Page 187 - Basic physical chemistry for the atmospheric sciences

Appendix VII . Answers t o exercises and hints and solutions to
selected exercises

CHAPTER 1
2 3
[NH (g)] . [N 2 (g)][Hz(g)]
1 . 8 . ( ) 3 3 ' 2
a
[N2(g)][H 2(g)] [NH3(g)]
2 112
[ NO(g)] [N 2 0(g)][Oz(g)]
(b) 2
[N 0(g)][02(g)] 112 ; [NO(g)]
2
2 6 4 3
[N 2(g)] [H 2 0(g)] [NH 3(g)] [0i{g)]
(c) 4 ; 2
[NH3(g)] [0 2 (g)] 3 [Nz(g)] [H 20(g)] 6

) ] [ l (
(d) [N H 3 (g)] [ HC l ( g)] ;
[NH 3 (g HC g)]
1 . 9. 0.09 M .
I . 1 0 . 0 . 3 3 atm.
I . I I . 278.
1 . 1 2 . 4. 8 x 1 0 - 3 . Hint: Do not forget to express the reactant and
product as concentrations before substituting into Eq. ( 1 .6) .
1 . 1 3 . 7 . 6 6 g of N20 (g) and 1 . 49 g of N02(g). Hint: Proceed as in
4
Exercise 1 . 6 , and solve resulting quadratic equation.
1 . 1 4 . These theorems follow in a fairly straightforward manner
from the definition of the equilibrium constant [see Eq.
( 1 .6)].
1 . 1 5 . 3 . 6 x 1 0 - 3 and 1 7 . Hint: Use the theorems stated in Exer­
cise 1 . 1 4 .
1 . 1 6 . 0 . 1 4 . Hint: U s e one of the theorems stated in Exercise 1 . 1 4 .
1 . 1 7 . 2 x 1 0 -4 atm each. Hint: The partial pressure of a solid can
be equated to unity in Eq. ( l .9b).
1 . 1 8 . 2 . 1 x 1 0 - 5 atm ; C02(g). Hint: The partial pressure exerted
by a gas in a mixture of gases is equal to the fractional
contribution that the gas makes to the total volume of the
mixture multiplied by the total pressure exerted by the
mixture.

1 7 3