Thermo - Gibbs Free Energy & Entropy

  • #1

Homework Statement


Consider fuel cell using methane as fuel. Reaction is

CH^4 + 2O_2 -> 2H2O+CO_2

Assume room temperature and atmospheric temperature
Determine values of delta H (Helmholtz) and delta G (Gibbs) for this reaction for one mole of methane.

Question instructed the use of the web to find thermodynamic tables with values of H and G for the chemicals in the reaction

Homework Equations


I haven't encountered this sort of question where substitution of "real" values is necessary. Hence I've used this site as a reference:

http://members.aol.com/profchm/gibbs.html [Broken]

I think I found H alright (-802.3kJ), but to find G I need entropy (delta S)


The Attempt at a Solution



dG = dH - T.dS

dS = Sum of products (RHS) - Sum of reactants (LHS)
= [2(188.7)+213.7]-[186.3 + 2(205)]
= -5.2 (but isn't an entropy of less than zero impossible?)

The problem lies in the uncertainty of me obtaining an negative delta S :confused:

Thanks in advance for any hints/tips
 
Last edited by a moderator:

Answers and Replies

  • #2
182
0
dS = Sum of products (RHS) - Sum of reactants (LHS)
= [2(188.7)+213.7]-[186.3 + 2(205)]
= -5.2 (but isn't an entropy of less than zero impossible?)

The problem lies in the uncertainty of me obtaining an negative delta S :confused:
Actually, that formula uses [itex] \ \Delta G \ = \Delta H \ - \ T\Delta S_{internal}[/itex]

and, [itex]\Delta S_{internal}+\Delta S_{surrounding}=\Delta S_{total} \geq 0[/itex]
See here
 
Last edited:
  • #3
Thanks for your reply.

So in this case the answer I obtained is correct?

dS_total is > 0, but the entropy of the 'external' system (the "universe"?) balances the negative entropy of the internal system (the reaction and its components in the engine)

^Is this line of thinking correct^

If it is, then:

dG = -802.3 - (300K * (-5.2/1000))
= -800.74 kJ

^Answer obtained^
 
Last edited:

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