics: Rate Laws Practice B. Use the steps above for these calculations. 1. For the data at the right, Time [A] 0 0.500 M 1.0 min. 0.460 M b. Does the data fit the prediction for zero-order in A? Why or why not? 2.5 min. 0.400 M c. Calculate the slope between two widely spaced points of data. 3.5 min. 0.360 M 5.0 min. 0.300 M 7.0 min. 0.220 M a. Graph the data, putting the independent variable (see Lesson 19C if needed) on the x-axis. d. Write a rate law that fits the data. Include values and units for the two constants in the rate law. e. Test your rate law: see if it correctly predicts [A] at t = 3.5 min. 2. If experimental data fits the linear equation ln[A] = ─kt + ln[A]0 a. What values would be plotted on the y-axis, and what on the x-axis, to produce the graph of a line? b. What term in the equation above would be the y-intercept? c. What ratio would be constant between any two points? d. What is the relationship between k and the slope? ANSWERS Practice A 1a. If y = mx + b, constant ratio? The slope: Δy / Δx m = slope = rise = change in y = Δ y = y2 ─ y1 run change in x Δx x2 ─ x1 m = slope = rise = Δy = Δ˚F = run Δx Δ˚C 2 ˚F2 ─ ˚F1 ˚F ˚C2 ─ ˚C1 a. Using the 2nd and last points, setting x1 as the lower x number, m= ˚F2 ─ ˚F1 = ( 212 ─ 32 ) ˚F = ˚C2 ─ ˚C1 ( 100 ─ 0 ) ˚C b. All slopes should calculate to the same result. 1.80 ˚F/˚C = m ˚C ─40. ─40. 32 0 68 20. 122 50. 212 100. c. y = mx +b is the equation for a line, and a line has a constant slope, so two equal slopes are consistent with the equation for a line. d. Equation for line: ©2011 ChemReview.net v. 2z ˚F = (m)( ˚C) + b Page 768 Module 27 — Kinetics: Rate Laws e. See Part A answer: m = 1.80 ˚F/˚C Since y = mx + b , if we choose the easy case, in which x = ˚C = 0 ˚C , y = ˚F = 32 ˚F, f. Substituting into the rate law: 32 ˚F = (m)( 0 ) + b , so 32 ˚F = b . g. y = mx +b , ˚F = (m)(˚C) + b ; ˚F = ( 1.80 ˚F/˚C ) ˚C + 32 ˚F h. If we choose 50 ˚C, ˚F = ( 1.80 ˚F/ ˚C ) ( 50 ˚C ) + 32 ˚F = 90˚F + 32 ˚F = 122 ˚F In the data table, 50˚C = 122˚F . The equation correctly predicted the ˚F result. i. You were able to derive an equation which explains the numbers in the table and predicts results for measurements not in the table. In science, that’s considered “cool.” The above relationship is generally remembered as ˚F = 9/5 (˚C ) + 32 which matches the equation for the DATA developed above. Practice B The independent (more controlled) variable in this experiment is the time at which the measurements were taken. Your results should be similar to the graph at the right. The points fall close to a straight line. 1b. 1c. If the points on a graph of [A] versus time fall close to a straight line, the data fits zero order behavior. [A] versus time 0.5 0.4 [A] in mol/L 1a. 0.3 0.2 0.1 0 0 1 2 3 4 minutes 5 6 7 8 If you use any two widely spaced points, your answer should be close to the slope found below. If the points chosen are at the left and right sides of the graph above, At x1 = t1 = 0 minutes (the lower time), y1 = 0.500 M Estimating that at x2 = t2 = 8 minutes, y2 ≈ 0.18 M. m = rise = Δ y = y2 ─ y1 = 0.18 M ─ 0.500 M = ─0.32 M = 0.040 M/min. = run Δ x x2 ─ x1 (8.0 ─ 0) min. 8.0 min. = ─ 4.0 x 10―2 M/min. 1d. Since the slope is constant, • the data is consistent with the equation y = m x + b , and since y = [A] and x = time for the graph, • the data is consistent with the zero-order rate law: [A]t = ─k t + [A]0 where k (the rate constant) = ─m = minus the slope; so k = + 4.0 x 10―2 M/min , and [A]0 = the [A] in the data at t = 0, which is 0.500 M,. Since [A]t = ─k t + [A]0 ; ©2011 ChemReview.net v. 2z Page 769 Module 27 — Kinetics: Rate Laws the proposed specific rate law is [A]t = ─ ( + 4.0 x 10―2 M/min)( t ) + 0.500 M [A]t = ─ ( + 4.0 x 10―2 M/min)( t ) + 0.500 M = 1e. = ─ ( + 4.0 x 10―2 M/min)( 3.5 min. ) + 0.500 M = = ─ 14 x 10―2 M + 0.500 M = ─ 0.14 M + 0.500 M = 0.36 M This calculated [A] matches the data at t = 3.5 minutes. The proposed rate law works. This equation is linear: 2a. ln[A]t = ─kt + ln[A]0 because it matches the form y = mx + b For linear data, the plot of y versus x is a line, so for the specific equation above, ln[A] vs. t as a line. plots b = ln[A]0 2b. The y-intercept? 2c. Constant for a line, in general: Δy / Δx ; For this specific equation: m = slope = rise = Δy = Δln[A] = ln[A]2 ─ ln[A]1 run Δx Δt t2 ─ t1 k = ─m 2d. (or m = ─ k ) ***** Lesson 27D: Base 10 Logarithms 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 that base 10 logarithm calculations are encountered in science classes. Pretest: If you think you know this topic, try the last 4 calculations in Practice C at the end of this 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 review the rules for powers and logarithms. As always, on the problems below, cover the answers below the * * * * * line and write your answer the questions that are above the line. I. Numbers, Bases, and Exponents Any positive value that is written in fixed decimal notation (as a regular number) can be expressed as a number (a base) to a power. ©2011 ChemReview.net v. 2z Page 770 Module 27 — Kinetics: Rate Laws For example: Computer science often calculates in base 2, such as 1,024 = 210 , and (fill in the blank) 24 = _______. ***** 24 = 16. However, values can be represented by any base to any power. Using your calculator, convert this base and power to a fixed decimal number: 3.52.7 = ________ (Use the calculator that you will use for quizzes and tests.) ***** • 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… A primary goal of science is to discover relationships that explain how our universe works. Those relationships are often most simply expressed as equations that include base e. Calculations involving these relationships may be done in both base 10 and base e, but 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 Numeric values can be expressed by a variety of methods: as fixed decimal numbers, as exponential notation, or as a number to a power. In scientific calculations, it is often useful to express numeric values as 10 to a power, where the power of 10 can be either an integer or a number with decimals. For example: 102 = the fixed decimal number 100 and 103 = the number ___________. Without a calculator, 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 a 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 ©2011 ChemReview.net v. 2z Page 771 Module 27 — Kinetics: Rate Laws 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: “estimate, then calculate.” On sf/rounding: when converting between numbers and exponentials or logarithms, the statistical justification for significant figures breaks down. We will add a systematic rule when we study acid-base pH. Until then, we will use the rules: • • Q. If an exponent is a whole number, assume math rules and exact numbers, if an exponent has a decimal, round answers to 3 significant figures. Use the estimation logic above and your calculator key sequence to write either numbers or numbers in scientific notation for the problems below. If you are unsure about an answer, check it below before doing the next part. a. 103.9 = (estimate): _____________________ (calculate): ________________ b. 1021.7798 = (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. 