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), =
Physical Chemistry
2nd Edition
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Ball, David W. (david Warren), BAER, Tomas
Chapter20: Kinetics
Section: Chapter Questions
Problem 20.54E
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Question
![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,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F654cac59-b8d9-47e3-8bb2-0398ad7587b0%2F0893053c-eec4-4aa3-beac-a469d21f6ed1%2Ffq12hz_processed.jpeg&w=3840&q=75)
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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