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$\Delta G=\Delta G^o+RT\ln Q?$ In this equation, what does $\Delta G$ mean? Is $\Delta G$ the change in gibbs free energy if we complete the reaction (convert all products to reactants)? That doesn't seem right as it depends on $Q$ which changes as the reaction goes forward My book doesn't seem of offer a better explanation, it just explains that if $\Delta G$ is negative then the reaction is thermodynamically possible

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The $\Delta G$ you quote is the slope of the free energy vs extent of reaction. When the reaction comes to equilibrium, which we will assume it does at some extent of reaction, then the slope of free energy vs extent of reaction is zero, $\Delta G=0$ and we can replace $Q$ with the equilibrium constant $K$, hence $\Delta G^o=-RT\ln(K)$. This means that if $K>1$ then $\Delta G^o$ is negative, i.e. more product than reactant and vice versa if $K<1$.

If the reaction is $A=B$ we start with A only and so to get to equilibrium $G$ must decrease, as equilibrium is the most stable place to be, i.e. lowest energy. The decrease is due largely to the increase in entropy due to the fact that once A reacts it forms B and so there is now a mixture of A and B and so entropy has increased. This always happens.

At equilibrium the enthalpy $H$ and entropy $S$ are also important as $\Delta G^o=\Delta H^o-T\Delta S^o$. The entropy change in the reaction itself could be small, for example the reaction might be an isomerisation or it could be large. These factors (H and S) will change the standard free energy $\Delta G^o$ making $K$ greater or less than $1$ and so change the position of the equilibrium as a fraction of the extent of reaction. If the equilibrium position lies with an extent of reaction more that $0.5$ then A has a bigger free energy than B and vice versa.

(Finally recall that time does not enter into thermodynamics, even if a reaction appears not to happen this does not matter, the position of equilibrium can be found. Thermodynamics only informs us about starting and ending points.)

re extent of reaction. The extent of reaction $\zeta$ is zero when only A is present (i.e. reaction totally to the left) and $1$ when one mole of reactant has reacted to form product, then $\displaystyle \Delta G=\left(\frac{\partial G}{\partial \zeta}\right)_{T,p}$. This defines the $\Delta G$ in your equation.

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  • $\begingroup$ Δ𝐺 is not the slope of the free energy vs extent of reaction, as you state. This term (extent of reaction) is not defined with enough precision. For me, Δ𝐺 is the result of the product - z·E·F where E is the potential difference measured when the reaction is going, i.e. when the cell is only partially discharged, F is the faraday, and z is the number of electrons exchanged in the redox equation of the reaction. When at equilibrium, the reaction stops, E is zero, and Δ𝐺 is also zero. So : Δ𝐺° = - RT ln K $\endgroup$
    – Maurice
    Apr 2 at 19:50
  • $\begingroup$ No, you are not correct, and this is not restricted to a cell it is quite general. The extent of reaction $\zeta$ is zero when only A is present (i.e. reaction totally to the left) and 1 when 1 mole of reactant has reacted to form product, then $\displaystyle \Delta G_{reaction}=\left(\frac{\partial G}{\partial \zeta}\right)_{T,p}$ $\endgroup$
    – porphyrin
    Apr 3 at 7:40
  • $\begingroup$ In fact, the reaction extent is much better defined than the measured potential difference that depends on other, badly defined, mainly kinetic factors. It can be even not measurable or not making sense for non-redox reactions. // The reaction delta G could be also expressed via signed stoichiometric coefficients, chemical potentials and amount differentials. $\endgroup$
    – Poutnik
    Apr 3 at 8:42

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