1021.7798 = (estimate): 1 x 1021 < 1021.7798 < 1022 = 10 x 1021 so the number will be between 1 x 1021 and 10.0 x 1021. On the calculator: 6.02 x 1021 ~8.0 x 1021 ?? 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. 103.7 = (in scientific notation): _______________________ (expos ± one?)___ d. 10─9.7 = (in scientific notation): _____________________ (expos ± one?) ____ e. 10─13.2 = (in scientific notation): ________________________ (expos ± 1 ?)___ ***** c. 103.7 = 5,010 or 5.01 x 103 . (expos ± one? 3.7 and 3 = √ ) 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 ©2011 ChemReview.net v. 2z 3.7 +/- +/- 10x 10x Page 772 Module 27 — Kinetics: Rate Laws d. 10─9.7 = 2.00 x 10─10 (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 = Practice A: 6.31 x 10─14 (expos ± 1?) √ 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 the bases used most often. The symbol for a base 10 logarithm 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 2) 1000 3) 0.001 ***** ©2011 ChemReview.net v. 2z Page 773 Module 27 — Kinetics: Rate Laws 1) The log of 102 is 2. 3) log 0.001 = log 10─3 = ─3 2) log 1000 = log 103 = 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. c. For numbers greater than one, a log value will be a positive number. Positive numbers between 0 and 1 have 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 may not have a log button. You can learn a work-around (log x = ln x/2.303) or buy an inexpensive calculator with a log button. log Did the calculator agree with the answers done in your head? It must. d. In cases where you cannot solve 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 for 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 20 = (estimate): ______________________ ***** (on the calculator): ________ Log 10 = 1, and log 100 = 2 , so log 20 is between 1 and 2; about 1.2 ? 1.30 ©2011 ChemReview.net v. 2z Page 774 Module 27 — Kinetics: Rate Laws 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 ? ___) ***** 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 Commit to Memory 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 ? __) ©2011 ChemReview.net v. 2z Page 775 Module 27 — Kinetics: Rate Laws 6. Log 4.6 = ____________________________________ (expo ± 1 ? __) 7. Log 0.0020 = ____________________________________ (expo ± 1 ? __) 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, or taking the inverse log, 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 log of a number is 2, the number is: ___________ ***** 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 of numbers, write 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: write the log as a power of 10, then convert that exponential term to a number (or to scientific notation). 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. ©2011 ChemReview.net v. 2z Page 776 Module 27 — Kinetics: Rate Laws 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): _______ ***** 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 = 501,00,000 or 5.01 x 108 4) Log x = ─10.7 , x = 2.00 x 10─11 (expo ± 1 ? √ ) (expo ± 1 ? √ ) In cases like 3) and 4), your calculator may answer in 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 of a number, to find the number, calculate the value for 10log. This is called “taking the antilog” or “finding the inverse log.” 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 , = . 9. When you encounter log calculations, it helps to write: • log 10x = x ; the log of 100 is 2 ; and • 10log x = x ; “10 to the log x equals x.” ©2011 ChemReview.net v. 2z Page 777 Module 27 — Kinetics: Rate Laws 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 ? __) 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 (expos ± 1 ?) √ (expos ± 1 ?) √ 4. 10─11.7 = 2.00 x 10─12 5. 10─0.7 = 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 ) = 12.8 5. Log (6.8 x 10─12 ) = ─11.2 6. Log 4.6 = 0.663 7. Log 0.0020 = ─2.70 Practice C 1. Log x = 12.4 , x = 2.51 x 1012 2. Log x = ─5.9 , x = 1.26 x 10─6 3. Log x = ─0.25 , x = 0.562 = 5.62 x 10─1 4. Log x = 1.1 , antilog = 12.6 = 1.26 x 101 5. 10─3.3 = 5.01 x 10─4 6. Log (2.0 x 10─9 ) = ─8.70 7. Log 0.50 = ─0.301 ©2011 ChemReview.net v. 2z Page 778 Module 27 — Kinetics: Rate Laws Lesson 27E: Natural Log Calculations Prerequisites: Complete Lesson 27D on base 10 logs before this lesson. Timing: This lesson must be done before first-order-integrated rate law calculations, but it may be helpful at any time that base e logarithm calculations are encountered in science classes. Pretest: If you think you know this topic, try the last 4 calculations in the Practice at the end of this lesson. If you can do those calculations, skip the lesson. ***** I. Base e Calculations a. The Symbol e In mathematical and scientific equations, the lower-case e is an abbreviation for a number: 2.7182818… For calculations, the value e = 2.718 must be memorized. The number e has many interesting mathematical properties. It is important in science because it is found in many equations that predict natural phenomena. In these equations, e is the base for values expressed in exponential notation in the form ex, and e is termed the natural exponential. For example, 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 . For a constant rate of decay, this rate law 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…. Now find a value for e1 on your calculator, using 1 and the ex function. Write the key sequence that produces the answer 2.718... ***** • A standard TI-type calculator might use: 1 ex . • On an RPN scientific calculator, try: 1 enter ex . Use your calculator-key sequence to do these, and then check your answers below. 1) e2 = 2) e2.5 = 3) e─1 = 4) e─2.5 = (Because the statistical basis for significant figures does not apply to logarithmic calculations, we will use this general rule: during e and ln calculations, round numbers in answers to 3 significant figures.) ***** ©2011 ChemReview.net v. 2z Page 779 Module 27 — Kinetics: Rate Laws 1) e2 = 7.39 2) e2.5 = 12.2 Recall that to enter a negative number, you usually use a +/- key. 3) e─1 ( = 1/e = 1/2.718.. ) = 0.368 4) e─2.5 ( = 1/e2.5 ) = 0.0821 II. Calculations Using ln a. 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 by definition, 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 = ______ ***** 1) ln e0 = 0 2.) ln e1 = 1 3. ln e─4 = ________ 2) ln e1 = ______ 3) ln e─4 = ─4 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 do the same calculation on your calculator: ln(2.718) = ___________ ***** Is the calculator answer close to the mental arithmetic answer? It must be. 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? ___ ***** ©2011 ChemReview.net v. 2z Page 780 Module 27 — Kinetics: Rate Laws 3) ln(6.02 x 1023) = 54.8 2) ln 0.0050 = ─ 5.30 4) ln(19.29 x 10─15) = ─ 31.6 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. Practice A: Use the calculator you will use on quizzes and tests to do these. 1. e2.0 = 2. e─4.7 = 3. e─11 = 4. ln 42 = 5. ln 0.020 = 6. ln(9 x 105) = 7. ln(5.0 x 10─4) = 8. ln(10─4) = b. 