Page 45 - Basic physical chemistry for the atmospheric sciences

Chemical thermodynamics � I

Exercise 2.6. Calculate A7fJ at 25°C and 1 atm for the reaction
H202(g)� H20(g) + !Oz(g)

Is H 2 0z(g) stable at 25°C and 1 atm?
Solution. From Appendix V we see that at 25°C and 1 atm
1
d0/[H20z(g)] = - 1 8 8 . 2 kJ mo1- and AG/[H20(g)] = - 228.6 kJ
1
mo1- and, since 02(g) is the stable form of oxygen at 25°C and 1 atm,
AG/[Oi{g)] = 0. Hence, applying Eq. (2.34) to the above reaction
ac0 = [( - 228.6)] - [( - 1 8 8 . 2 )) kJ
or,
A7fJ = - 4 0.4 kJ

Since SC° is negative, the reaction will proceed spontaneously in
the forward direction and will be unstable. However, thermodynamic
calculations give no information on the speed of chemical reactions.
We will consider this subject in Chapter .
3

2 . 7 Free energy change and the equilibrium constant
It is only under standard conditions ( 1 atm and 25°C) that a(}l is
sufficient to determine whether a reaction is spontaneous or not. We
will now derive a relation for the molar free energy change AG under
any conditions in terms of aeo and the equilibrium constant for the
chemical reaction.
)
For a reversible (equilibrium) transformation, Eqs. (2.6), (2. 16 , and
(2.27) can be combined to give
dg = a d p - s dT (2.35)

or at constant temperature,
dg = a d p (2.36)

Applied to n moles of a substance, Eq. (2.36) becomes
G
dG = n d = n Vdp (2.37)
where V is the volume of 1 mole of the substance. If we now consider
I mole of an ideal gas, from the gas equation [see Eq. 1 . 8 d)]
(
p V = R " T (2.38)

From Eq . (2.37) and (2. 3 8)
s