Page 138 - Physical chemistry understanding our chemical world

ENTHALPY 105

−1
reaction is 118.5J K −1 mol . Assume this value is temperature independent over this
small temperature interval.

Inserting data into Equation (3.21):

−1
H O =−3223.7kJ mol −1 + 118.5J K −1 mol (298 − 293) K
r 298 K
H O =−3223.7kJ mol −1 + 592.5J mol −1
r 298 K
so
−1
H O =−3223.1kJ mol or − 3.2231 MJ mol −1
r 298 K
SAQ 3.7 Ethane burns completely in oxygen to form carbon dioxide and
O
waterwithanenthalpyof H =−1558.8kJ mol −1 at 25 C. What is H c O
◦
c
at 80 C? First calculate the change in heat capacity C p from the data in
◦
the following table and Equation (3.22).
Substance C 2 H 6(g) O 2(g) CO 2(g) H 2 O (l)
◦
C p at 80 C/JK −1 mol −1 52.6 29.4 37.1 75.3

C p = νC p − νC p (3.22)
products reactants

where the upper-case Greek letter Sigma means ‘sum of’, and the
lower-case Greek letters ν (nu) represent the stoichiometric number of
each species, which are the numbers of each reagent in a fully balanced
equation. In the convention we adopt here, the values of ν are positive for
products and negative for reactants.

Justification Box 3.2

Starting with the definition of heat capacity in Equation (3.20):

∂H

= C p
∂T
p
This equation represents C p for a single, pure substance. Separating the variables yields
dH = C p dT

Then we integrate between limits, saying the enthalpy is H 1 at T 1 and H 2 at T 2 :

H 2 T 2
dH = C p dT
H 1 T 1