A Zephyr in Time: An Equation for Gibbs Free Energy

One very useful equation in second-year thermodynamics class is

$\Delta G = \Delta G^\circ + RT\ln{Q}$

where ΔG is the Gibbs free energy for the process, ΔG^o is the standard Gibbs free energy for the process, Q is the reaction quotient. The process takes place under isothermal and isobaric conditions.

Over at Matter Waves is a request for the proof. So here it is. Feel free to point out the mistakes.

For a k-component system of ideal gases, the total differential for G can be written as

$\begin{eqnarray*} dG & = & \left(\frac{\partial G}{\partial T}\right)_{P,\{ n_{j}\}}dT+\left(\frac{\partial G}{\partial P}\right)_{T,\{ n_{j}\}}dP+\sum_{i}\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,\{ n_{j}\}_{j\not=i}}dn_{i}\end{eqnarray*}$

where the cut off part is $\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,\{ n_{j}\}_{j\not=i}}$ *dn_i, n_j is the number of moles of component j, $\{ n_{j}\}$ is shorthand for $\{ n_{1},\ldots,n_{k}\}$ , and $\{ n_{j}\}_{j\not=i}$ is shorthand for $\{ n_{j}\}\backslash\{ n_{i}\}$ ( $\{ n_{i}\}$ being a singleton). Because we are too lazy to write $\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,\{ n_{j}\}_{j\not=i}}$ too many times, define the chemical potential of species i, $\mu_{i}$ , as

$\begin{eqnarray*} \mu_{i} & = & \left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,\{ n_{j}\}_{j\not=i}}\end{eqnarray*}$

Under isothermal and isobaric conditions, we get

$\begin{eqnarray*} dG & = & \sum_{i}\mu_{i}dn_{i}\end{eqnarray*}$

Let us go to a diversion and consider just a one-component system (say the ith one) for now. If we write $G_{i,m}$ as the molar Gibbs free energy of this ith component (as a function of P,T
only since this is a one-component system), we have

$\begin{eqnarray*} G & = & n_{i}G_{i,m}(P,T)\\ \mu_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P} & = & \frac{d}{dn_{i}}[n_{i}G_{i,m}(P,T)]=G_{i,m}(P,T)\end{eqnarray*}$

Using the Gibb's Equation dG = SdT - VdP under isothermal conditions, (and dropping the subscript i meaning "i-th component" for laziness' sake)

$\begin{eqnarray*} dG & = & -SdT+VdP\\ dG_{m} & = & -S_{m}dT+V_{m}dP=V_{m}dP\textrm{ since }dT=0\\ d\mu & = & V_{m}dP=\frac{RT}{P}dP\\ \int_{P_{i}}^{P_{f}}d\mu & = & \int_{P_{i}}^{P_{f}}\frac{RT}{P}dP\\ \mu(T,P_{f})-\mu(T,P_{i}) & = & RT\ln\frac{P_{f}}{P_{i}}\\ \mu(T,P) & = & \mu(T,P_{i})+RT\ln\frac{P}{P_{i}}\end{eqnarray*}$

where I wrote P instead of $P_{f}$ because I don't know what $P_{f}$ will be for a particular situation (the cut-off part just says "since dT = 0). For convenience, set $P_{i}$ (which, because I am abusing notation, means initial pressure, by the way) to some standard pressure. The Powers that Be who defined "standard state" says 1 bar sounds good, so we'll go with that. Let $P^{\circ}=1\textrm{ bar}$ . Thus

$\begin{eqnarray*} \mu(T,P) & = & \mu(T,P^{\circ})+RT\ln\frac{P}{P^{\circ}}=\mu^{\circ}+RT\ln\frac{P}{P^{\circ}}\end{eqnarray*}$

where $\mu^{\circ}$ is the chemical potential when the single component is at standard state.

