Modern Engineering Thermodynamics

15.11 Entropy of Formation and Gibbs Function of Formation 625
s CH4
= s CH4 ° − R ln ðp CH4 in MPaÞ/ð0:100 MPaÞ
= 186:256 − 8:3143ðln 0:0951Þ
= 205:8 KJ/ ðkgmole . KÞ
s O2 = 205:138 − 8:3143ðln 0:190Þ
= 218:9 kJ/ ðkgmole . KÞ
s N2 = 191:610 − 8:3143ðln 0:715Þ
= 194:4 kJ/ ðkgmole . KÞ
Then,
∑ðn i /n fuel Þs i = 205:8 + 2ð218:9Þ + 7:52ð194:4Þ
R
= 2100 kJ/ðkgmole . KÞ
Also, for the products,
∑ðn i /n fuel Þs i = s O2 + 2ðs H2OÞ + 7:52ðs N2 Þ
P
where, again from Eq. (15.26) with T = 473 K and using Table 15.7,
s CO2 = s° CO2 + ðc p Þ CO2
ln 473 − R
ln ðp CO2 in MPaÞ/ð0:100 MPaÞ
298
= 213:795 + 37:19 ln 473
− 8:3143ð ln 0:0951Þ
298
= 250:5 kJ/ðkgmole . KÞ
s H2O = 188:833 + 33:64 ln 473
− 8:3143ð ln 0:190Þ
298
= 218:2 kJ/ðkgmole . KÞ
and
s N2 = 191:610 + 29:08 ln 473
− 8:3143ðln 0:715Þ
298
= 207:8 kJ/ðkgmole . KÞ
Then,
∑
p
ðn i /n fue1 Þs i = 250:5 + 2ð218:2Þ + 7:52ð207:8Þ
= 2250 kJ/ðkgmole . KÞ
Finally, the desired result is
ðs p Þ r = 2250 − 2100 −ð−134,158/473Þ
= 434 kJ/ðkgmole . KÞ
Only about one third of the value of ðs p Þ r
here is due to the reaction itself; the remaining two thirds come from the associated
heat transfer. Note that ðs p Þ r
> 0, as required by the second law of thermodynamics.
Exercises
34. Determine the molar specific entropy produced in Example 15.12 if the temperature of the surface of the system
where the heat transfer occurs T b is 20.0°C rather than 200.°C. Assume all the remaining variables are unchanged.
Answer: ðs p Þ r = 602 kJ/(kgmole ·K).
35. If the temperature of the products in Example 15.12 is 500.°C instead of 200.°C, determine the specific molar entropy
production for this reaction. Assume all the remaining variables are unchanged (including T b ). Answer: ðs p Þ r
= 586 kJ/
(kgmole·K).
36. Determine the specific molar entropy production in Example 15.12 if the pressure of the products of combustion is
0.200 MPa instead of 0.100 MPa. Answer: ðs p Þ r
= 488 kJ/(kgmole·K).
15.11 ENTROPY OF FORMATION AND GIBBS FUNCTION
OF FORMATION
Since the specific molar Gibbs function, g = h − T s, depends upon both h and s, it does not have an absolute
zero reference state. But, since s does have an absolute zero reference state, we cannot arbitrarily set the
molar specific entropies of the elements of the reaction equal to zero at the standard reference state, as