I'm a beginner, learning things like entropy and gibbs free energy for the first time. I'm trying to derive the famous equation: $$ΔG=ΔH-TΔS$$ but it looks like I'm getting nowhere. I start by imagining a closed system that undergoes a chemical reaction in time $t$ during which it's pressure $P$ and $T$ only change negligibely. I assume that the system is constantly close enough to a thermodynamic equilibrium for entropy to be definable. So I tried calculating $ΔG$. I know that it's defined to be the total non-mechanical work that gets done on the system during the reaction (such that the work is considered negative if it being done by the system on the surrounding). Using the 1st law of thermodynamics, I know that the total work done on the system $ΔU$ is just equal to $Q+W$. And since $W$ is the mechanical work done on the system,$$ ΔG=ΔU-W=Q$$ Now let's calculate $ΔH$ (note that symbols like $H_0$ just mean the enthalpy at time '0'). $$ΔΗ=H_t-H_0=(U_t+PV_t)-(U_0+PV_0)=ΔU+PΔV=ΔU-W=Q$$ so $$ΔΗ=Q$$, and substituting this back into our formula that we derived for $ΔG$, we get $ΔΗ=ΔG$! which totally excludes the TΔS present in the actual formula. I tried to figure this out using the Wikipedia article on Gibbs energy but the derivation there was not understandable to me. To clarify my level, I can easily do basic calculus (including things like vector calculus and understanding Volume integrals) but little more. So I'm really looking forward to a correction of my mistake, and derivation of the true formula at my level.
2 Answers
I'm trying to derive the famous equation: ΔG=ΔH−TΔS
This isn’t a derivable equation but the Gibbs free energy definition ($G\equiv H-TS=U+PV-TS$) applied to a scenario of constant temperature.
So what went wrong in the later reasoning? This part:
Using the 1st law of thermodynamics, I know that the total work done on the system ΔU is just equal to Q+W.
The total work done on a system is $W$. The nonmechanical work done on a system at constant pressure is $W_\text{nonmechanical}=W+P\Delta V=\Delta U-Q+P\Delta V$. At constant temperature as well, $Q=T\Delta S$. So one obtains $W_\text{nonmechanical}=\Delta G$ in your scenario, as expected.
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The formula you are looking to derive is actually an immediate consequence of the definition of the Gibbs free energy. To give some background context and motivation, I will show why the enthalpy and Gibbs free energy are defined the way they are via some differential forms. The Gibbs free energy is defined via the differential Legendre transform of the enthalpy, which is itself a Legendre transform of the internal energy. The fundamental differential relation, which is intimately related to the 1st law of thermodynamics, says that the internal energy for a system with a fixed amount of substance varies as, \begin{gather*} dE = TdS - PdV \end{gather*} meaning that internal energy is a natural function of entropy and volume. We can define the enthalpy via the relations, \begin{align*} H &= E + PV \\ \implies dH &= dE + PdV + VdP \\ &= TdS - PdV + PdV + VdP \\ &= TdS + VdP \end{align*} which is manifestly a function of entropy and pressure. Notice the definition was carefully chosen so that some terms would cancel out when we wrote the differential! The Gibbs free energy is then, \begin{align*} G &= H - TS \\ \implies dG &= dH - TdS - SdT \\ &= TdS + VdP - TdS - SdT \\ &= VdP - SdT \end{align*} which is naturally a function of pressure and temperature. If we consider a system that has a constant temperature, we could have written the differential form for the Gibbs free energy instead as, \begin{align*} dG &= dH - TdS \\ \implies \int dG &= \int dH - T\int dS \\ \implies \Delta G &= \Delta H - T \Delta S \end{align*} where we have integrated the differential to produce the formula you were looking for. Note that this formula is strictly only valid when the temperature of the system is fixed! The study of thermodynamics is fundamentally based on multivariable calculus, so if this seems like a leap above where you were at before, that's the reason!
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