Standard Gibbs' energy change

  1. My thermodynamics textbook says that chemical potential can take any of the following symbols: μ (condition-dependent), μ (standard), μΘ (standard) (superscript plimsoll line) and μO (standard), with which one of the latter 3 being used depending on the temperature chosen to define standard state.

    I have seen that the notation typically used for ΔH under standard conditions of 298.15 K, 1 bar pressure is the ΔHΘ symbol (and ΔSΘ similarly). However, ΔGO is used to denote ΔG under standard conditions of 298.15 K, 1 bar pressure. Should this more correctly be ΔGΘ (so we write ΔGΘ=ΔHΘ-T·ΔSΘ, so the conditions are the same)? Or perhaps they are all actually O?

    Also, does ΔGΘ for a reaction actually refer to the value of ΔG for that reaction under standard conditions of 298.15 K, 1 bar pressure? If so, wouldn't that mean that the equilibrium constant must be 1 under standard conditions (which is not necessarily true), as a result of ΔG=ΔGΘ+R·T·loge(Q) and now ΔG=ΔGΘ under such conditions? If not, what is the actual definition of ΔGΘ, and with regards to ΔG itself?
  2. jcsd
  3. DrDu

    DrDu 4,639
    Science Advisor

    To your second question. Yes, K=1 at the standard state, but this is not an equilibrium situation and the standard state is often only a theoretical construct which is impossible to realize experimentally.
    E.g. you can't have a mixture of hydrogen and oxygen with partial pressure 1 bar of each component and total pressure being 1 bar, too.
  4. I'm getting confused now. How exactly is ΔGr° to be defined? I know that ΔGr°=ΔHr°-T*ΔSr°. But doesn't the ° symbol mean "standard conditions of 298.15 K and 1 bar pressure" or not - because if so then we may as well write ΔGr°=ΔHr°-298.15*ΔSr° and I'm sure this is wrong, because now let's look for the temperature at which K=1 (which should, according to what I showed in my OP, be 298.15 K, but is as I will show now indeterminate) through ΔGr°=-R*T*loge(K) and now ΔHr°-298.15*ΔSr°=-R*T*loge(K) and our system crashes: T=(ΔHr°-298.15*ΔSr°)/(-R*loge(K)), and far from 298.15 K, our answer will be undefined for K=1.

    So the question becomes - what actually is ΔGr° with relation to ΔGr? Evidently it is incorrect to say "standard conditions of 298.15 K and 1 bar pressure" alone because clearly ΔGr° does exhibit temperature dependency due to the T variable in ΔGr°=ΔHr°-T*ΔSr°.
  5. DrDu

    DrDu 4,639
    Science Advisor

    I would have to look up the details myself.
    A very good reference on standard states is:

    Klotz, Irving M. / Rosenberg, Robert M.
    Chemical Thermodynamics
    Basic Concepts and Methods
  6. ΔG° = ΔH°-TΔS° This is the free energy change with standard conditions @ 298 K from elements
    in their standard states. Where ΔG = ΔH-TΔS is for calculating when ΔG becomes negative with
    increasing temperature. Also regarding solving for T when K =1 , its my understanding that its
    when ΔG = 0 that K =1
  7. DrDu

    DrDu 4,639
    Science Advisor

    If K=1 the logarithm vanishes and ##\Delta G=\Delta G^0##. K=1 in general does not correspond to an equilibrium situation (even less at fixed temperature so ##\Delta G## won't be 0. In contrast, ##\Delta G^0=-RT \ln K## holds only in equilibrium.
    Last edited: Apr 11, 2013
  8. In post # 3 Big-Daddy is solving for T with ΔG° = ΔH° - TΔS° and ΔG°(298K) = -RTlnK
    Standard free energy changes @ 298K. And in ΔG° = -RTlnK 'K' can have different values
    and when K = 1 , ΔG° = 0 Not questioning your last post - just for clarification
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