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 value, 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. ©2011 ChemReview.net v. 2z Page 781 Module 27 — Kinetics: Rate Laws 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] = 0.0300 , [A] = e = (nbr. and unit): _________________ ***** 2) If ln = ─4.5 , number = e─4.5 = 0.0111 = 1.11 x 10─2 3) ln = 57.2 , number = e57.2 = 6.94 x 1024 4) If ln [A] = 0.0300 , [A] = e0.0300 = 1.03 M c. Units and Logarithms Note that in 4) above, the unit expected for a concentration has been added. From a strict mathematical perspective, logarithms cannot be taken of values with units, and logarithm values do not have units. All precisely stated scientific relationships obey these rules. However, some of these precise scientific equations can be very complex, or they involve quantities that are difficult to measure. In those cases, the equations we write in chemistry are often “shortcuts” which simplify the complex relationships in order to speed problem solving. When using these shortcut equations, the rules for dimensional homogeneity and unit cancellation may not apply, and special rules may be needed to assign units to answers. To make shortcut equations work, two of our rules will be When taking the logarithm of a value with units, write the result as a value without units. If a WANTED quantity is calculated using a logarithm value, convert all of the supplied units in the calculation to consistent units if needed, then add the appropriate consistent unit to the answer. The shortcut equations we will encounter frequently in upcoming lessons will involve taking a logarithm of a concentration in moles per liter. The rule will be: if a [x] is wanted, add moles/liter (M) to the answer. Apply that rule to the following problems. Write the answer as a fixed decimal number or in scientific notation. When in doubt, check answers as you go. ©2011 ChemReview.net v. 2z Page 782 Module 27 — Kinetics: Rate Laws 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 (add M) 8) ln(0.250 M) = ─ 1.39 (drop the unit) 7) [D] = 0.249 M 9) [A] = e─ 2.63 = 0.0721 M d. 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 = 1,620 If you get lost on a natural log calculation, a good strategy is to do a similar, simplified 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. e. Converting between base 10 and natural logs A general rule for the logarithm 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 ln(x) = 2.303 log(x) “The natural log of a number is always 2.303 times higher than the base 10 log.” Try using the above equation to solve this without a calculator: Q1. ln (10) = ***** ©2011 ChemReview.net v. 2z Page 783 Module 27 — Kinetics: Rate Laws A1. ln (10) = 2.303 log (10) = 2.303 log (101) = 2.303 (1) = 2.303 Check this answer by solving Q1 using your calculator. Try the following without a calculator in a manner similar to A1. Q2. ln (100) = Q3. ln (1) = Q3. ln (0.10) = ***** A2. ln (100) = 2.303 log (100) = 2.303 log (102) = 2.303 (2) = 4.606 A3. ln (1) = 2.303 log (1) = 2.303 log (100) = 2.303 (0) = 0 A4. ln (0.10) = 2.303 log (0.10) = 2.303 log (10─1) = 2.303 (─1) = ─ 2.303 Now check these answers by solving Q2-4 using your calculator. Summary: Add these rules to your in-memory 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 value, 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. ln(x) = 2.303 log(x) Practice B: Solve the odd-numbered problems. Use the evens for later review or additional practice. 1. e5.2 = 4. ln 1066 = 7. ln(14.92 x 10─6) = ©2011 ChemReview.net v. 2z 2. e─1.7 = 5. ln 0.0050 = 8. ln e6.2 = 3. e─20.75 = 6. ln(3 x 108) = 9. eln(─42) = Page 784 Module 27 — Kinetics: Rate Laws 10. If ln = ─6.8 , number = e = (number in sci. notation):___________________ 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 = 17. ln [B] = ─13.7 , [B] = 18. e ─11.7 = 19. ln(0.050 M) = 20. e ─0.693 = 21. If log(x) = 5.0 , ln(x) = 22. if ln(x) = 34.5 , log(x) = 23. If ln(x) = ─ ( 0.075 day─1)(4.0 days) ; x = ? 24. Given that ln[A]t = ( ─ 0.0173 s─1) ( t ) + 0.693 a. If t = 20. s, [A] = ? 25. b. If [A] = 0.710 M, t = ? ln[A]t = ( ─ 0.0241 yrs.─1 ) ( t ) ─ 4.61 Given that a. If [A] = 0.0025 M, t = ? b. If t = 28.8 years, [A] = ? ANSWERS Practice A 2. e─4.7 = 9.10 x 10─3 1. e2.0 = 7.39 5. ln 0.020 = ─3.91 6. 8. ln(10─4) = 3. e─11 = 1.67 x 10─5 ln(9 x 105) = 13.7 ln(1 x 10─4) = 4. ln 42 = 3.74 7. ln(5.0 x 10─4) = ─7.60 ─9.21 Practice B 1. e5.2 = 181 2. e─1.7 = 0.183 4. ln 1066 = 6.97 5. ln 0.0050 = 7. ln (14.92 x 10─6) = ─11.1 ─5.30 8. ln e6.2 = 6.2 3. e─20.75 = 9.74 x 10─10 6. ln (3 x 108) = 19.5 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 = 1780 12. If ln = ─12.5 , antilog = 3.73 x 10─6 13. If log [A] = ─9 , [A] = 10─9 M (When finding a concentration, add the unit) 14. If log x = 13.7 , x = 5.01 x 1013 15. Log A = ─13.7 , A= 2.00 x 10─14 17. ln [B] = ─13.7 , [B] = 1.12 x 10─6 M ©2011 ChemReview.net v. 2z 16. 10─11.7 = 2.00 x 10─12 18. e ─11.7 = 8.29 x 10─6 Page 785 Module 27 — Kinetics: Rate Laws 20. e ─0.693 = 0.500 19. ln(0.050 M) = ─3.00 (When taking a log, drop the unit) 21. If log(x) = 5.00 , ln(x) = ? ln(x) = 2.303 log(x) ; ln(x) = (2.303)(5.00) = 11.5 22. If ln(x) = 34.5 , log(x) = ? ln(x) = 2.303 log(x) ; ; log(x) = 34.5/2.303 = 15.0 23. If ln(x) = ─ ( 0.075 day─1)(4.0 days) ; x = ? ln(x) = ─ 0.300 (day─1)(days) = ─ 0.300 day0 = ─ 0.300 (1) = ─ 0.300 x = eln(x) = e─0.300 = 0.741 24. Given that 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 = 25. Given that = ─ 1.035 ─1 ─ 0.0173 s 59.9 s = t = ( 1/ s─1 = (s─1) ─1 = s ) ln[A]t = ( ─ 0.0241 /yr. ) ( t ) ─ 4.61 a. Strategy: Solve for t in symbols first. t = ln[A] + 4.61 ─ 0.0241 /yr. = = ln[0.0025 M] + 4.61 = ─ 5.99 + 4.61 = ─ 0.0241 yr ─1 ─ 0.0241 yr ─1 ─ 1.38 = 57.3 years = t ─ 0.0241 yr ─1 b. Strategy: to find [A]t , first solve for ln[A]t ? = ln[A]t = ( ─ 0.0241 yr ─1 ) (28.8 yr. ) ─ 4.61 = ─ 0.694 ─ 4.61 = ─ 5.30 WANTED is [A]. Known is: ln[A] = ─ 5.30 ***** [A] = eln[A] = e─ 5.30 = 0.00500 M Solve for [A]. If a [ ] is wanted, add M. ***** ©2011 ChemReview.net v. 2z Page 786 Module 27 — Kinetics: Rate Laws Lesson 27F: Integrated Rate Law – First Order Prerequisites: Complete Lessons 27D and 27E before this lesson. ***** For first-order reactants, one form that is written for 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/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 /day = 0.181 day─1 = k DATA: ( for the unit: 1/x ≡ x─1 ) 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] ***** ©2011 ChemReview.net v. 2z Page 787 Module 27 — Kinetics: Rate Laws ? = ln [Rn] = ─kt + ln [Rn]0 = ─ (0.181 day─1)(14 days) + ln(2.5 x 10─4 M) = ─ 2.53 ─ 8.29 = ─ 10.82 = ln [Rn]14 days (Math help? See Lesson 27E, section II.) If needed, finish solving for the WANTED unit. ***** WANT: [Rn]after 14 days = eln [Rn] = e(─ 10.82) = 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) are generally expressed as as “1/time” units, such as “sec─1.” However, in solving first-order rate equations, solving for concentration may not produce a unit as part of the answer. This is because the form we use for the first-order rate equation is a “shortcut.” The terms ln[A]t and ln[A]0 are easy to calculate values for, but taking a logarithm of a value with units is not legal and is one indication that our equation is a simplification of a more complex relationship. To solve first-order rate calculations, our rule will be: 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 (or the unit proportional to concentration) used elsewhere in the problem. If no unit for concentration is supplied, assume the unit is moles/liter. ©2011 ChemReview.net v. 2z Page 788 Module 27 — Kinetics: Rate Laws Practice A 1. The earth’s atmosphere has a small amount of carbon dioxide that contains 14C (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 the remains of the plant, 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. 