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    I read everywhere that for a reaction to be spontaneous, total entropy must increase.

    But surely it must be possible for a reaction to be spontaneous even with a decrease in entropy, provided it is sufficiently exothermic?

    Since

    \Delta G=\Delta H-T\Delta S

    so even with a negative \Delta S as long as \Delta H is more negative than T\Delta S the reaction should be spontaneous?

    Can anyone clarify for me?
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    A reaction is spontaneous when the Gibbs free energy for the reaction is negative \Delta G < 0, entropy as you say is only part of that :yes:
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    (Original post by EierVonSatan)
    A reaction is spontaneous when the Gibbs free energy for the reaction is negative \Delta G < 0, entropy as you say is only part of that :yes:
    Right. That's what I thought. So why does it say everywhere eg here, that delta S has to be positive?

    According to the second law of thermodynamics, when a spontaneous process occurs, there must be an increase in total entropy.
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    (Original post by DonnieBrasco)
    Right. That's what I thought. So why does it say everywhere eg here, that delta S has to be positive?
    Because they are talking about the total entropy of the Universe, not just the entropy of the system in question.
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    (Original post by DonnieBrasco)
    Right. That's what I thought. So why does it say everywhere eg here, that delta S has to be positive?
    The total entropy is a combination of the entropy of the system (the reaction) and that of the surroundings - so it's

    The most annoying thing about thermodynamics (for me) is all the detail contained in the definitions of each of the terms used - you have to be very careful :p:
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    (Original post by charco)
    Because they are talking about the total entropy of the Universe, not just the entropy of the system in question.
    so in the gibbs free energy equation I quoted at the top, is that just \Delta S_{system}?

    How do we define what is the system and what is the surroundings?
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    (Original post by DonnieBrasco)
    so in the gibbs free energy equation I quoted at the top, is that just \Delta S_{system}?

    How do we define what is the system and what is the surroundings?
    It's like EVS said:

    S(universe) = S(system) + S(surroundings)

    Therefore:

    ΔS(universe) = ΔS(system) + ΔS(surroundings)

    The universe can be thought of as being composed of the system under study and the surroundings (everything else).

    Gibbs free energy refers to the system under study. The system effectively means the matter of the chemical compounds.

    The derivation of GFE involves the total energy of the universe and effectively GFE measures that part of the system, which could be used to do useful work. Hence if GFE decreases it means that the entropy of the whole universe increases.

    So if GFE is negative the entropy change in the universe must be positive, i.e. feasible.

    If you consider the consequence of a change in enthalpy...

    Where does the energy go to in an exothermic reaction?

    It goes to the surroundings, changing the number of possible ways that the energy can be arranged over the particles and increasing the entropy of the suroundings. This is why exothermic reactions are (usually) feasible.

    The only other way of increasing the entropy of the universe is to increase the actual number of particles in the system.

    Hence GFE involves two terms, ΔH and TΔS

    The first term relates to the entropy change of the surroundings (and hence the universe) and the second term relates to the entropy change of the system (and hence the universe)
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    (Original post by charco)
    It's like EVS said:

    S(universe) = S(system) + S(surroundings)

    Therefore:

    ?S(universe) = ?S(system) + ?S(surroundings)

    The universe can be thought of as being composed of the system under study and the surroundings (everything else).

    Gibbs free energy refers to the system under study. The system effectively means the matter of the chemical compounds.

    The derivation of GFE involves the total energy of the universe and effectively GFE measures that part of the system, which could be used to do useful work. Hence if GFE decreases it means that the entropy of the whole universe increases.

    So if GFE is negative the entropy change in the universe must be positive, i.e. feasible.

    If you consider the consequence of a change in enthalpy...

    Where does the energy go to in an exothermic reaction?

    It goes to the surroundings, changing the number of possible ways that the energy can be arranged over the particles and increasing the entropy of the suroundings. This is why exothermic reactions are (usually) feasible.

