- #1
calculus_jy
- 56
- 0
I would like to know if someone here can help sort out my confusion...
It is easy to derive that
dU=TdS-pdV (no particle exchange)
Then enthalpy H=U+pV
will imply dH=TdS+Vdp
That means under a constant temperature (dT=0) and constant pressure (dp=0) situation
dH=TdS>(=) dQ
Why is it then that if we look at [tex] C_6 H_{12} +6O_2 + 6O_2\rightarrow O_2+6H_2O[/tex]
[tex] \Delta H=-2803kJ/mol [/tex]
[tex] \Delta G=-2879kJ/mol [/tex]
[tex] T\Delta S=77.2 kJ/mol [/tex]
(all above data are obtained at constant temperature and pressure)
now WHy is it that [tex] \Delta H =/= T\Delta S [/tex]...
but that [tex] \Delta G= \Delta H- T\Delta S [/tex] which i know will hold generally but should reduce to [tex] \Delta G= 0 [/tex] under (dT=0, dp=0) when no other work done excluding expansion work...
I thought that [tex] \Delta H [/tex] should be the heat flow in a constant temperature and pressure process...
but in this case why is the heat flow [tex] T\Delta S=77.2 kJ/mol [/tex] instead...
It is easy to derive that
dU=TdS-pdV (no particle exchange)
Then enthalpy H=U+pV
will imply dH=TdS+Vdp
That means under a constant temperature (dT=0) and constant pressure (dp=0) situation
dH=TdS>(=) dQ
Why is it then that if we look at [tex] C_6 H_{12} +6O_2 + 6O_2\rightarrow O_2+6H_2O[/tex]
[tex] \Delta H=-2803kJ/mol [/tex]
[tex] \Delta G=-2879kJ/mol [/tex]
[tex] T\Delta S=77.2 kJ/mol [/tex]
(all above data are obtained at constant temperature and pressure)
now WHy is it that [tex] \Delta H =/= T\Delta S [/tex]...
but that [tex] \Delta G= \Delta H- T\Delta S [/tex] which i know will hold generally but should reduce to [tex] \Delta G= 0 [/tex] under (dT=0, dp=0) when no other work done excluding expansion work...
I thought that [tex] \Delta H [/tex] should be the heat flow in a constant temperature and pressure process...
but in this case why is the heat flow [tex] T\Delta S=77.2 kJ/mol [/tex] instead...
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