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. ©2011 ChemReview.net v. 2z [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 789 Module 27 — Kinetics: Rate Laws 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] 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 ln[A] = ln(2.00 M) = 0.693 = 0.693 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. ***** ©2011 ChemReview.net v. 2z Page 790 Module 27 — Kinetics: Rate Laws 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= ©2011 ChemReview.net v. 2z Page 791 Module 27 — Kinetics: Rate Laws 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. ***** ©2011 ChemReview.net v. 2z Page 792 Module 27 — Kinetics: Rate Laws ? = 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: _________________________________________ ©2011 ChemReview.net v. 2z Page 793 Module 27 — Kinetics: Rate Laws 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 = (+ 23.14 ─ 23.72) ─k ─ 1.21 x 10─4 yr─1 ©2011 ChemReview.net v. 2z = ─ 0.58 yr. = 4,800 yr. ─4 ─ 1.21 x 10 Page 794 Module 27 — Kinetics: Rate Laws 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 consistent concentration unit 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. ©2011 ChemReview.net v. 2z Page 795 Module 27 — Kinetics: Rate Laws 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 units proportional to the molarity 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 ***** ©2011 ChemReview.net v. 2z Page 796 Module 27 — Kinetics: Rate Laws Lesson 27G: Reciprocal Math Timing: This lesson should be done before calculations using the second-order integrated rate law. Pretest: If you think you know this topic, try the last lettered part of each calculation in Practice B at the end of the lesson. If you can do those calculations, skip the lesson. Prerequisites: Complete Lesson 17C (Complex Unit Cancellation) before this lesson. ***** Rules For Reciprocals Second-order rate laws will require work with reciprocals of both numbers and units. Some rules for reciprocals are as follows. 1. The reciprocal of X is 1/X. Three equivalent ways to write the reciprocal of X are: 1 = 1/X = X─1 X 1 = 1/X3 = X─3 Also: X3 2. The reciprocal of 1/X is X. The reciprocal of the reciprocal of X is X. In equation format: 1 = 1/(1/X) = (X─1)─1 = X 1 X 3. On a calculator, to convert a reciprocal to a number, either a. divide the denominator into one, Example: 1/8 = (1 divided by 8) = 0.125 b. or, take the reciprocal of the denominator using the reciprocal 1/x or x─1 key. Example: to convert 1/8 to a number, input 8 1/x or x─1 . Try it. 4. If you know the value of the reciprocal of a number, and you want the number, take the reciprocal of the reciprocal value. Example: Since 1/(1/x) = x if 1/A = 0.25, A = 1/0.25 = 4 Practice A. Learn the rules, then apply them below. If you are unsure of an answer, check at the end of the lesson before doing the next problem. Doing every other problem today, and the rest tomorrow, will help in remembering the rules. 1. Trying these without a calculator, convert these reciprocals to numbers. a. 1/(3/4) = b. 1/(1/8.2) = c. 1/e─1 = 2. Using a calculator as needed, convert these reciprocals to numbers. a. d. 1/7.9 = (0.75)─1 = ©2011 ChemReview.net v. 2z b. 1/0.40 = e. 1/e3 = c. (2,000)─1 = f. (1/6.2)─1 = Page 797 Module 27 — Kinetics: Rate Laws 3. If 1/A = 0.0625, A = 4. If 1/D = 12.5, D = 5. Convert these to numbers in scientific notation. Use a calculator. a. 1/(4.7 x 103) = b. 1/(9.2 x 10─11) = c. (2.5 x 10─2)─1 = d. (1/(2.5 x 10─2))─1 = More Reciprocal Rules 5. To take the reciprocal of a unit of measurement, change the sign of its exponent. 1/(mol ● s─1) = mol─1 ● s 1/s = 1/s1 = s─1 Examples: 6. To take the reciprocal of a number to a power, change the sign of the power. Examples: 1/1010 = 10─10 1/10─10 = 1010 1/e─6 = e6 7. To multiply exponentials, add the exponents. To take an exponential to a power, multiply the exponents. Examples: (104)3 = 1012 s ● s─1 = s0 = 1 (e─6)─1 = e6 Practice B. If you are unsure of an answer, check the answer at the end of the lesson before doing the next problem. Do the math for both the number and its units. Do every other part, and more if you need more practice. 1. Convert these to numbers without denominators. Do not use a calculator. a. 1/102 = d. (3─2)─1 = b. 1/10─10 = c. 1/log(100) = e. 1/(4─2) = 2. Convert these to numbers in scientific notation. Do not use a calculator. a. 1/(5.0 x 105) = b. 1/(5.0 x 10─5) = Simplifying Reciprocals 8. To take the reciprocal of a fraction, invert the fraction. 1 = 1/(B/C) = (B ● C─1)─1 = B─1 ● C = C/B B C 9. When a term has two fraction lines ( either _____ or / ), separate the terms that are fractions. To do so, apply these steps in this order (from Lesson 17C). a. If a term has a fraction in the denominator, separate the terms into a reciprocal of the fraction ( 1/fraction in the denominator ) multiplied by the remaining terms. b. If there is a fraction in the numerator, separate that fraction from the other terms in the numerator or denominator. ©2011 ChemReview.net v. 2z Page 798 Module 27 — Kinetics: Rate Laws c. To simplify the fraction, invert any reciprocal fractions, cancel units that cancel, and multiply the terms. Practice C. If you are unsure of an answer, check the answer at the end of the lesson before doing the next problem. Do the math for both the number and its units. Do every other part, and more if you need more practice. 1. Write the following without a denominator by using positive and negative exponents. a. b. 1/(mol ● s─2) = mol L ● sec. c. (meters/s2) ─1 = e. d. (meters/s─2) ─1 = f. 1/s─1 = = 1 mol L●s 2. Write these units as simple fractions, with a numerator and denominator, in which all of the exponents are positive (for additional review, see Lesson 17C). b. 1/(sec./L) b. 1 mol c. M─1 M─1 ● s─1 L●s 3. Convert mol/L to M for molarity, and then write these units without a denominator. a. 1/(mol/L) b. 1 mol L●s 4. In these, convert M to moles/liter, then express the unit without a denominator by using positive and negative exponents. a. M● s─1 = b. 1/(M ● s─1) = c. M─1 s d. (M/s─1)─1 = = 5. Simplify: convert to a number and unit on one line, without a denominator. a. 1/(8.00 mol) b. 1/(0.25 mol/L) c. (0.450 M─1•s─1 ) ( 25.0 s ) d. (0.500 M─1•s─1 )( 15.0 s) + 1.20 M─1 6. Solve 1/[A] = +kt + 1/[A]0 7. Given this equation: for a. k b. t c. 1/[A]0 d. [A] 1/[A] = (0.250 M─1•s─1 ) ( t ) + 2.00 M─1 and t = 30.0 s, solve for [A]. 