    The only other way of increasing the entropy of the universe is to increase the actual number of particles in the system.

    Hence GFE involves two terms, ?H and T?S

    The first term relates to the entropy change of the surroundings (and hence the universe) and the second term relates to the entropy change of the system (and hence the universe)
    Brilliant! Thanks SO much.

    So, basically, my system can have a negative entropy change and still be spontaneous, provided it's sufficiently exothermic (because that will release heat to the surroundings and raise the overall entropy)?

    (I tried to rep you, but it won't let me)
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    (Original post by DonnieBrasco)
    Brilliant! Thanks SO much.

    So, basically, my system can have a negative entropy change and still be spontaneous, provided it's sufficiently exothermic (because that will release heat to the surroundings and raise the overall entropy)?
    :borat:
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    (Original post by EierVonSatan)
    A reaction is spontaneous when the Gibbs free energy for the reaction is negative \Delta G < 0, entropy as you say is only part of that :yes:
    When delta G is negative reaction is feasible ! is there any reason for it!!!??:rolleyes:
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    (Original post by arvin_infinity)
    When delta G is negative reaction is feasible ! is there any reason for it!!!??:rolleyes:
    Did you read this whole thread carefully?
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    (Original post by DonnieBrasco)
    Brilliant! Thanks SO much.

    So, basically, my system can have a negative entropy change and still be spontaneous, provided it's sufficiently exothermic (because that will release heat to the surroundings and raise the overall entropy)?

    (I tried to rep you, but it won't let me)
    There's a pretty simple reasoning behind it all too:

    dS_\text{sys} = \frac{\delta q_\text{rev,sys}}{T}

    This means that a small change in entropy of the system is equal to the change in heat flowing reversibly into the system divided by the temperature. The reversible bit just means that it happens really slowly.

    Now all the heat flowing into or out of a system has to come from the surroundings, because of conservation of energy. Our system is also much smaller than the surroundings, since they're the rest of the Universe. This means that the change in energy is pretty much reversible as far as the surroundings are concerned, since it's such a small change compared to the overall heat in the Universe.

    So, we get:

    \delta q_\text{sys} = -\delta q_\text{rev,surr}

    So, the change in the entropy of the surroundings is given by:

    \Delta S_\text{surr} = -\frac{\delta q_\text{sys}}{T}

    This means that the entropy change of the whole Universe, including both system and surroundings, is:

    \Delta S_\text{Univ} = \frac{\delta q_\text{sys,rev}}{T} -\frac{\delta q_\text{sys}}{T}

    These are subtly different, since one is reversible and the other one isn't. But, reversible changes in heat are always bigger than irreversible changes, so the overall change must be greater than or equal to zero.

    This links to Gibbs energy because:

    \Delta G = \Delta H - T \Delta S

    We already have the expression for the entropy change of the system:

    \Delta G = \Delta H - T \frac{\delta q_\text{rev,sys}}{T}

    Next, we use that enthalpy change is actually just a change in heat (when you keep the pressure constant, like in a normal experiment in the lab):

    \Delta H = \delta q_\text{sys}

    Subbing this in, we get:

    \Delta G = \delta q_\text{sys} - T \frac{\delta q_\text{rev,sys}}{T}

    If we look carefully, you can see that this is a factor of -T different from our expression for the total entropy of the Universe:

    \Delta G = -T \frac{\delta q_\text{rev,sys}}{T} + T \frac{\delta q_\text{sys}}{T} = -T \Delta S_\text{Univ}

    So, for our spontaneous reaction, we have that:

    - The entropy change of the Universe must be greater than or equal to zero.
    - Temperature must be greater than or equal to zero (measured in Kelvins).
    - The minus sign.

    This means that for a spontaneous reaction, \Delta G \le 0

    I know this is a bit beyond your course, but if you ignore the stuff about reversibility and the subtleties of different types of deltas, I think it's quite a nice derivation that explains why we relate Gibbs free energy of the system and entropy of the Universe together.
 
 
 
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