Standard Gibbs' energy change

In summary, the thermodynamics textbook states that the chemical potential can be represented by μ (condition-dependent), μ∇ (standard), μΘ (standard) (superscript plimsoll line), or μO (standard), with the latter three being used depending on the chosen temperature for defining the standard state. The notation typically used for ΔH under standard conditions is ΔHΘ, while for ΔG it is ΔGO. However, it may be more correct to use ΔGΘ, as ΔGΘ=ΔHΘ-T·ΔSΘ and the conditions are the same. The equilibrium constant is not necessarily 1 under standard conditions, as shown by
  • #1
Big-Daddy
343
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?
 
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  • #2
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.
 
  • #3
DrDu said:
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.

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°.
 
  • #4
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
 
  • #5
Big-Daddy said:
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°.
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
 
  • #6
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:
  • #7
DrDu said:
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.
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
 

1. What is Standard Gibbs' energy change?

Standard Gibbs' energy change, also known as standard free energy change, is the measure of the change in free energy that occurs in a chemical reaction under standard conditions. It is denoted by the symbol ΔG° and is a key factor in determining the spontaneity of a reaction.

2. How is Standard Gibbs' energy change calculated?

The standard Gibbs' energy change is calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change, T is the temperature in Kelvin, and ΔS° is the standard entropy change. This equation takes into account both the enthalpy and entropy changes that occur in a reaction.

3. What are standard conditions in relation to Standard Gibbs' energy change?

Standard conditions refer to a specific set of conditions that are used as a basis for comparison in thermodynamic calculations. These conditions include a temperature of 298 K (25°C), a pressure of 1 bar, and a concentration of 1 mol/L for all reactants and products in solution.

4. How does Standard Gibbs' energy change relate to the spontaneity of a reaction?

The sign of the standard Gibbs' energy change (ΔG°) indicates the spontaneity of a reaction. A negative ΔG° value indicates that the reaction is spontaneous, meaning it will proceed in the forward direction without the need for external energy. A positive ΔG° value indicates that the reaction is non-spontaneous and will only occur with the input of external energy.

5. Can Standard Gibbs' energy change be used to predict the extent of a reaction?

Yes, the magnitude of the standard Gibbs' energy change can indicate the extent to which a reaction will proceed. The larger the negative ΔG° value, the further the reaction will proceed towards completion. If ΔG° is close to zero, the reaction is in equilibrium, meaning the forward and reverse reactions are occurring at equal rates. A positive ΔG° value indicates that the reaction will not proceed to any significant extent under standard conditions.

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