Aleks 360 Access Card (1 Semester) For Chemistry: Atoms First
Aleks 360 Access Card (1 Semester) For Chemistry: Atoms First
2nd Edition
ISBN: 9781259207013
Author: Julia Burdge; Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.126QP
Interpretation Introduction

Interpretation:

The rate constant and half-life for the given radium decay and the activity after 500yr have to be determined.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

  • Half-life of a reaction is represented by the symbol as t12
  • Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
  • The half-life period is then shortened as half-life in early 1950s.
  • The formula which is used to calculate the half-life is t12=0.693k

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Expert Solution & Answer
Check Mark

Answer to Problem 19.126QP

The rate constant and half-life for the given radium decay and the activity after 500yr is determined.

The rate constant of the given reaction is k =1.4×10-11/s

The half-life of the given reaction is t12=5.0×1010s

The activity after 500yr is activity=3.1×1010disintigarastions/s

Explanation of Solution

First of all, calculate the number of radium nuclei in 1.0g

1.0g×1molRa226.03gRa×6.022×1023Ranuclei1 mol Ra=2.7×1021Ranuclei

From the given information and the above value we can easily calculate the rate constant and then from the rate constant we can calculate the half-life of the given reaction as follows.

The rate constant is:

activity = kN

k = activityN=3.70×1010nucleardisitegrations/s2.7×1021Ranuclei=1.4×10-11/s

k =1.4×10-11/s

The half-life is:

t12=0.693k=0.6931.4×10-11/s=5.0×1010s

t12=5.0×1010s

The activity after 500yr is:

Let us convert 500 years into seconds.  Then we can calculate the number of nuclei remaining after 500 years.

500yr×365days1yr×24h1day×3600s1h=1.58×1010s

Now, use the first order rate law for the calculation of the number of nuclei remaining after 500 years.

lnNtN0 = -kt

ln(Nt2.7×1021 )= -(1.4×10-1/s)(1.58×1010s)

Nt2.7×1021=e0.22

Nt=2.2×1021Ranuclei

Finally, after 500 years the number of nuclei remains and the rate constants, we can determine the activity as follows.

activity = kN

activity = (1.4×10-11/s)(2.2×1021Ranuclei)=3.1×1010disintigarastions/s

activity=3.1×1010disintigarastions/s

Conclusion

The rate constant and half-life for the given radium decay and the activity after 500yr were determined.

