Help Sort My Confusion on dH, TdS and dQ

[calculus_jy]

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...

 

[DrDu]

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)

... and no chemical reaction. What is the definition of quantities like [tex]\Delta H[/tex]?

 

[calculus_jy]

"... and no chemical reaction. "

why is that...
H=U+PV
[tex]\Delta H=T\Delta S+V \Delta p[/tex]?

 

[DrDu]

[tex]\Delta[/tex] does not mean difference, here but [tex] \Delta X=\sum_i \nu_i \partial X /\partial n_i|_{p,T}[/tex], for an arbitrary quantity X where [tex]n_i[/tex] is the amount of reactant i and [tex] \nu_i [/tex] is the stochiometric coefficient in the reaction equation, e.g. 6 for H20 and -1 for C6H12 in the reaction you consider.
You have to start from the general equation
[tex]
dU=TdS-pdV+\sum_i \mu_i dn_i[/tex].
When all changes of n_i are due to one chemical reaction taking place, this can be written as:

[tex] dU=TdS -pdV +\Delta G d\xi [/tex]
where [tex] d\xi=dn_i/\nu_i[/tex] is the change of reaction number.
The [tex]\Delta G[/tex] follows from [tex] \mu_i=\partial U/\partial n_i|_{V,S}=\partial G/\partial n_i|_{p,T}[/tex].
Now with U, S and V being considered as depending on variables T, p and [tex]\xi[/tex],
we obtain:
[tex] \Delta U=\partial U/\partial \xi|_{p,T}=T \Delta S -p\Delta V +\Delta G[/tex] or
[tex] \Delta G =\Delta U+p\Delta V -T\Delta S=\Delta H-T\Delta S[/tex] .

 

[Studiot]

Hello, calculus_jnr.

Whenever you quote figures such as you have done you should always also state the reaction temperature.

This is particularly important in this case as one of the products is water.
Depending upon the final state of the products, (ie temperature) you may need to allow for latent heat in your calculations.

 

[DrDu]

Also I suppose that the values of [tex] \Delta H[/tex], [tex]\Delta G[/tex] etc refer to standard conditions where the concentrations (or better activities) are 1 mol/l and partial pressure is 1 atm. This does not correspond to chemical equilibrium whence [tex]\Delta G \ne 0[/tex].

 

[Studiot]

A small point first.

T[tex]\Delta[/tex]S is energy flow, not heat flow

In your reaction the LHS has 13 molecules aand the RHS has 7. So there is a definite entropy change.

As I said previously the actual change will also depend upon the state of the reaction products.

 

[DrDu]

A small point first.

T[tex]\Delta[/tex]S is energy flow, not heat flow

In your reaction the LHS has 13 molecules aand the RHS has 7. So there is a definite entropy change.

As I said previously the actual change will also depend upon the state of the reaction products.

Whatever flow T[tex]\Delta[/tex]S is, it is not heat flow, as you said.
[tex] \delta Q+\delta W=dU=TdS -pdV+\Delta G d\xi[/tex]
If T=const, p=const then [tex] dS(T,P,\xi)=\Delta S d\xi[/tex].
If furthermore
[tex]\delta W=-pdV [/tex]
then
[tex] \delta Q =dU+pdV=dH=TdS+\Delta G d\xi=T\Delta S d\xi +(\Delta H -T\Delta S)d\xi=\Delta H d\xi.[/tex]