Temperature Dependence of Enthalpy

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Temperature Dependence of Enthalpy Thermodynamic
data are often tabulated and can even be located
with some effort at 25.0 oC, but are often needed
at other temperatures, e.g., for processes
occuring in boilers or during the winter in
colder climates. Heat capacities are the key
data that allow thermodynamic values to be
calculated at temperatures other than that at
which they are tabulated. Suppose you wanted to
know the enthalpy change at 800 K and 1.00 bar
for the reaction DHo800K ? H2 (g)
1/2 O2 (g) -----------------gt H2O (g)
800 K, 1.00 bar Tabulated data that
you might find related to this reaction are the
standard enthalpy of formation of liquid water at
298 K DHof, 298K - 285.830
kJ / mole H2 (g) 1/2 O2 (g)
-----------------gt H2O (l)
298 K, 1.00 bar and the standard enthalpy of
vaporization of water at the normal boiling of
water at 373 K DHovap, 373K 40.670
kJ / mole H2O (l) -------------------gt H2O
(g) 373 K, 1.00 bar
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The alternate path involves a total of 6
steps. The above method of mapping out an
alternate hypothetical path along which the
change in a state function can be calculated is
very powerful and we will use this approach again
and again.
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1. In the 1st step 1.00 moles of H2 (g) are
cooled at constant pressure from 800 K to 298 K.
The enthalpy change in this step is calculated by
integrating the temperature dependent constant
pressure heat capacity for H2 (g) (obtained from
tables) over this temperature range DHo1 800
K ? 298 K n Cp, H2 (g) dT (1.00) 800 K ? 298
K3.496 -0.1006x10-3 T 2.419x10-7 T2 dT
3.496 (298 K - 800 K) - 0.1006x10-3 (298 K)2 -
(800 K)2/2 2.419x10-7 (298 K)3 - (800
K)3/3 (- 1755) (27.725) (-39.15)
- 1766 J Is the sign on DHo1 consistent with the
fact that the H2 (g) is cooling? 2. In the 2nd
step 1/2 mole of O2 (g) are cooled at constant
pressure from 800 K to 298 K DHo2 800 K ?
298 K n Cp, O2 (g) dT (1/2) 800 K ? 298
K3.0673 1.637x10-3 T 5.118x10-7 T2 dT -
954 J
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3. In step 3 the hydrogen and oxygen are
reacting to form liquid water at 298 K and 1.00
bar. This is the standard formation reaction for
liquid water at 298 K and the standard enthalpy
of formation of liquid water is readily found in
tables DHof, 298K - 285.830
kJ / mole H2 (g) 1/2 O2 (g)
-----------------gt H2O (l)
298 K, 1.00 bar DHo3 - 285.830 kJ 4. In
step 4 the liquid water is warmed at constant
pressure from 298 K to its normal boiling point
at 373 K or 100 oC. DHo4 298 K ? 373 K n
Cp, H2O (l) dT (1.00 mole) 298 K ? 373 K(75.40
J / mole K) dT 5655 J Why is this enthalpy
change positive?
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5. In step 5 the liquid water is vaporizing to
form water vapor at the normal boiling point of
298 K and 1.00 bar. The enthalpy change is the
standard enthalpy of vaporization water at the
normal boiling point is readily found in
tables DHovap, 373K 40.670 kJ /
mole H2O (l) -------------------gt H2O (g) 373
K, 1.00 bar DHo5 40.670 kJ Why is
this enthalpy change positive? 6. In step 6 the
water vapor is warmed at constant pressure from
373 K to 800 K. DHo6 373 K ? 800 K n Cp,
H2O (g) dT (1.00 mol) 373 K ? 800 K 3.633
1.195x10-3 T 1.34x10-7 T2 dT 2,175
J Since enthalpy is a state function, the
enthalpy change at 800 K and 1.00 bar is equal to
the sum of the enthalpy changes along this
alternate path DHo800 K DHo1 DHo2 DHo3
DHo4 DHo5 DHo6 (- 1.766 kJ) (- 0.954
kJ) (-285.830 kJ) (5.655 kJ) (40.670
kJ) (2.175 kJ) - 240.05 kJ
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In certain situations the calculation can be
simplified Consider the sublimation of iodine at
75.0 oC and 1.00 atm
348.2 K I2 (s) ----------- gt I2 (g)
DHosub, 348.2 K ? 1.00 atm The
standard enthalpy of sublimation of iodine vapor
at 298.2 K can be found in thermodynamic
tables 298.2 K I2 (s) ----------- gt I2
(g) DHosub, 298.2 K 14.923 kcal 1.00
atm An alternate path between the inital and
final states via which we could calculate the
enthalpy change at 348.2 K is
348.2 K, 1.00 atm I2 (s)
----------------------------- gt I2 (g) DHosub,
348.2 K ?
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The enthalpy of sublimation at 348.2 K and 1.00
atm can be calculated as DHosub, 348.2 K
DHo1 DHosub, 298.2 K DHo3
348.2 K ? 298.2 KCp, I2 (s) dT DHosub, 248.2 K
298.2 K ? 348.2 K
Cp, I2 (g) dT DHosub, 298.2 K 298.2 K
? 348.2 K(Cp, I2 (g) - Cp, I2 (s)) dT

Why is
there a minus sign in front of Cp, I2 (s) in the
above integral? DHosub, 298.2 K 298.2
K ? 348.2 K DCp dT where DCp is the difference
in the constant pressure heat capacity of the
products minus the constant pressure heat
capacity of the reactants Cp, I2 (g) (cal /
mole K) 8.82 - Cp, I2 (s) (cal / mole K)
- (9.589 0.01190 T)
----------------------------
----------------------------
DCp (cal / mole K) - 0.92 - 0.01190 T
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The enthalpy change is now DHosub, 348.2 K
DHosub, 298.2 K 298.2 K ? 348.2 K(- 0.92 -
0.01190 T) dT 14.923
kcal (10-3 kcal/cal) ( -0.92 (348.2
K - 298.2 K) - (0.01190 / 2) (348.2
K)2 - (298.2)2) ( 14.923 kcal) (-
0.046 kcal) (-0.192 kcal) 14.685
kcal This latter example had advantages over the
previous example in that it required only a
single integration and a tabular approach, less
likely to lead arithematic error, could be used
to calculate DCp. Why wouldnt this simpler
approach have worked in the previous example?