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Thermo - Gibbs Free Energy & Entropy |
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| Apr1-07, 06:53 AM | #1 |
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Thermo - Gibbs Free Energy & Entropy
1. The problem statement, all variables and given/known data
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 2. Relevant 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 I think I found H alright (-802.3kJ), but to find G I need entropy (delta S) 3. 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 ![]() Thanks in advance for any hints/tips |
| Apr1-07, 03:13 PM | #2 |
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and, [itex]\Delta S_{internal}+\Delta S_{surrounding}=\Delta S_{total} \geq 0[/itex] See here |
| Apr1-07, 09:44 PM | #3 |
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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^ |
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