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In Example 20.7, we showed that there will be a "momentary" buildup of the intermediate product, 210PO. (a) Use the [B]t expression in equation 20.47 to derive an expression for the time it takes for the maximum amount of 210P0 to be present. Here's what to do: Take the derivative of the expression for [B]t with respect to time, set it equal to zero (because if the amount is at a maximum, the plot of the amount versus time has a slope of zero), and solve for time t. (b) Use this value for time and equations 20.47 to determine the specific amounts of 210BI, 210Po, and 206Pb when the amount of 210Po is at a maximum. Example 20.7 Kinetics of consecutive reactions are easily applicable to nuclear decay processes, in which a parent isotope produces a radioactive daughter isotope that also decays. (In fact, in the early twentieth century, such sequential processes were a major complicating factor in trying to understand this new phenomenon.) One such example is faa Bi Po Pb Which are the last two steps in the radioactive decay series starting with 238/92 U and ending in the nonradioactive isotope of Pb. (It is sometimes called the 4n + 2 series because all of the mass numbers of the isotopes involved can be represented by that general equation.) The half- lives, t1/2,1 and t1/2,2, are 5.01 days and 138.4 days, respectively. Comment on the relative amounts of 210Bİ, 210Po, and 206pb over time. Equations 20.47 [A], = [A]•e k, [A] [B], = k - k, [C), =

In Example 20.7, we showed that there will be a "momentary" buildup of the intermediate product, 210PO. (a) Use the [B]t expression in equation 20.47 to derive an expression for the time it takes for the maximum amount of 210P0 to be present. Here's what to do: Take the derivative of the expression for [B]t with respect to time, set it equal to zero (because if the amount is at a maximum, the plot of the amount versus time has a slope of zero), and solve for time t. (b) Use this value for time and equations 20.47 to determine the specific amounts of 210BI, 210Po, and 206Pb when the amount of 210Po is at a maximum. Example 20.7 Kinetics of consecutive reactions are easily applicable to nuclear decay processes, in which a parent isotope produces a radioactive daughter isotope that also decays. (In fact, in the early twentieth century, such sequential processes were a major complicating factor in trying to understand this new phenomenon.) One such example is faa Bi Po Pb Which are the last two steps in the radioactive decay series starting with 238/92 U and ending in the nonradioactive isotope of Pb. (It is sometimes called the 4n + 2 series because all of the mass numbers of the isotopes involved can be represented by that general equation.) The half- lives, t1/2,1 and t1/2,2, are 5.01 days and 138.4 days, respectively. Comment on the relative amounts of 210Bİ, 210Po, and 206pb over time. Equations 20.47 [A], = [A]•e k, [A] [B], = k - k, [C), =

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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In Example 20.7, we showed that there will be a "momentary" buildup of the intermediate product, 210PO. (a) Use the [B]t expression in equation 20.47 to derive an expression for the time it takes for the maximum amount of 210PO to be present. Here's what to do: Take the derivative of the expression for [B]t with respect to time, set it equal to zero (because if the amount is at a maximum, the plot of the amount versus time has a slope of zero), and solve for time t. (b) Use this value for time and equations 20.47 to determine the specific amounts of 210Bİ, 210Po, and 206pb when the amount of 210Po is at a maximum. Example 20.7 Kinetics of consecutive reactions are easily applicable to nuclear decay processes, in which a parent isotope produces a radioactive daughter isotope that also decays. (In fact, in the early twentieth century, such sequential processes were a major complicating factor in trying to understand this new phenomenon.) One such example is 21 Bi fa, 210Po fa, 20pb Which are the last two steps in the radioactive decay series starting with 238/92 U and ending in the nonradioactive isotope of Pb. (It is sometimes called the 4n + 2 series because all of the mass numbers of the isotopes involved can be represented by that general equation.) The half- lives, t1/2,1 and t1/2.2, are 5.01 days and 138.4 days, respectively. Comment on the relative amounts of 210Bİ, 210Po, and 206Pb over time. Equations 20.47 [A], = [A]•e k, [A], [B], = k - k, (e-t - ek4) (k, ekt - k, e k, - k,
Transcribed Image Text:In Example 20.7, we showed that there will be a "momentary" buildup of the intermediate product, 210PO. (a) Use the [B]t expression in equation 20.47 to derive an expression for the time it takes for the maximum amount of 210PO to be present. Here's what to do: Take the derivative of the expression for [B]t with respect to time, set it equal to zero (because if the amount is at a maximum, the plot of the amount versus time has a slope of zero), and solve for time t. (b) Use this value for time and equations 20.47 to determine the specific amounts of 210Bİ, 210Po, and 206pb when the amount of 210Po is at a maximum. Example 20.7 Kinetics of consecutive reactions are easily applicable to nuclear decay processes, in which a parent isotope produces a radioactive daughter isotope that also decays. (In fact, in the early twentieth century, such sequential processes were a major complicating factor in trying to understand this new phenomenon.) One such example is 21 Bi fa, 210Po fa, 20pb Which are the last two steps in the radioactive decay series starting with 238/92 U and ending in the nonradioactive isotope of Pb. (It is sometimes called the 4n + 2 series because all of the mass numbers of the isotopes involved can be represented by that general equation.) The half- lives, t1/2,1 and t1/2.2, are 5.01 days and 138.4 days, respectively. Comment on the relative amounts of 210Bİ, 210Po, and 206Pb over time. Equations 20.47 [A], = [A]•e k, [A], [B], = k - k, (e-t - ek4) (k, ekt - k, e k, - k,
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