Schaum's outline of theory and problems of physical chemistry by Clyde R. Metz

Author:Clyde R. Metz
Language: eng
Format: epub
Tags: Chemistry, Physical and theoretical -- Problems, exercises, etc.
ISBN: 4895015076
Publisher: McGraw-Hill
Published: 1989-09-05T16:00:00+00:00

40 60

f/(min)Fig. 12.7

12.15. How will the plot of the concentrations given by (12.26) look for A,> 100A2 and for 100A, < A2?

In both cases NA will decrease exponentially with a rate constant of A,; NB will essentially be givenby NA0 e~*2' and (A,NA0/A2) e-A|', respectively; and Nc will essentially be given by NAfi(l -e"A;') and^a.o(1 — *~A,')» respectively.

12.16. The half-life of 2^|Pb, which decays to 2^Bi, is 10.6 h. The half-life of 2gBi is 60.5 min. Describethe activity in a sample of 2g2Pb after equilibrium has been established.

The total activity of the sample will decay with a half-life equal to 10.6 h, and the ratio of parent todaughter will be given by (12.33) as

N, _ (0.693//,/2,2) - (0.693//,/2,,) = 60/60.5 -1/10.6N2 (0.693/f1/2>1) 1/10.6

256

RATES OF CHEMICAL REACTIONS

[CHAP. 12

12.17. What mass of 2jlRn having tl/2 = 54.5 s is equivalent to 1 mCi?Using (12.31) to determine A gives

0.693 , .

A = =1.27xl0"2s_l

54.5 s

The definition of a curie gives for the millicurie

dN , .

- = 3.7x 107s_1

dt .

Solving (12.27) for N and substituting values gives

3.700 x 107s-1

1.27xl0"2s"1

n = ; __ 2.9i xio9

which can be converted to mass by

2 91 x 109

-(220gmol"1) = 1.06 x 10"15 kg

6.022 xlO23 moK

12.18. A sample of wood from an Egyptian tomb gave a 14C activity per unit mass of 7.3 min~' g-1.What is the age of the wood?

Using (12.31) gives

0.693 . .

A= = 1.21xl0~4 yr-1

5 730 9

and using (12.30) gives

7.3

In -0.546 = -1.21 x 10-4r

12.6

which upon solving gives t = 4 510yr.

12.19. A current nuclear theory suggests that 235U/238U was nearly unity at the time of the formationof the elements. If the current ratio is 7.25 x 10~3, calculate the age of the elements.

Using (12.31) for the isotopes gives

0.693 m .

A,35 = : = 9.76 x 10"'° yr"1

-35 7.1 x 10s

and A238= 1.54 x 10_10yr_1. The exponential form of (12.28) gives for the current ratio7.25 x 10~3 = (1) exp[-(A23S - A238)f] = exp(-8.22 x 10_10f)

Solving for t gives

7.25xlO"J

t = -In s = 6.0 x 109 yr

8.22 xlO"10

Supplementary Problems

RATE EQUATIONS FOR SIMPLE REACTIONS12.20. For the reaction

H2(g) + I2(g) * 2HI(g)

RATES OF CHEMICAL REACTIONS

257

the rate of reaction has been found experimentally to be first-order wth respect to H2 and to I2. Write theexpression for the rate of reaction, and express the rates of change of the concentrations of the reactantsand product in terms of the rate of reaction. Could this be an elementary reaction?

Ans. ( V-'){d£/dt) = -dC(H2)/dt = -dC{\2)/dt =\dC{H\)/dt;

Although the reaction-order data suggest that this reaction could be elementary, other data suggestthat it is not.

12.21. Discuss the following overall reactions with respect to order, catalysts, etc.

(a) 2A > 4B + C, —r^=fcCA (b) A + B * C, —£=kCACB~l

dt dt

dC dC

(c) 2A+B * 2C + D + B, -=kCACB (d) 2A+B > 2C, —£=feCACc'/2

dt dt

(e) 2A + B > 2C, ^=kCA2CB (f) A + 2B ► C + D, ^£=kCACB2Cc

dt dt

Ans. (a) First-order in A, first-order overall

(b) First-order in A, negative first-order in B (reactant is inhibitor), zero-order overall

(c) First-order in A, first-order in B (catalyst), second-order

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