Page 189 - Basic physical chemistry for the atmospheric sciences

Answers, hints, and solutions to exercises

gases released by reaction (A) exert the same pressur . 1 1 1
e
i
the case of NH3(g), however, there s an additional pressure
p; due to the o.3 g of this gas that are already present in the
flask. Therefore, we can write
'
P 1 = pz + P ; (D)
We can calculate p; from the gas equation in the form
, R *
p v = m T
;
M
where v i s the volume occupied by mass m of a gas with
molecular weight M at temperature T. For the NH (g) al­
3
3
ready present in the flask, v = 2.00 x 0 - 3 m ,
1
m = 0.300 x 1 0 - 3 kg, M = 1 7 , T = 298K and R* = 8. 3 1 J deg- 1
mo1 - 1 • Hence, p; = 0 .2 1 9 x 1 0 5 Pa = 0 . 2 1 9 atm.
From Reactions (B) and (D), with p; 0 .21 9 atm we get
=
0
(p + . 2 1 9 )p2 = 1 . 08 x 10 - 1
2
or,
2
(p 2) + 0.2 l9p2 - 0 . 1 0 8 = 0
Solution of this quadratic equation yields p2 = 0.237 atm.
From Reactions (C) and (D), with p; 0 . 2 1 9 atm and
=
p2 = 0.237 atm , we get
p = 2p2 + p; = 0 . 6 93 atm
Therefore, the total pressure in the flask after chemical
0
equilibrium is established is . 6 93 atm.
1 . 23 . Since the reaction is endothermic it "moves to the right"
with increasing temperature.
1 . 24. At lower temperatures the reaction "shifts back to the lef , "
t
o
b u t the rate f the reaction s now slow.
i

CHAPTER 2
2. 1 1 . Hint: Remember that In x = 2 . 3 03 log 1oX .

2 . 1 2 . 2 . 5 7 x 1 0 - 6 · Hint: Use e xpression given i n Exercise 2 . 1 1 .
2. 1 3 . (a) 1 8 1 kJ; endothermic. (b) - 1 99 kJ; exothermic. (c)