Let us come back to our k-component system. If we have the chemical reaction

$\begin{eqnarray*} \sum_{i=1}^{l}n_{i}A_{i} & \rightleftharpoons & \sum_{i=l+1}^{k}n_{i}A_{i}\end{eqnarray*}$

(or, sligtly less abstractly, something like
$\begin{eqnarray*} aA+bB & \rightleftharpoons & cC+dD\end{eqnarray*}$ ) we can write the equation in a really weird way:

$\begin{eqnarray*} 0 & = & \sum_{i=l+1}^{k}n_{i}A_{i}-\sum_{i=1}^{l}n_{i}A_{i}\end{eqnarray*}$

(or cC + dD - aA - bB = 0, so to speak). To simplify notation, define

$\begin{eqnarray*} \nu_{i} & = & \left\{ \begin{array}{rc} n_{i} & \textrm{if component }i\textrm{ is a product}\\ -n_{i} & \textrm{if component }i\textrm{ is a reactant}\end{array}\right.\end{eqnarray*}$

so that the weird chemical equation becomes

$\begin{eqnarray*} 0 & = & \sum_{i=1}^{k}\nu_{i}A_{i}\end{eqnarray*}$

Notice that $\int_{n_{i}}^{0}dn_{i}=\nu_{i}$ (for a reactant), $\int_{0}^{n_{f}}dn_{i}=\nu_{i}$ (for a product). Coming back to the expression $dG={\displaystyle \sum_{i}\mu_{i}dn_{i}}$ , and noting that the "final pressure" of component i is $P_{i}$ , the partial pressure of component i at equilibrium:

$\begin{eqnarray*} dG & = & \sum_{i}\mu_{i}dn_{i}\\ \Delta G & = & \sum_{i}\nu_{i}\mu_{i}(T,P)\\ & = & \sum_{i}\nu_{i}\left(\mu_{i}^{\circ}+RT\ln\frac{P_{i}}{P^{\circ}}\right)\\ & = & \sum_{i}\nu_{i}\mu_{i}^{\circ}+\sum_{i}\nu RT\ln\frac{P_{i}}{P^{\circ}}\\ & = & \Delta G^{\circ}+RT\ln\prod_{i}\left(\frac{P_{i}}{P^{\circ}}\right)^{\nu_{i}}\end{eqnarray*}$

The product $\prod_{i}\left(\frac{P_{i}}{P^{\circ}\ }\right)^{\nu_{i}}=Q_{p}$ is defined as the reaction quotient of the reaction. And now we have what we desire:

$\begin{eqnarray*} \Delta G & = & \Delta G^{\circ}+RT\ln Q_{p}\end{eqnarray*}$

Some might say that $Q_{p}$ is supposed to be some ratio of partial pressures, not that weird product thing. Well. Read the left side of the definition of $Q_{p}$ really carefully. The fraction $\frac{P_{i}}{P^{\circ}}$ ,
because $P^{\circ}=1\textrm{ bar}$ , is basically $P_{i}$ without the unit "bar''. Also, $\nu_{i}$ is positive for a product, and negative for a reactant. So if our reaction is the simpler
$\begin{eqnarray*} aA+bB & \rightleftharpoons & cC+dD\end{eqnarray*}$
we get

$\begin{eqnarray*} Q_{p} & = & \left(\frac{P_{A}}{P^{\circ}}\right)^{-a}\left(\frac{P_{B}}{P^{\circ}}\right)^{-b}\left(\frac{P_{C}}{P^{\circ}}\right)^{c}\left(\frac{P_{D}}{P^{\circ}}\right)^{d}\\ & = & \frac{\left(\frac{P_{C}}{P^{\circ}}\right)^{c}\left(\frac{P_{D}}{P^{\circ}}\right)^{d}}{\left(\frac{P_{A}}{P^{\circ}}\right)^{a}\left(\frac{P_{B}}{P^{\circ}}\right)^{b}}\end{eqnarray*}$

Ah, there we go. That looks better.

It's pretty straightforward to convert $Q_{p}$ to other reaction quotients (e.g. in terms of mole fractions $\chi_{i}$ , or concentrations $C_{i}$ , instead of $P_{i}$ ), assuming ideality. We can also complain
that in real life gases aren't ideal, and fix this with an experimental "fudge factor", γ :

$\begin{eqnarray*} f_{i} & = & \gamma_{i}P_{i}\end{eqnarray*}$

We call $f_{i}$ the fugacity of component i.

For liquids and solids, the argument is usually that the activity of a liquid or solid is always 1, so they don't show up in the reaction quotient. Now if you ask me why can we adapt this equation for Q when we have solutions, well, I don't know. My guess would be a really handwavy explanation that molecules in (dilute) solutions behave much like ideal gas particles (i.e. satisfying the postulates of the kinetic molecular model). Or we can use that $V_{i,m}={\displaystyle \frac{V}{n_{i}}=\frac{1}{C}}$ , but that's missing the RT part.

A Zephyr in Time

Thursday, July 24, 2008

An Equation for Gibbs Free Energy

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