8. Given the same equation as in problem 7, if [A] = 0.174 M, solve for t. ©2011 ChemReview.net v. 2z Page 799 Module 27 — Kinetics: Rate Laws ANSWERS Practice A There are many ways to do these calculations. Use any legal methods that get the same answer. 1. a. 1/(3/4) = 4/3 = 1.33 2. a. 1/7.9 = 0.13 d. (0.75)─1 = 1.33 b. 1/(1/8.2) = 8.2 c. 1/e─1 = (e─1)─1 = e1 = 2.718… c. (2,000)─1 = 0.0005 or 5 x 10─4 b. 1/0.40 = 2.5 f. (1/6.2)─1 = 1/(1/6.2) = 6.2 e . 1/e3 = 1/20.09 = 0.0498 2. If 1/A = 0.0625, A = 1/(1/A) = 1/0.0625 = 16.0 4. a. 1/(4.7 x 103) = 2.1 x 10─4 [ keys might be: 4.7 E or EE 3 1/x ] b. 1/(9.2 x 10─11) = 1.1 x 1010 c. (2.5 x 10─2)─1 = 40. 3. If 1/D = 12.5, D = 1/12.5 = 0.0800 [ keys might be: 9.2 E or EE 11 +/- 1/x ] [ enter the number, then, to take a ─1 power, use 1/x ] d. (1/2.5 x 10─2)─1 = [(2.5 x 10─2) ─1]─1 = (2.5 x 10─2) +1 = 2.5 x 10─2 Practice B 1. a. 1/102 = 10─2 b. 1/10─10 = 10+10 d. (3─2)─1 = 32 = 9 c. 1/log(100) = 1/2 = 0.5 e. 1/(4─2) = (4─2)─1 = 42 = 16 2. a. 1/(5.0 x 105) = 1/5.0 x 1/105 = 0.20 x 10─5 = 2.0 x 10─6 b. 1/(5.0 x 10─5) = 1/5.0 x 1/10─5 = 0.20 x 105 = 2.0 x 104 Practice C 1. a. mol = mol ● L─1 ● sec─1 L ● sec b. 1/(mol ●s─2) = (mol ●s─2)─1 = mol─1 ● s2 c. (meters/s2)─1 = (meters ● s─2)─1 = meters─1 ● s2 d. (meters/s─2)─1 = (meters ● s2)─1 = meters─1 ● s─2 e. 1 mol L●s = 1/( mol ● L─1 ● s─1) = mol─1 ● L ● s f. 1/s─1 = (s─1) ─1 = s 2. One way to do these is to use the rule: to take the reciprocal of a fraction, invert the fraction. a. 1/(sec./L) = L sec. b. 1 mol = L●s mol c. = 1/s─1 = s M─1 M─1 ● s─1 L●s 3. a. 1/(mol/L) = 1/M = M─1 ©2011 ChemReview.net v. 2z b. 1 = mol L●s 1 M s = (M ● s─1) ─1 = M─1 ● s Page 800 Module 27 — Kinetics: Rate Laws M● s─1 = mol ● s─1 = mol ● L─1 ● s─1 L b. 1/(M ● s─1) = (mol ● L─1 ● s─1) ─1 = 4. a. M─1 s c. = (mol ● L─1) ─1 ● s─1 = mol─1 ● L ● s─1 d. (M/s─1) ─1 = (mol ● L─1 ● s) ─1 = 5. a. 1/(8.00 mol) = 0.125 mol─1 mol─1 ● L ● s mol─1 ● L ● s─1 b. 1/(0.25 mol/L) = (1/0.25)( 1/(mol ● L─1) = 4.0 mol─1 ● L c. (0.450 M─1 •s─1 ) ( 25.0 s ) = 11.2 M─ 1 d. (0.500 M─1 • s─1 )( 15.0 s ) + 1.20 M─1 = 7.50 M─1 + 1.20 M─1 = 8.70 M─1 b. t = 1/[A] ─ 1/[A]0 k 6. a. k = 1/[A] ─ 1/[A]0 t c. 1/[A]0 = [A] ─1 ─ kt d. [A] = ( kt + 1/[A]0 )─1 1/[A] = (0.250 M─ 1•sec─ 1 ) ( t ) + 2.00 M─ 1 and t = 30.0 s, solve for [A]. 7. Given Strategy: ***** Equation: DATA: Since the equation solves for 1/[A], first solve the equation for 1/[A], then take the reciprocal of 1/[A] to find [A]. 1/[A] = ( 0.250 M─1 • sec─1 ) ( t ) + 2.00 M─1 Use the equation’s symbols for variables to make the data table. 1/[A] = ? t = 30.0 s Solve: ? = 1/[A] = (0.250 M─1 • s─1 )( 30.0 s ) + 2.00 M─1 = = 7.50 M─1 + 2.00 M─1 = 9.50 M─ 1 = 1/[A] [A] = 1/ ( 1/[A] ) = 1/ (9.50 M─1 ) = 0.105 M = [A] 8. Given the same equation and [A] = 0.174 M, solve for t. Equation: 1/[A] = ( 0.250 M─1 • s─1 ) ( t ) + 2.00 M─1 DATA: Use the equation’s variables in the data table. 1/[A] = 1/(0.174 M) = 5.75 M─1 t=? t= 1/[A] ─ 2.00 M─1 ( 0.250 M─1 • s─1 ) Solve in symbols before substituting numbers. = 5.75 M─1 ─ 2.00 M─1 = ( 0.250 M─1 • s─1 ) 3.75 M─1 = 15.0 s ( 0.250 M─1 • s─1 ) ***** ©2011 ChemReview.net v. 2z Page 801 Module 27 — Kinetics: Rate Laws Lesson 27H: Integrated Rate Law – Second Order The following table summarizes the rate law equations used for zero-, first-, and secondorder reactants. You need to be able to write these 9 rows from memory. ***** Summary: Rate Laws 1 Order Zero First Second 2 (Differential) Rate Law rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 3 If [A] doubles, the rate: Stays the same Doubles Quadruples 4 At double the first halflife, [A] remaining is: None 25% 33% 5 Integrated Rate Law [A] = ─kt + [A]0 ln[A] = ─kt + ln[A]0 1 = +kt + 1 [A] [A]0 6 To graph a line, plot: [A] on y, t on x ln[A] on y, t on x 1/[A] on y, t on x 7 This slope is constant between any 2 points: Δ[A] / Δt Δ ln[A] / Δt Δ 1/[A] / Δt 8 Rate constant (k) = Minus the slope Minus the slope The slope 9 Half-life and k t1/2 = [A]0 2k t1/2 = ─(ln ½) = 0.693 k k t1/2 = 1 k[A]0 To do the problems below, begin by memorizing the table’s first 8 rows for second-order reactants. Practice until you can write those 8 rows from memory. Then, cover below the * * * * * lines below with a cover sheet and answer the questions above the line. 1. Write two forms of the rate law for a reaction A B that is second-order in A. ***** Second-order differential rate law: Rate = k [A]2 Second-order integrated rate law: 1/[A] = +kt + 1/[A]0 2. Compare the second-order integrated rate law: to the equation for a line on a graph: 1/[A] = +kt + 1/[A]0 y = mx + b Write the symbols in the second-order integrated rate law next to the matching symbols in the equation for a line. y= m= x= b= ***** ©2011 ChemReview.net v. 2z Page 802 Module 27 — Kinetics: Rate Laws y = 1/[A] m = +k x= t b = 1/[A]0 3. Fill in the blanks on these. For a reaction that is second order in A, a. What ratio that uses [A] and time will be constant? _____________________ b. What will be the y-intercept term in the integrated rate law? __________ c. If [A] versus time data is collected, i. Will a plot of [A] versus time have points on a line? ________ ii. What plot will produce points on a line? _________________________ ***** a. Δ 1/[A] /Δt b. 1/[A]0 c. i. No ii. 1/[A] versus t To learn the method for graphical analysis of second-order data, try this example. Q. For the reaction: D measured. E , [D] and time are a. Based on the data at the right, estimate the first half-life of D in the reaction. b. Estimate [D] after double the first half-life; c. Determine the order of reactant D. d. Write the differential rate law. Time [D] 0 0.400 M 10.0 s 0.250 M 18.0 s 0.192 M 28.0 s 0.149 M 36.0 s 0.127 M 44.0 s 1/[D] 0.110 M ***** a. The original concentration is cut in half, to 0.200 M, after about 16 s. b. Double the first half-life is about 32 s. [D] at about 32 s is about 0.130 M. c. Since 0.130 M/0.400 M = 0.32; about 32% of the original concentration remains at double the first half-life, which is close to 33%. This fits the profile for a second-order reactant. d. rate = k[D]2 e. If this data represents a second-order reactant, what ratio using [D] and t should be constant? ***** e. The ratio Δ (1/[D])/Δt , which is the slope component in the second–order integrated rate law, should be constant. f. Using a calculator, calculate values for 1/[D] in the table above. Enter the results in the last column, then check your answers below. ***** ©2011 ChemReview.net v. 2z Page 803 Module 27 — Kinetics: Rate Laws f. Time [D] 1/[D] 0 0.400 M 2.50 M─1 10.0 s 0.250 M 4.00 M─1 18.0 s Sample calculations: 0.192 M 5.21 M─1 28.0 s 0.149 M 6.70 M─1 36.0 s 0.127 M 7.87 M─1 44.0 s 0.110 M 9.10 M─1 For 1/[D]: If [D] = 0.400 M, 1/[D] = 1/0.400 M = 2.50 M─1 ***** Your values should match those at the right. g. On the two grids below, graph the data: first [D] vs. time, then 1/[D] vs. time. 1/[D] versus time [D] versus time 8 1/[D ] in L/m ol 10 0.4 [D ] in m o l/L 0.5 0.3 0.2 0.1 0 6 4 2 0 0 10 20 30 seconds 40 50 0 10 20 30 seconds 40 50 h. For which graph is the data closer to linear behavior? i. For