Page 197 - Physical chemistry understanding our chemical world

164 REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

Worked Example 4.11 Consider the reaction between ethanol and
Care:we must always
ethanoic acid to form a sweet-smelling ester and water:
convert from kJ to J
before calculating with CH 2 CH 2 OH + CH 3 COOH −−−→ CH 2 CH 2 CO 2 CH 3 + H 2 O (4.58)
Equation (4.57).
What is the equilibrium constant K at room temperature (298 K) if
−1
the associated change in Gibbs function is exogenic at −3.4kJ mol ?
Care:ifwecalculate Inserting values into eq. (4.57):
avalue of K that is −1
extremely close to K = exp −(−3400 J mol )
one, almost certainly 8.314 J K −1 mol −1 × 298 K
we forgot to convert
from kJ to J, mak- K = exp(+1.372)
ing the fraction in the K = 3.95
bracket a thousand
times too small.
A value of K greater than one corresponds to a negative value of
G , so the esterification reaction is spontaneous and does occur
O
to some extent without adding addition energy, e.g. by heating.
A few values of G O are summarized as a function of K in
Table 4.1 and values of K as a function of G O are listed in
Table 4.2. Clearly, K becomes larger as G becomes more neg-
O
ative. Conversely, G is positive if K is less than one.
O

Justification Box 4.7
We start with Equation (4.55):
O
G =−RT ln K
Both sides are divided by −RT , yielding

− G O
= ln K (4.59)
RT
Then we take the exponential of both sides to generate Equation (4.57).

Table 4.1 The relationship between G O and equilib-
rium constant K:valuesof G O as a function of K
O
K G /kJ mol −1
We see from Table 4.1
that every decade 1 0
increase in K causes 10 2 −5.7
G O to become more 10 3 −11.4
negative by 5.7kJ mol −1 10 4 −17.1
10 −22.8
per tenfold increase 10 −1 +5.7
in K. −2
10 +11.4
10 −3 +17.1