Page 191 - Basic physical chemistry for the atmospheric sciences

Answers, hints, and solutions to exercises 1 7 7

where da is the work done over and above any pda work .
In this case, this extra work is electrical. Therefore,
AG = - (electrical work done)
- (voltage difference) (charge transfer in coulombs)
- AE (nF)

or,
AG = - nF AE

CHAPTER 3
3 . 8 . m = 3 , n = , k = 8 . 3 x 1 0 - 4 M - 3 s - 1 ; reaction rate =
1
6 . 6 x 1 0 -7 M s - 1 •
3 . 1 0 . (a) kpseudo = 1 . 8 X 1 0 - 20 [Oig)] . (b) pseudo = 7 . 3 X 1 0 - 3 S - 1 .
k
(
3 . 1 1 . Hint: Combine Eqs. (3.5) and l .8d).
3 . 1 2 . (a) 203(g) - 30 z(g). (b) O(g). (c) For step (i): Rate = k1
[0 3 (g)] . For step (ii): Rate = k2 [O J(g)][O(g)] . (d) Step (ii). (e)
[O(g)] C( [03(g)][Oz(g)] - I .
d[0 2 (g)]
.
c
h
·
3 . 1 3 . Hmt: n eac case, wnte an expression ior - c
·
ior
I
dt
(
the rate-determining step ; then apply Eq. 1 . 6 ) to the fast re­
n
actio .
kc 1 . 5 x 1 0 - 1 2
3 . 1 4 . K - - - 5 . 1 0 - 2 3 o• x 1 0 - 1 1
-
c
- kr 0 x
3 . 1 5 . (a) Reaction mechanism (iii).
k4cksr
(b) k =
k4r
k4cksck6r
c
( c) K =
k4rksrk6 r
Hint: Apply the principle of detailed balancing. See, for
example, Exercise . 4.
3
3 . 1 6 . 4 .8 x 1 0 7 mol L 1 s - 1 ; 5 . 0 x 1 0 - 10 .
s
-
3 . 1 8 . Ea = 5 3 . 3 k J mo - 1 •
l
3 . 1 9 . Rate increases by a factor of . 8 .
1
2
3 . 20. Ea = l 1 5 kJ mol - 1 ; A = l 0 5 s - 1 •
3 . 2 1 . 1 . 25 g. Hint: The amount of the reactant will halve at the
end of each period of time equal to one half-life.
3 . 23 . 1 . 9 x 1 0 3 years. Solution: Since radioactivity is a first-order
chemical reaction, we have