Page 190 - Basic physical chemistry for the atmospheric sciences

1 7 6 Basic physical chemistry

- 1 96.2 kJ; exothermic. Hint: The enthalpies of formation
of elements in their standard states are zero.
1
k
2 . 1 4 . - 74.8 J mol - . Hint: If a set of chemical reactions can be
algebraically combined to yield a net reaction, the standard
molar enthalpy of the net reaction can be obtained by simi­
larly combining the standard molar enthalpies of the constit­
uent reactions .
i
2 . 1 5 . Hint: Show that f a small quantity of heat does flow f r om a
cold body to a hot body the total entropy of the system is
decreased. Since this violates the second law of thermody­
namics [see Eq. (2.20)] , heat cannot flow from a cold to a
hot body in an isolated system.
2 . 1 6 . In all three cases the entropy decreases.
2 . 1 7 . (a) Spontaneous at all temperatures. (b) Not spontaneous at
any temperature. (c) Spontaneous at temperatures below
a certain valu . (d) Spontaneous at temperatures above a
e
certain value.
l
2 . 1 8 . Hint: n addition o the two relations given, you will need to
t
use Eqs . (2.6), (2. 1 6 ) , and (2.27).
2 . 1 9 . Hint: Follow analogous steps to those leading from Eq.
(2.25) to Eq. (2. 3 0).
2
2 . 2 1 . !!.. G}l = - 6 9 . 6 kJ ; K P = l . 6 x 1 0 1 ; products are favored .
7
2 . 22. Solution: From values of 5° and SHY given in Appendix V ,
u
i t i s fo n d that as0 and smx are both negative fo r the stated
reaction. Therefore , from the Gibbs-Helmholtz equation
(2.29), the forward reaction is more likely to produce a
decrease in the Gibbs free energy, which favors dissociation
of NO(g), at low temperatures. At the high temperatures
n
present in combustio , the reverse reaction, which forms
)
the pollutant NO(g , will be favored.
2 . 2 4. K P(298K) = 3 . 92 x 1 0 2 1 ; K P (500K) = 1 . 27 x 1014• Hint: U s e the
0
relation given in Exercise . 1 1 to find K at 50 K .
2
P
2.25. Decrease of 3 x 1 0 - 2 1 J .
2 . 2 7 . Hint: Start with Eq. (2.8). For solids and liquids (but not
gases) dV and dp are small near I at m .
2 . 2 8. Hint: Combine the gas equation for a unit mass [see Eq.
q
( I . S c)] with E . (2.6) and the definition of c P .
G
2 . 2 9. l!.. = - n F!!..E
2
Solution: From Exercise . 1 8 we have
da = - d g