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118 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

3.3 Indirect measurement of enthalpy

How do we make ‘industrial alcohol’?

Enthalpy Cycles from Hess’s Law

Industrial alcohol is an impure form of ethanol made by hydrolysing ethene,
=CH 2 :
CH 2
H
H H H H
H 2 O
(3.33)
H H H H OH

We pass ethene and water (as a vapour) at high pressure over a suitable catalyst,
causing water to add across the double bond of the ethene molecule. The industrial
alcohol is somewhat impure because it contains trace quantities of ethylene glycol
(1,2-dihydroxyethane, III), which is toxic to humans. It also contains unreacted water,
and some dissolved ethene.

H
OH H
H
H OH
(III)
We may rephrase
Hess’s law, saying ‘The
standard enthalpy of an But what is the enthalpy of the hydration reaction in Equation
overall reaction is the (3.33)? We first met Hess’s law on p. 98. We now rephrase it by
sum of the standard saying ‘The standard enthalpy of an overall reaction is the sum of
enthalpies of the indi- the standard enthalpies of the individual reactions into which the
vidual reactions into
which the reaction may reaction may be divided.’
be divided’. Accordingly, we can obtain the enthalpy of reaction by drawing
a Hess cycle, or we can obtain it algebraically. In this example, we
will use the cycle method.

Worked Example 3.14 What is the enthalpy change H r of the reaction in Equation
(3.33)?

We start by looking up the enthalpies of formation H f for ethene,
Notice that each of ethanol and water. Values are readily found in books of data; Table 3.1
these formation reac- contains a suitable selection.
tionsishighly exother-
mic, explaining why =CH 2 ] =−52 kJ mol −1
H f(1) [CH 2
energy is needed to −1
obtain the pure ele- H f(2) [CH 3 CH 2 OH] =−235 kJ mol
ments. −1
H f(3) [H 2 O] =−286 kJ mol