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As I understood $\Delta G^0$ is just $\Delta G$ but at standard conditions i.e. $\pu{1 bar}$ and $\pu{298 K}$. But formulas like:

$\Delta G^0 = \Delta H^0 - T \Delta S^0$

$\Delta G^0 = - RT \ln K$

make me uncertain about that as these are dependent on $T$ and don't seem to be used at standard conditions only i.e. $T = \pu{298 K}$.

Could someone explain to me what the difference is between $\Delta G$ and $\Delta G^0$?

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$\Delta G^0$ is change in free energy change in going reversibly from molar stoichiometric quantities of pure reactants, each at temperature T and pressure 1 bar, to corresponding stoichiometric quantities of pure products, each at temperature T and pressure 1 bar. So $\Delta G^0$, $\Delta H^0$, and $\Delta S^0$ are all functions of T, but not pressure (which is fixed in the initial and final states as 1 bar). The values of these function changes tabulated in your text book are all for 298 K. But values at other temperatures can be obtained from integration, for example, in the case of $\Delta G^0(T)$ using the van't Hoff equation.

$\Delta G$ is change in free energy change in going reversibly from molar stoichiometric quantities of pure reactants, each at temperature T and its own arbitrary pressure, to corresponding stoichiometric quantities of pure products, each at temperature T and its own arbitrary pressure. If these arbitrary pressures happen to match the partial pressures of the same reactants and products in an equilibrium mixture, then $\Delta G$ will turn out equal to zero. Otherwise, $$\Delta G=\Delta G^0+RT\ln(Q)=RT\ln{(Q/K)}$$and, for equilibrium, Q=K.

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A wonderful explanation by All difference between follows:

1. $\Delta$ G relate to these two quantities ΔH° and ΔS° while $\Delta$ G° does not relate to the quantities:

$\Delta$ G definitely does relate to those two quantities. ΔH° and ΔS° represent the change in enthalpy and entropy between product and reactant but they do not mean a “100% complete reaction.” They determine the energetic difference (at a given temperature). This difference in energy determines the composition at equilibrium

2. $\Delta$ G is not a non-zero value while $\Delta$ G° is a non-zero value:

$\Delta$ G° is a non-zero value and it can’t use ΔH° or ΔS° to find $\Delta$ G because either ΔH° or ΔS° represents 100% complete reaction.

3. At equilibrium $\Delta$ G is equal to zero while $\Delta$ G° is negative:

If $\Delta$ G° is negative at equilibrium, then we will have lots of products at equilibrium, which means that Q must be greater and also greater than 1 to approximate K. When Q gets bigger, it means more product is accumulated. The term ‘$RT \ln Q$’ gets increasingly positive, and eventually adding that term to a negative $\Delta$ G°, will make $\Delta$ G = 0, equilibrium will be established and no further change occurs.

4. Equilibrium is established when $\Delta$ G is at zero while lots of reactant at equilibrium when $\Delta$ G° is positive:

If $\Delta$ G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K. As Q gets smaller (i.e., as we get more reactants), the term ‘$RT \ln Q$’ gets increasingly negative and eventually adding that term to a positive $\Delta$ G°, will make $\Delta$ G = 0, equilibrium will be established and no further change occurs.

5. When $\Delta$ G is zero at equilibrium, it will define which way the reaction proceeds while $\Delta$ G° does not:

Note that it is $\Delta$ G and not $\Delta$ G° that will be zero at equilibrium and the sign of it generated by the combination of $\Delta$ G° and $RT \ln Q$, will define which way the reaction proceeds.

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  • $\begingroup$ Next time when you quote an answer from a site, atleast make sure the notation used is correct "in $\neq$ $\ln$".. $\endgroup$ Aug 30 '20 at 20:03
  • $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ Aug 30 '20 at 20:03

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