The rate constant of the given reaction was k =1.4×10-11/s

The half-life of the given reaction was t12=5.0×1010s

The activity after 500yr was activity=3.1×1010disintigarastions/s

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Chapter 19 Solutions

Aleks 360 Access Card (1 Semester) For Chemistry: Atoms First

Ch. 19.4 - The gas-phase reaction of nitric oxide with...Ch. 19.4 - Prob. 3PPACh. 19.4 - Prob. 3PPBCh. 19.4 - Prob. 3PPCCh. 19.4 - Prob. 19.4.1SRCh. 19.4 - Prob. 19.4.2SRCh. 19.4 - Prob. 19.4.3SRCh. 19.4 - Prob. 19.4.4SRCh. 19.4 - Prob. 19.4.5SRCh. 19.5 - Prob. 19.4WECh. 19.5 - Prob. 4PPACh. 19.5 - Prob. 4PPBCh. 19.5 - Prob. 4PPCCh. 19.5 - Prob. 19.5WECh. 19.5 - Prob. 5PPACh. 19.5 - Prob. 5PPBCh. 19.5 - Prob. 5PPCCh. 19.5 - Prob. 19.6WECh. 19.5 - Prob. 6PPACh. 19.5 - Calculate the rate constant for the first-order...Ch. 19.5 - Prob. 6PPCCh. 19.5 - Prob. 19.7WECh. 19.5 - The reaction 2A B is second order in A with a rate...Ch. 19.5 - Prob. 7PPBCh. 19.5 - Prob. 7PPCCh. 19.5 - Prob. 19.5.1SRCh. 19.5 - Prob. 19.5.2SRCh. 19.5 - Prob. 19.5.3SRCh. 19.5 - Prob. 19.5.4SRCh. 19.6 - Prob. 19.8WECh. 19.6 - Prob. 8PPACh. 19.6 - Prob. 8PPBCh. 19.6 - Prob. 8PPCCh. 19.6 - Prob. 19.9WECh. 19.6 - Prob. 9PPACh. 19.6 - Prob. 9PPBCh. 19.6 - Prob. 9PPCCh. 19.6 - Prob. 19.10WECh. 19.6 - Prob. 10PPACh. 19.6 - Prob. 10PPBCh. 19.6 - Prob. 10PPCCh. 19.6 - Prob. 19.6.1SRCh. 19.6 - Prob. 19.6.2SRCh. 19.7 - Prob. 19.11WECh. 19.7 - Prob. 11PPACh. 19.7 - Prob. 11PPBCh. 19.7 - Prob. 11PPCCh. 19.7 - Consider the gas-phase reaction of nitric oxide...Ch. 19.7 - Prob. 12PPACh. 19.7 - Prob. 12PPBCh. 19.7 - Prob. 12PPCCh. 19.7 - Prob. 19.7.1SRCh. 19.7 - Prob. 19.7.2SRCh. 19.7 - Prob. 19.7.3SRCh. 19.7 - Prob. 19.7.4SRCh. 19 - The rate of a reaction in which the reactant...Ch. 19 - The rate of a reaction in which the reactant...Ch. 19 - The rate of a reaction in which the reactant...Ch. 19 - Increasing the temperature of a reaction increases...Ch. 19 - Define activation energy. 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The...Ch. 19 - The decomposition of N2O to N2 and O2 is a...Ch. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - The integrated rate law for the zeroth-order...Ch. 19 - Prob. 19.90QPCh. 19 - Prob. 19.91QPCh. 19 - Prob. 19.92QPCh. 19 - The reaction of G2 with E2 to form 2EG is...Ch. 19 - Prob. 19.94QPCh. 19 - Prob. 19.95QPCh. 19 - Prob. 19.96QPCh. 19 - Strictly speaking, the rate law derived for the...Ch. 19 - Prob. 19.98QPCh. 19 - The decomposition of dinitrogen pentoxide has been...Ch. 19 - Prob. 19.100QPCh. 19 - Prob. 19.101QPCh. 19 - Prob. 19.102QPCh. 19 - To prevent brain damage, a standard procedure is...Ch. 19 - Prob. 19.104QPCh. 19 - Prob. 19.105QPCh. 19 - Prob. 19.106QPCh. 19 - Prob. 19.107QPCh. 19 - Prob. 19.108QPCh. 19 - Prob. 19.109QPCh. 19 - Prob. 19.110QPCh. 19 - (a) What can you deduce about the activation...Ch. 19 - Prob. 19.112QPCh. 19 - Prob. 19.113QPCh. 19 - Prob. 19.114QPCh. 19 - Prob. 19.115QPCh. 19 - Prob. 19.116QPCh. 19 - Prob. 19.117QPCh. 19 - Prob. 19.118QPCh. 19 - Prob. 19.119QPCh. 19 - Prob. 19.120QPCh. 19 - Prob. 19.121QPCh. 19 - Prob. 19.122QPCh. 19 - Consider the following potential energy profile...Ch. 19 - Prob. 19.124QPCh. 19 - Prob. 19.125QPCh. 19 - Prob. 19.126QPCh. 19 - Prob. 19.127QPCh. 19 - Prob. 19.128QPCh. 19 - The following expression shows the dependence of...Ch. 19 - Prob. 19.130QPCh. 19 - The rale constant for the gaseous reaction H2(g) +...Ch. 19 - Prob. 19.132QPCh. 19 - Prob. 19.133QPCh. 19 - At a certain elevated temperature, ammonia...Ch. 19 - Prob. 19.135QPCh. 19 - The rate of a reaction was followed by the...Ch. 19 - Prob. 19.137QPCh. 19 - Prob. 19.138QPCh. 19 - Prob. 19.1KSPCh. 19 - Prob. 19.2KSPCh. 19 - Prob. 19.3KSPCh. 19 - Prob. 19.4KSP
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