Page 92 - Physical chemistry eng

4.2 INTERNAL ENERGY AND ENTHALPY CHANGES ASSOCIATED WITH CHEMICAL REACTIONS 69

are very small. The enthalpy of reaction, ¢H R , at specific values of T and P is defined
as the heat exchanged between the system and the surroundings as the reactants are
transformed into products at conditions of constant T and P. By convention, heat flow-
ing into the system is given a positive sign. ¢H R is, therefore, a negative quantity for
an exothermic reaction and a positive quantity for an endothermic reaction. The
standard enthalpy of reaction, ¢H° R , refers to one mole of the specified reaction at a
pressure of 1 bar, and unless indicated otherwise, to T = 298.15 K.
How can the reaction enthalpy and internal energy be determined? We proceed
in the following way. The reaction is carried out at 1 bar pressure, and the tempera-
ture change ¢T that occurs in a finite size water bath, initially at 298.15 K, is meas-
ured. The water bath is large enough that ¢T is small. If ¢T is negative as a result
of the reaction, the bath is heated to return it, the reaction vessel, and the system to
298.15 K using an electrical heater. By doing so, we ensure that the initial and final
states are the same and therefore the measured ¢H is equal to ¢H° R . The electrical
work done on the heater that restores the temperature of the water bath and the system
to 298.15 K is equal to ¢H° R . If the temperature of the water bath increases as a
result of the reaction, the electrical work done on a heater in the water bath at
298.15 K that increases its temperature and that of the system by ¢T in a separate
experiment is measured. In this case, ¢H° R is equal to the negative of the electrical
work done on the heater.
Although an experimental method for determining ¢H° R has been described, to
tabulate the reaction enthalpies for all possible chemical reactions would be a monu-
mental undertaking. Fortunately, ¢H° R can be calculated from tabulated enthalpy
values for individual reactants and products. This is advantageous because there are
far fewer reactants and products than there are reactions among them. Consider ¢H° R
for the reaction of Equation (4.1) at T = 298.15 K and P = 1 bar. These values for
P and T are chosen because thermodynamic values are tabulated for these values.
at other values of P and T can be calculated as discussed in Chapters 2
However, ¢H R
and 3. In principle, we could express ¢H° R in terms of the individual enthalpies of
reactants and products:
¢H° = H° products - H° reactants
R
= 3H° (Fe,s) + 4H° (H O,l) - H° (Fe O ,s) - 4H° (H ,g) (4.2)
3 4
2
m
m
2
m
m
The m subscripts refer to molar quantities. Although Equation (4.2) is correct, it does
not provide a useful way to calculate ¢H° R . There is no experimental way to determine
the absolute enthalpy for any element or compound because there is no unique refer-
ence zero against which individual enthalpies can be measured. Only ¢H and ¢U , as
opposed to H and U, can be determined in an experiment.
Equation (4.2) can be transformed into a more useful form by introducing the
enthalpy of formation. The standard enthalpy of formation, ¢H° f , is defined as the
enthalpy change of the reaction in which the only reaction product is 1 mol of the species
of interest, and only pure elements in their most stable state of aggregation under the stan-
dard state conditions appear as reactants. We refer to these species as being in their
standard reference state. For example, the standard reference state of water and carbon
at 298.15 K are H O(l) and solid carbon in the form of graphite. Note that with this defi-
2
nition, ¢H° = 0 for an element in its standard reference state because the reactants and
f
products are identical.
We next illustrate how reaction enthalpies can be expressed in terms of formation
enthalpies. The only compounds that are produced or consumed in the reaction
O (s) and H O(l). All ele-
Fe O (s) + 4 H (g) ¡ 3 Fe(s) + 4 H O(l) are Fe 3 4 2
2
2
3 4
ments that appear in the reaction are in their standard reference states. The formation
reactions for the compounds at 298.15 K and 1 bar are
1
H (g) + O (g) ¡ H O(l)
2
2
2
2
1
¢H° =¢H° (H O, l) = H° (H O, l) - H° (H , g) - H° (O , g) (4.3)
2
2
m
R
f
m
2
2
m
2