the graph that is linear, calculate the slope of the line between two widely spaced points. j. Based on the graph, does the data better fit the behavior of a reaction that is zero order in D, or second order in D? Check your answers below. ***** h. The second graph should be very close to linear behavior (see next page). i. For slope between 0 and 50 s, at t1 = the lowest time = 0 s, 1/[D] = 2.50 M─1 At t2 = 50 s, 1/[D] ≈ 10.0 M─1 m = slope = Δ 1/[D] = 1/[D]2 ─ 1/[D]1 = (10.0 ─ 2.50) M─1 = 0.150 M─1 Δt t 2 ─ t1 (50.0 ─ 0) s s For real experimental data, the slope between points will vary, and the slope of the “best line” among the points will be a judgment call with uncertainty. ©2011 ChemReview.net v. 2z Page 804 Module 27 — Kinetics: Rate Laws [D] versus time 1/[D] versus time 0.5 10 1 /[D ] in L /m o l [ D ] i n m o l/L 0.4 0.3 0.2 0.1 0 8 6 4 2 0 0 j. 10 20 30 seconds 40 50 0 10 20 30 seconds 40 50 When a graph is linear, the equation for the line is y-axis variable = (constant slope)(x-axis variable) + y-intercept Since the first graph above is not linear, but the second is, the data in the table fits the equation 1/[D] = (constant slope)(time) + y-intercept The value of the y-intercept is the value of y when time = 0, which is represented in symbols as 1/[D]0 . The equation for the line above is therefore 1/[D] = (constant slope)(time) + 1/[D]0 which matches the form of the second-order integrated rate law: 1/[D] = +kt + 1/[D]0 For data that is second-order in D, the slope ratio Δ(1/[D])/Δt should be constant. In the second graph above it is. k. For this experiment, the numeric values and units for the constants are: k = ___________________________ and 1/[D]0 = _________________________ ***** k. For the above reaction, in symbols, the rate law is: Since the rate law is in the form for this data: 1/[D] = +kt + 1/[D]0 y = mx + b k = m = the constant slope = 0.150 M─1•s─1 from part i, and 1/[D]0 = 1[D] at 0 s = 1/0.400 M = 2.50 M─1 l. Write the specific rate law for this data, keeping the same variable symbols but substituting the values and units of the two constants. ***** ©2011 ChemReview.net v. 2z Page 805 Module 27 — Kinetics: Rate Laws 1. In symbols: 1/[D] = +kt + 1/[D]0 Substituting the constants, the rate law for this data is: 1/[D] = (0.150 M─1•s─1 ) ( t ) + 2.50 M─1 m. Test the rate law: choose a time value in the original data table that you did not use to calculate a slope. Plug that time into the rate law. See if the law, with your calculated constants, accurately predicts the [D] in the data at that time. (Use t = 36.0 s to match the answer below.) ***** m. Rate Law: 1/[D] = (0.150 M─1•s─1 ) ( t ) + 2.50 M─1 DATA: 1/[D] = ? t = 36.0 s WANTED: [D] Strategy: To find [D], use the rate law to find 1/[D], and then take the reciprocal to find [D]. If needed, adjust your work and finish. ***** ? = 1/[D] = (0.150 M─1•s─1 )( 36.0 s ) + 2.50 M─1 = [D] = 5.40 M─1 + 2.50 M─1 = 1/( 1/[D] ) = 1/(7.90 M─1 ) = 7.90 M─1 = 1/[D] 0.127 M = [D] Compare that answer to the original data table for t = 36.0 s. Does the rate equation predict the experimental result? n. Test your equation again: use the equation to calculate the time at which [A] will equal 0.192 M. ***** n. Rate Law: DATA: 1/[D] = +kt + 1/[D]0 [D] = 0.192 M 1/[D] = 1/(0.192 M) = 5.21 M─1 k = (0.150 M─1•s─1 ) t= ? 1/[D]0 = 1/0.400 M = 2.50 M─1 Strategy: Solve the equation for t, first in symbols, then plug in values. ***** t = 1/[D] ─ 1/[D]0 = 5.21 M─1 ─ 2.50 M─1 = k ©2011 ChemReview.net v. 2z k 2.71 M─1 = 0.150 M─1•s─1 18.1 s Page 806 Module 27 — Kinetics: Rate Laws Compare this calculated time to the time in the original table at [A] = 0.192 M. The answers agree within one doubtful digit. Once an integrated rate law with its calculated constants is known, the concentration of the reactant at any time, and the time required to reach any concentration, can be calculated. Practice: If you are unsure of the answer to any part, check it before doing the next part. Additional second-order calculations will be found in the next lesson. 1. In the reaction: A B , time and [A] are measured as a reaction proceeds. Time [A] a. Using the “double the first half-life” method, determine the order of the reaction. 0 0.500 M 60.0 s 0.444 M b. Write the differential and the integrated rate law that fits this data, in symbols. 150. s 0.381 M 225 s 0.340 M 450. s 0.255 M 900. s 0.174 M c. What two variables will need to be plotted to produce a line with a constant slope? d. Graph the data to determine the rate constant for the reaction. You may use the grid below, your own graph paper, or graphing software. e. Write the integrated rate law with values for the two constants. f. Calculate the predicted [A] at t = 1450 sec. g. How long will it take for [A] to equal 0.100 M? 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0 ©2011 ChemReview.net v. 2z 200 400 600 800 1000 Page 807 Module 27 — Kinetics: Rate Laws ANSWERS 1 a. The first half-life is about 460 s. Double the first half-life is about 920 s. At t = 920 s, [A] ≈ 0.170/0.500 = about 34% of the original concentration. This fits the behavior of second order in A. b. Second-order differential rate law: Second-order integrated rate law: Rate = k [A]2 1/[A] = +kt + 1/[A]0 c. Calculate values for 1/[A], Time [A] 1/[A] in L/mol (see values at right) 0 0.500 M 2.00 60.0 s 0.444 M 2.25 150. s 0.381 M 2.62 225 s 0.340 M 2.94 450. s 0.255 M 3.92 900. s 0.174 M 5.75 then graph 1/[A] versus t. d. For second order reactions, the slope of the graph of 1/[A] versus t is the value of the rate constant. ***** Using the second-order integrated rate law and the estimated data points from the graph at 0 s and 900 s, the rate constant is: 6.00 1/[A] in L/mol To find the rate constant, calculate the slope of the line. 1/[A] versus time 5.00 4.00 3.00 2.00 1.00 0.00 0 200 400 600 seconds 800 1000 k = m = Δ 1/[A] = 1/[A]2 ─ 1/[A]1 = (1/0.174 ─ 1/0.500) M─1 = 5.75 ─ 2 = 0.00416 M─1 Δt t 2 ─ t1 (900 ─ 0) s 900 s Your slope may differ slightly. Taking the slope of a line that is a “best fit” involves uncertainty. e. Integrated Rate Law: f. 1/[A] = (0.00416 M─1 • s─1 ) ( t ) + 2.00 M─1 To find [A], use the rate law to find 1/[A] and then take its reciprocal. 