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Quick Entropy / Enthalpy Question

  1. Jan 31, 2012 #1
    Hello All,

    I've been struggling to figure out how to solve for change of enthalpy and entropy. My reference text makes it look so easy but then gives an example where it provides totally different information. Here is the reference from the text:

    "A Lead Acid Battery's measurments were taken at 27 deg C yielding the following data:

    Open Circuit Voltage (OCV) = 2.05 V
    n = 2
    dE/dT = 0.2 mV/K

    DeltaG, DeltaH, and DeltaS can be found using Gibbs Free Energy Equation and the formula DeltaG=DeltaH-T*DeltaS."

    I have no idea how to find DeltaH or DeltaS given this data!

    Here is how I solved for DeltaG:

    DeltaG=-nFE=-2(98485)(2.05)=-403/788 J/mol (dived by 3600 s to convert to Wh/mol)
    DeltaG = -112.163 Wh/mol

    Where can I go from here?

    Any and all help is greatly appreciated! --Brim
  2. jcsd
  3. Feb 1, 2012 #2
    If you know [itex]E[/itex], then you can calculate [itex]\Delta G[/itex], as you have done already. You also know that [itex]\Delta G = \Delta H - T \Delta S[/itex]. But, this leaves you with two unknowns.

    Perhaps you can calculate some more values of [itex]\Delta G[/itex], and then solve for [itex]\Delta H[/itex] and [itex]\Delta S[/itex]? Hint: they've given you a piece of information that you haven't needed to use yet :)
  4. Feb 1, 2012 #3
    What do you mean by more values of deltaG ?
  5. Feb 1, 2012 #4
    Because there are two unknowns, [itex]\Delta H[/itex] and [itex]\Delta S[/itex], you need more than one value of [itex]\Delta G[/itex] (at least two), to get the solution.

    [itex]\Delta G = nFE[/itex], so you need to work out [itex]\Delta G[/itex] over a range of values of [itex]E[/itex]. Now, what does [itex]E[/itex] depend on?
  6. Feb 1, 2012 #5
    I don't understand how I would attempt solving for deltaG again once I've already found it. All of its variables are known, so how can I deviate?

    Any advice on deltaS and deltaH ?
  7. Feb 1, 2012 #6
    [itex]E[/itex] (and, therefore, [itex]\Delta G[/itex]) depends on temperature. So work out [itex]\Delta G[/itex] at some different temperatures, and you can solve the equation to get [itex]\Delta H[/itex] and [itex]\Delta S[/itex]
    Last edited: Feb 1, 2012
  8. Feb 1, 2012 #7
    I do not follow this logic.

    E is known
    Delta G is known

    Delta H and Delta S are functions of Delta G (Which is down).

    Why find another Delta G? You would find Delta G and Delta G whose delta (difference) seems to be irrelevant.

    Perhaps you can provide your proof?
  9. Feb 1, 2012 #8
    You need to solve the following equation: [itex]\Delta G = \Delta H - T \Delta S[/itex]. You know [itex]\Delta G[/itex] and [itex]T[/itex], which leaves two unknowns: [itex]\Delta H[/itex] and [itex]\Delta S[/itex].

    You cannot solve for two unknowns with a single equation: you need to have two sets of values. The way to do this is to solve the following set of simultaneous equations, using the value of [itex]\Delta G[/itex] and two different temperatures.

    [itex]\Delta G_{(1)} = \Delta H - T_{(1)} \Delta S[/itex]
    [itex]\Delta G_{(2)} = \Delta H - T_{(2)} \Delta S[/itex]

    The difference between these two [itex]\Delta G[/itex] values isn't negligible- [itex]\Delta G[/itex] is a temperature dependent quantity!
  10. Feb 1, 2012 #9
    Okay, wait a minute - I found a hint in the text that says to utilize the derivative of deltaG = deltaH - T*deltaS

    [itex]d \Delta G/dT = - \Delta S [/itex]
    [itex] d (nFE)/ dT = - \Delta S [/itex]
    [itex] nF * dE/dT = - \Delta S [/itex]
    [itex] 2(98485)(0.2*10^-3) = - \Delta S [/itex]
    [itex] - \Delta S = -39.394 [/itex]

    Does that seem feasible?
  11. Feb 1, 2012 #10
    Yes, that's another way of doing it :)
  12. Feb 1, 2012 #11

    I lost track of my units, do you have any insite as to what units my [itex] \Delta S [/itex] is in?
  13. Feb 1, 2012 #12
    You can work that out by looking at the units on the left hand side.

    The units of [itex]2F\frac{dE}{dT}[/itex] are [itex]C mol^{-1} V K^-1[/itex]. Can you simplify that a bit to get the more standard unit?
  14. Feb 1, 2012 #13
    Hello niallj, I am unable to simplify that unit. I'm supposed to wind up in either Joules or Watt hours.

    Can you assist with the conversion process? I wiki'd Joule and Watt (and Watt Hours) for their equivalent units with no luck.
  15. Feb 1, 2012 #14
    [itex]CV[/itex] is [itex]J[/itex] :)
  16. Feb 1, 2012 #15
    So I have J/(K mol) but when I calculated [itex]\Delta G[/itex] I only had units of J/mol --> Wh/mol.

    Does this K get cancelled out anywhere?
  17. Feb 1, 2012 #16
    That's right- remember that [itex]\Delta G = \Delta H - T \Delta S[/itex]. So it get's cancelled when you multiply entropy by temperature.
  18. Feb 2, 2012 #17


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    How about using Gibbs Helmholtz equation?