Thermo - Gibbs Free Energy & Entropy

In summary, the conversation discussed the use of a fuel cell using methane as fuel and determining the values of delta H and delta G for the reaction. The equations used involved finding the entropy of the products and reactants, which led to a negative delta S. However, this was balanced by the positive entropy of the external system, resulting in a correct answer for delta G.
  • #1
Ivegottheskill
11
0

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
 
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  • #2
Ivegottheskill said:
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
 
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  • #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^
 
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1. What is Gibbs Free Energy?

Gibbs Free Energy is a thermodynamic quantity used to measure the amount of energy available to do useful work in a system. It takes into account both the change in enthalpy (heat content) and the change in entropy (disorder) of a system.

2. How is Gibbs Free Energy related to spontaneity?

Gibbs Free Energy is used to determine whether a chemical reaction or physical process will occur spontaneously. If the Gibbs Free Energy is negative, the process is spontaneous and can occur without any external influence. If it is positive, the process is non-spontaneous and will only occur with the input of energy.

3. What is the relationship between Gibbs Free Energy and equilibrium?

At equilibrium, the Gibbs Free Energy of a system is at its minimum value. This means that the system is in its most stable state and there is no longer any driving force for a spontaneous change to occur.

4. How does entropy affect Gibbs Free Energy?

Entropy is a measure of the disorder or randomness in a system. As entropy increases, the Gibbs Free Energy also increases. This means that systems tend to move towards a state of higher disorder and the process will become more spontaneous.

5. Can Gibbs Free Energy be used to predict the direction of a chemical reaction?

Yes, Gibbs Free Energy can be used to predict the direction of a chemical reaction. If the value of ΔG is negative, the reaction is spontaneous in the forward direction. If it is positive, the reaction is non-spontaneous in the forward direction and will instead proceed in the reverse direction. A ΔG value of zero indicates that the reaction is at equilibrium.

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