1/[A] = (0.00416 M─1 • s─1 )( 1450 s ) + 2.00 M─1 = 8.04 M─1 [A] = 1/(1/[A]) = 1/8.04 M─1 = 0.124 M g. Rate Law: 1/[A] = +kt + 1/[A]0 t = 1/[A] ─ 1/[A]0 = (1/0.100 ─ 2.00) M─1 = k k 8.00 M─1 = 1,920 s 0.00416 M─1 • s─1 ***** ©2011 ChemReview.net v. 2z Page 808 Module 27 — Kinetics: Rate Laws Lesson 27I: Half-Life Calculations Timing: IF you are asked to solve half-life calculations as part of your unit on kinetics, do this lesson now. You will also need to download and complete Lesson 39D as part of the following lesson. If you are not asked to solve calculations that include half-lives at this point in your course, do not do this lesson. IF you are asked to solve only radioactive decay half-life calculations at this point in your course, do not complete this lesson, but instead complete Lesson 39D. ***** Half-Lives: Simple Multiple Cases The half-life of a reactant (symbol t1/2) is the time required for a [reactant] to reach 50% of its initial concentration. After one half life, half of a sample remains and half is used up. Recall that for a • Zero-order reactant, at double the first half-life, all of the reactant is used up. • First-order reactant (including radioactive isotopes), the half-life is constant. o o • At double the first half-life, 1/4 of the original concentration remains; At triple the first half-life, 1/8 of the original concentration remains. Second-order reactant, o At double the first half-life, 1/3 of the original concentration remains; o At triple the first half-life, 1/4 of the original concentration remains. o Each successive half-life requires twice the time of the preceding half-life. Though the rate laws and the above rules are stated in terms of concentration, they will also be true for any measure that is proportional to concentration. If the reacting particles are in a sample that has a constant volume (which should be assumed unless other conditions are stated), these rules will apply to particle counts as well as concentration. For half-life calculations involving these easy multiples, problems can often be solved quickly using the rules above. These easy multiple rules can also be used to estimate answers as a check for problems that are not easy multiples. Practice A 1. The nucleus of carbon-14 undergoes first-order radioactive decay with a half-life of 5,730 years. In a sample of constant volume containing C-14 , a. After how many years will 75% of the original C-14 nuclei decay? b. After how many half-lives will the [C-14] be 1/16th of its original concentration? c. What percentage of the C-14 will be present after 3 half lives? ©2011 ChemReview.net v. 2z Page 809 Module 27 — Kinetics: Rate Laws 2. For a second-order reactant, if the first half-life is 25 seconds, a. How long is the second half-life? b. After how much time the reactant have one-third of its original concentration? c. What percentage of the original reactant concentration will remain after 175 s? 3. For zero-order reactants, a. if the first half-life for a reactant is 15 seconds, how much remains after 30. s? b. If all of a different reactant is used up after 80. seconds, what is its half-life? Half-Life Calculations for Non-Simple Multiples These equations relate half-life and rate constants. 1 Order Zero First 9 Half-life and k t1/2 = [A]0 2k t1/2 = ─(ln ½) = 0.693 k k Second t1/2 = 1 k[A]0 These formulas can be derived by substituting t1/2 for t and 1/2 [A]0 for [A]t in each integrated rate law, but it will likely speed your work to simply memorize them. It helps to note that all of these equations solve for t1/2 and have k in the denominator. If a half-life is known, these equations will find values for the rate constant k without calculating slopes. First-Order Reactants First-order reactants are a special case. Note that in the three half-life equations above, first-order reactants are the only type for which the half-life does not depend on the original concentration of the reactant. The integrated rate law for first-order reactants is ln[A]t = ─kt + ln[A]0 which can be re-written as ln[A]t ─ ln[A]0 = ─kt Using the rule that “the log of a quotient is the subtraction of the logs,” the second equation above can be written as ln [A]t = ─kt which can be written ln(fraction remaining) = ─kt [A]0 The four equations above are equivalent ways to write the first-order integrated rate law. First order rate calculations often involve fractions or percentages, and in such cases the fraction form above will be the most convenient to use. The above equations apply to any reactants that are first-order, but the decay of radioactive nuclei is the first-order process encountered most frequently. ©2011 ChemReview.net v. 2z Page 810 Module 27 — Kinetics: Rate Laws One difference between chemical reactions that are first order and radioactive decay (a nuclear reaction) is that for chemical reactions, the rate constant for a reactant will vary with temperature, but in radioactive decay, at the range of temperatures encountered in the earth’s atmosphere or interior, the rate constant remains constant: a characteristic of the particular isotope. However, other aspects of first order half-life calculations are the same for both chemical reactions and radioactive decay. Because calculations involving the decay of radioactive isotopes are the type of first-order kinetics encountered most often, in these lessons we cover the math of first-order in Lesson 39D as part of Nuclear Chemistry. If you are assigned calculations involving either radioactive decay or first-order half-lives for chemical reactions at this time, during the study of kinetics in your course, you should complete Lesson 39D now, then return to this point in these lessons. Zero- and Second-Order Reactants Zero- and second-order half-life calculations apply the two half-life equations: Zero-order half-life: t1/2 = [A]0/2k Second-order half-life: t1/2 = 1/(k[A]0) Try the following example in your notebook. Q. For a second-order reactant D, at 25°C, 50% of the original 0.180 M D remains after 3.00 minutes. Calculate a. The rate constant for the reaction. b. [D] after 9.0 minutes. c. The time required for [D] to reach 0.025 M. ***** Answer a. WANT: k Part A may be done in several ways, but the quickest is to use the half-life equation. By definition, the first half-life is the time when 50% of the original reactant remains. DATA: t1/2 = 3.00 min. The equation that relates k and t1/2 for second-order reactants is t1/2 = 1/(k[A]0) Solve in symbols for the WANTED k first, and then find a value for k. ***** k= 1 = 1 = t1/2 ([A]0) (3.00 min) (0.180 M) 1.85 M─1•min─1 ***** b. WANT: DATA: [D]9.0 min = [D]t t = 9.0 min ©2011 ChemReview.net v. 2z Page 811 Module 27 — Kinetics: Rate Laws The equation that uses t, k from Part a and [A]t for second-order reactants is the second-order integrated rate law. 1/[D]t = +kt + 1/[D]0 To find [D]t , find 1/[D]t and then take its reciprocal. ***** Substituting the constants: 1/[D]t = (1.85 M─ 1•min─1 )( t ) + (1/0.180) M─1 Substituting t: 1/[D]9.0 min = (1.85 M─1•min─1 )( 9.0 min ) + 5.56 M─1 = 16.7 M─1 + 5.56 M─1 = 22.3 M─1 = 1/[D]9.0 min [D]9.0 min = ? ***** [D]9.0 min = ? = 1/( 1/[D]9.0 min ) = 1/( 22.3 M─1 ) = 0.045 M Is this answer reasonable? Use what you know about half-lives to estimate the result. ***** Note that 9.0 minutes is triple the first half-life. For second-order kinetics, after triple the first half-life, 25% of the [original] remains. For [D], 25% of the original 0.180 M = 0.045 M. The “double the first half-life” rules can be a check or an alternate way to solve Part b. ***** c. WANT: DATA: t 0.025 M = [D]t The equation that uses [D] and t for this second-order reactant is In symbols: 1/[D]t = +kt + 1/[D]0 With constants: 1/[D]t = (1.85 M─1•min─1 )( t ) + 5.56 M─1 Solving for t: t = 1/[D] ─ 1/[D]0 = 40.0 M─1 ─ 5.56 M─1 = k Practice B: k 34.4 M─1 = ─1•min─1 1.85 M 19 min. If you are unsure of the answer to a part, check it before doing the next part. 1. For a reaction A D that is second order in A, k = 6.0 x 10─3 L•mol─1•s─1. a. If after 150. s, [A] = 0.200 M, what was the original [A]? b. Calculate the first half-life of A. ©2011 ChemReview.net v. 2z Page 812 Module 27 — Kinetics: Rate Laws 2. For a catalyzed reaction D E, a graph of [D] versus time results in a straight line with a slope of ─ 8.0 x 10─6 M/s. a. What is the order in D? b. Write the differential and integrated rate laws for the reaction. c. If the initial [D] = 0.250 M, calculate the half-life for the reaction. 