Physical Chemistry , 1st ed.

(Darren Dugan) #1
Rather, a forward process is balanced by a reverse process. To emphasize that
forward and reverse reactions are occurring simultaneously, the double arrow
sign is typically used when writing a reaction, instead of a single arrow.
Equilibrium constants can be used to determine extents of reactions, as
shown in the following example.

Example 5.5
For the gas-phase reaction
CH 3 COOC 2 H 5 H 2 OCH 3 COOHC 2 H 5 OH
ethyl acetate water acetic acid ethanol
the equilibrium constant is 4.00 at 120°C.
a.If you start with 1.00 bar of both ethyl acetate and water in a 10.0-L con-
tainer, what is the extent of the reaction at equilibrium?

b.What is (^) rxnGat equilibrium? Explain.
c.What is (^) rxnG° at equilibrium? Explain.
Solution
a.The following chart shows initial and equilibrium amounts of the sub-
stances involved in the equilibrium:
Pressure (bar) CH 3 COOC 2 H 5  H 2 OCH 3 COOH  C 2 H 5 OH
Initial 1.00 1.00 0 0
Equilibrium 1.00 x 1.00 x x x
The expression for the equilibrium constant can be constructed from the
chemical reaction, and the values from the final row of the chart are substi-
tuted. We get
K4.00
4.00 (1.00
x

2
x)^2

This expression can be expanded and solved algebraically using the quadratic
formula. When we do this, we get two numerical answers for x, which are
x0.667 bar or x2.00 bar
We examine each of those roots, keeping in mind the reality of the situation.
If we are starting with only 1.00 bar of reactant, we cannot lose 2.00 bar.
Therefore we reject x2.00 bar as not a physically real answer. So in terms
of final amounts of reactants and products, we use the x0.667 bar as the
change in amount to get the equilibrium amounts
pCH 3 COOC 2 H 5 0.333 bar pH 2 O0.333 bar
pCH 3 COOH0.667 pC 2 H 5 OH0.667 bar
The extent of the reaction at equilibrium can be calculated using any of the
reaction species, after converting the amounts to moles. Using H 2 O and the
ideal gas law:
nH 2 O,init0.306 mol nH 2 O,e quil0.102 mol
(x)(x)

(1.00 x)(1.00 x)
(pCH 3 COOH/p°)(pC 2 H 5 OH/p°)

(pCH 3 COOC 2 H 5 /p°)(pH 2 O/p°)


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126 CHAPTER 5 Introduction to Chemical Equilibrium

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