Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 20, Problem 20.56E
Interpretation Introduction

Interpretation:

The concentration of 210Bi, 210Po, and 206Pb at t=1.00×106s are to be determined using the half-lives from the reaction  83210Bik1t1/2,1 84210Pok2t1/2,2 82206Pb and equations 20.47.

Concept introduction:

The rate of a reaction is defined as the speed by which the reaction is proceeding. The rate of reaction depends on several factors such as the concentration of reactant and temperature.

Expert Solution & Answer
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Answer to Problem 20.56E

The concentration of 210Bi, 210Po, and 206Pb at t=1.00×106s are 0.201897M, 0.76970M and 0.02994M respectively.

Explanation of Solution

The reaction in Example 20.7 is shown below.

 83210Bik1t1/2,1 84210Pok2t1/2,2 82206Pb

The half-lives t1/2,1 and t1/2,2 are 5.01 days and 138.4 days, respectively.

Conversion of half-lives from days to seconds is shown below.

1day=60×60×24seconds…(1)

Substitute t1/2,1 in equation (1).

5.01days=5.01×60×60×24seconds=4.33×105seconds

Substitute t1/2,2 in equation (1).

138.4days=138.4×60×60×24seconds=1.196×107seconds

The half-lives, t1/2,1 and t1/2,2 are 4.33×105s and 1.196×107s, respectively.

The rate constant for first order reaction is given by the formula shown below.

k=0.693t1/2…(2)

Where,

t1/2 is the half-life.

Substitute t1/2,1 in equation (2).

k1=0.6934.33×105s=1.60×106s1

Substitute t1/2,2 in equation (2).

k2=0.6931.196×107s=5.79×108s1

The value of k1 and k2 are 1.60×106s1 and 5.79×108s1 respectively.

The Equation 20.47 is given as follows:

[A]t=[A]0ek1t[B]t=k1[A]0k2k1(ek1tek2t)[C]t=[A]0[1+1k1k2(k2ek1tk1ek2t)]

Where,

k1 and k2 are the equilibrium constant.

[ A ]t is the concentration of A at time t.

[ B ]t is the concentration of B at time t.

[ C ]t is the concentration of C at time t.

[ A ]0 is the initial concentration of A.

The above expression in terms of concentration of 210Bi, 210Po, and 206Pb at t=1.00×106s is written as follows.

[210Bi]t=[210Bi]0ek1t[210Po]t=k1[210Bi]0k2k1(ek1tek2t)[206Pb]t=[210Bi]0[1+1k1k2(k2ek1tk1ek2t)]…(3)

Substitute t=1.00×106s, k1=1.60×106s1 and [210Bi]0=1.000M in the first expression of equation (3).

[210Bi]t=(1.000M)e(1.60×106s1)(1.00×106s)=0.201897M

Substitute t=1.00×106s, k1=1.60×106s1, k2=5.79×108s1 and [210Bi]0=1.000M in second expression of equation (3).

[210Po]t=(1.60×106s1)(1.000M)(5.79×108s1)(1.60×106s1)(e(1.60×106s1)(1.00×106s)e(5.79×108s1)(1.00×106s))=(1.03755M)(0.2018970.943744)=(1.03755M)(0.74185)=0.76970M

Substitute t=1.00×106s, k1=1.60×106s1, k2=5.79×108s1 and [210Bi]0=1.000M in third expression of equation (3).

[206Pb]t=(1.000M)[1+1(1.60×106s1)(5.79×108s1)((5.79×108s1)e(1.60×106s1)(1.00×106s)(1.60×106s1)e(5.79×108s1)(1.00×106s) )]=(1.000M)[1+(0.648×106s)(1.169×108s11.509×106s1)]=(1.000M)[1+(0.648×106s)(1.497×106s1)]=0.02994M

Therefore, the concentration of 210Bi, 210Po, and 206Pb at t=1.00×106s are 0.201897M, 0.76970M and 0.02994M respectively.

Conclusion

The concentration of 210Bi, 210Po, and 206Pb at t=1.00×106s are 0.201897M, 0.76970M and 0.02994M respectively.

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Chapter 20 Solutions

Physical Chemistry

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