Concise Physical Chemistry

c10 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

PROBLEMS AND EXAMPLES 163

Problem 10.5

The half-time of radioactive decay of^14 C is 5730 years. A wood sample from an

Egyptian tomb had^14 C radioactivity of 7. 3 ± 0 .1 cpm (counts per minute) per gram of

sample. Freshly harvested wood has a^14 C radioactivity of 12. 6 ± 0 .1 cpm per gram.

How old is the wood from the tomb and what is the uncertainty of your answer?

Problem 10.6

Given that

A=A 0 e−k^1 t

and

B(t)=k 1 A 0

1

k 1 −k 2

(

e−k^1 t−e−k^2 t

)

how does C vary with time for the reaction

A

k 1

→B

k 2

→C

Remember thatA+B+C=A 0 at all timest, and the initial conditions areA=A 0

andB=C=0.

Problem 10.7

Show that if

A 0 −A

A 0 A

=ktfor a second-order reaction, thenkt 12 A 0 =1.

Problem 10.8

A sequence of irreversible first-order chemical reactions was observed for 20 hours.

The sequence of reactants and products was

A

k 1 = 0. 6

︷︸︸︷

→ B

k 2 = 0. 7

︷︸︸︷

→ C

k 3 = 0. 06

︷︸︸︷

→ D...

The first three rate constants are 0.6, 0.7, and 0.06 h−^1 in the sequence given above,

and the rate constant for decrease in component D going to another product E was

0.02. Plot the concentrations of A, B, and C. The first two equations can be found in

Sections 10.1 and 10.3.2. The third equation is a rather intimidating:

C(t):= 0. 6 · 0. 7 ·

[ e− 0. 6 ·t

(0. 7 − 0 .6)·(0. 06 − 0 .6)

+ e

− 0. 7 ·t

(0. 6 − 0 .7)·(0. 06 − 0 .7)

+ e

− 0. 06 ·t

(0. 6 − 0 .06)·(0. 7 − 0 .06)

]