Page 169 - Physical Chemistry

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Chapter 5 In step (a), we isothermally reduce the pressure of the real gas from 1 bar to
Standard Thermodynamic zero. In step (b), we wave a magic wand that eliminates intermolecular interactions,
Functions of Reaction
thereby changing the real gas into an ideal gas at zero pressure. In step (c), we
isothermally increase the pressure of the ideal gas from 0 to 1 bar. The overall
process converts the real gas at 1 bar and T into an ideal gas at 1 bar and T.For this
process,

¢H H 1T, P°2 H 1T, P°2 ¢H ¢H ¢H c (5.14)
re
a
b
id
The enthalpy change H for step (a) is calculated from the integrated form of
a
Eq. (4.48), (
H/
P) V TVa [Eq. (4.63) with dT 0]:
T
¢H H 1T, 0 bar2 H 1T, P°2 0 1V TVa2 dP
a
re
re
P°
For step (b), H H (T, 0 bar) H (T, 0 bar). The quantity U U (both at
id
re
id
b
re
the same T ) is U intermol (Sec. 2.11), the contribution of intermolecular interactions
to the internal energy. Since intermolecular interactions go to zero as P goes to zero
in the real gas, we have U U in the zero-pressure limit. Also, as P goes to zero,
id
re
the equation of state for the real gas approaches that for the ideal gas. Therefore (PV) re
equals (PV) in the zero-pressure limit. Hence H U (PV) equals H in the
re
re
re
id
id
zero-pressure limit:
H 1T, 0 bar2 H 1T, 0 bar2 and ¢H 0 (5.15)
re
id
b
For step (c), H is zero, since H of an ideal gas is independent of pressure.
c
Equation (5.14) becomes
H 1T, P°2 H 1T, P°2 P° cTa 0V b Vd dP const. T (5.16)
re
id
0 0T P
1
where a V (
V/
T) was used. The integral in (5.16) is evaluated using P-V-T
P
data or an equation of state (Sec. 8.8) for the real gas. The difference H m,re H m,id is
quite small at 1 bar (since intermolecular interactions are quite small in a 1-bar gas)
N 2 , 25ºC
but is included in precise work. Some values of H m,re H m,id at 298 K and 1 bar are
7 J/mol for Ar, 17 J/mol for Kr, and 61 J/mol for C H . Figure 5.6 plots H m,re
2
6
and H m,id versus P for N (g) at 25°C, with H m,id arbitrarily set equal to zero. Steps (a)
2
and (c) of the process (5.13) are indicated in the figure. Intermolecular attractions
make U and H slightly less than U and H , respectively, at 1 bar.
id
re
re
id
Conventional Enthalpies. Instead of tabulating H°values and using these to find
f
H° of reactions, one can construct a table of conventional (or relative) standard-state en-
thalpies of substances and use these to calculate H° of reactions from H° n H° ,
m,i
i
i
where H° is the conventional standard-state molar enthalpy of substance i. To construct
m,i
such a table, we begin by arbitrarily assigning the value zero to the conventional molar
enthalpy at 25°C and 1 bar of the most stable form of each pure element:
H° m,298 0 for each element in its stable form (5.17)
Although the actual absolute enthalpies of different elements do differ, the convention
Figure 5.6 (5.17) cannot lead to error in chemical reactions because elements are never interconverted
in chemical reactions. Knowing the conventional H° of an element, we can use H
Change in H with P for the m,298
m
2
isothermal conversion of real-gas C dT at constant P to find the conventional H° of an element at any temperature T. If
1
P
m
N to ideal-gas N at 25°C. any phase changes occur between 298.15 K and T, we make separate allowance for them.
2
2