3. By definition, the half life t1/2 is the time required to reach 1/2 [A]0 . Substitute t1/2 for t and 1/2 [A]0 for [A]t , then solve for t1/2 in the integrated rate laws for a. The zero order law b. The first order law c. The second order law Compare these answers to the equations in Row 9 of the Kinetics Rules chart. ANSWERS Practice A 1a. First-order half-life is constant. If 75% has decayed, 25% remains. Half remains after one half-life, half of that half (25%) remains after two half lives. Two half lives = 2 x 5,730 years = 11,460 years. 1b. Half remains after one half-life, 1/4th after two, 1/8th after three, 1/16th after four half-lives. 1c. Half remains after one half-life, 1/4th after two, 1/8th after three; 1/8 = 0.125 = 12.5% 2a. For second-order reactants, double the first: 50. sec. 2b. One-third of the [original] remains after double the first half-life: after 50 seconds of reaction. 2c. For second-order reactants, each successive half-life is double the preceding. If half remains after the first 25 sec, 1/4th remains 50 seconds after that, and 1/8th remains 100 seconds after that. 25+50+100 = after 175 seconds, 1/8th = 0.125 = 12.5% remains. 3a. None. 3b. 40 sec. Practice B 1a. WANT: DATA [A]0 150. s = t 6.0 x 10─3 L • mol─1 • s─1 = k 0.200 M = [A]t The second-order law that uses t , k , [A]0 and [A]t is: 1/[A]t = +kt + 1/[A]0 1/[A]0 = ─ kt + 1/[A]t Substituting : 1/[A]0 = ─ (6.0 x 10─3 L • mol─1 • s─1 )( 150. s ) + (1/0.200) M─ 1 = ─ 0.90 M─1 + 5.00 M─1 = 4.10 M─1 = 1/[A] Solving for the term with the WANTED symbol: 0 [A]0 = 1/( 1/[A]0 ) = 1/( 4.1 M─1 ) = 0.244 M ©2011 ChemReview.net v. 2z Page 813 Module 27 — Kinetics: Rate Laws 1b. WANTED: t1/2 k = 6.0 x 10─3 L • mol─1 • s─1 DATA: The second-order equation that includes t1/2 and k is t1/2 = 1/(k[A]0) . [A]0 = 0.244 M t1/2 = 1/(k[A]0) `= 1/{(6.0 x 10─3 L • mol─1 • s─1 )( 0.244 mol/L )} = = 1/( 1.464 x 10─3 s─1 ) = 683 s = t1/2 2a. If concentration vs. time is linear, the reaction is zero order. 4b. rate = k and [A]t = ─ kt + [A]0 2c. WANTED: t1/2 For zero-order half-life: t1/2 = [D]0/2k [D]0 = 0.250 M DATA: k = minus the slope of [D] vs. t = + 8.0 x 10─ 6 M/sec. t1/2 = [D]0/2k = SOLVE: 3a. [A]t = ─ kt + [A]0 0.250 M = 1.6 x 104 s 16 x 10─ 6 M/sec 3b. ln[A]t = ─ kt + ln[A]0 3c. 1/[A]t = + kt + 1/[A]0 1/2[A]0 = ─ k t1/2 + [A]0 ln (1/2[A]0 ) = ─ k t1/2 + ln[A]0 1/[(1/2[A]0) = +k t1/2 +1/[A]0 ─1/2[A]0 = ─ k t1/2 k t1/2 = ln[A]0 ─ ln (1/2[A]0 ) 2/[A]0 = +k t1/2 +1/[A]0 t1/2 = [A]0/2k k t1/2 = ln( [A]0 /1/2[A]0 ) 2/[A]0─1/[A]0 = +k t1/2 k t1/2 = ln(2) (2 ─1) /[A]0 = +k t1/2 t1/2 = 1/k/[A]0 t1/2 = 0.693/k ***** Summary: Kinetics 1. Average Reaction Rate = change in [A] change in time = Δ[A] = [A]2 ─ [A]1 Δt t2 ─ t1 2. If the rate of reaction of one component is known, the rates of appearance and disappearance of other reactants and products can be calculated from the coefficients. 3. The Rate Laws 1 Order Zero First Second 2 (Differential) Rate Law rate = k[A]0 = k rate = k[A]1 = k[A] rate = k[A]2 3 If [A] doubles, the rate: Stays the same Doubles Quadruples 4 At double the first halflife, [A] remaining is: None 25% 33% ©2011 ChemReview.net v. 2z Page 814 Module 27 — Kinetics: Rate Laws ln[A] = ─kt + ln[A]0 Integrated Rate Law 5 [A] = ─kt + [A]0 or ln [A]t [A]0 () = ─kt or ln(fraction) = ─kt 6 To graph a line, plot: 7 This slope is constant: 8 Rate constant (k) = 9 Half-life and k 1 = +kt + 1 [A] [A]0 [A] on y, t on x ln[A] on y, t on x 1/[A] on y, t on x Δ[A] / Δt Δ ln[A] / Δt Δ 1/[A] / Δt Minus the slope Minus the slope The slope t1/2 = [A]0 2k t1/2 = ─(ln ½) k t1/2 = 1 k[A]0 Log Rules 1. A logarithm is 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 number to a power, what is the power? 3. On a calculator, • The log button calculates the power of a number written as 10 to a power. • 1 0 x or 2 nd or INV log buttons convert from “10 to a power” to a number. 4. Checking log answers: when a number is written in scientific notation, its power of 10 must agree with its base 10 logarithm within ± 1. 5. The definition of a log is log 10x = x ; the log of 100 is 2 . 6. Knowing the log of a number, to find the number, calculate the value for 10log. This is called “taking the antilog” or “finding the inverse log.” 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. Knowing the log, to find the number, take the antilog. On a calculator, • input the log value, then press INV LOG ; or 2nd LOG ; or • Input the log value, then press 10x . A log is simply an exponent. 9. The symbol e is an abbreviation for a number that has special properties: 2.718… 10. The ln (natural log) function answers the question: if a number is written as e to a power, what is the power? 11. Knowing the ln, to find the number, take the antilog: • input the log, then press INV ln ; or • Input the log, then press ex . ©2011 ChemReview.net v. 2z An ln is simply an exponent of e. Page 815 Module 27 — Kinetics: Rate Laws 12. When you encounter log calculations, it helps to write: log 10x = x ln ex = x 13. and 10log x = x . “The log of 10 to the x is x; 10 to the log x is x.” and eln x = x . Write the base 10 rules, then substitute e and ln. 2.303 log(x) = ln(x) Reciprocal Rules 1. Three equivalent ways of representing the reciprocal of X are: 1 = 1/X = X─1 X 2. The reciprocal of 1/X is X . The reciprocal of the reciprocal of X is X . In equation format: 1 = 1/(1/X) = (X─1)─1 = X 1 X 3. To convert a reciprocal to a number, using a calculator, either divide the number into one or use the reciprocal 1/x or x─1 key. 4. If you know the value of the reciprocal of a number, and you want the number, take the reciprocal of the reciprocal value. 5. To take the reciprocal of a unit of measurement, change the sign of its exponent. 6. To take the reciprocal of a number to a power, change the sign of the power. 7. To multiply exponentials, add the exponents. To take an exponential to a power, multiply the exponents. 8. To take the reciprocal of a fraction, invert the fraction. 9. When a term has two fraction lines ( either _____ or / ), separate the terms that are fractions. To do so, apply these steps in this order. b. If a term has a fraction in the denominator, separate the terms into a reciprocal of the fraction ( 1/fraction in the denominator ) multiplied by the remaining terms. b. If there is a fraction in the numerator, separate that fraction from the other terms in the numerator or denominator. c. To simplify fractions, invert reciprocal fractions, cancel units, and multiply terms. 1 = 1/(B/C) = (B ● C─1)─1 = B─1 ● C = C/B B C ##### ©2011 ChemReview.net v. 2z Page 816 ...
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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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