Consider a chemical reaction like

$$ (-\nu_1) R_1 + (-\nu_2) R_2 \Leftrightarrow \nu_3 R_3 + \nu_4 R_4 $$

with the Gibbs free energy

$$ dG = -SdT + VdP + \sum_i \mu_i dN_i \, . $$

The amounts of each reagent are given by $$ N_i(\xi) = N_i^{(0)} + \nu_i \xi $$

yielding an overall Gibbs free Energy like, $$ dG = -SdT + VdP + \left(\sum_i \mu_i \nu_i\right) d\xi \, . $$

$ \xi $ is the extent to which the reaction has taken place which I expect to have a temperature and pressure dependence (eg. water dG vs. T)

How do you solve for this dependence?

  • 1
    $\begingroup$ Suppose that the temperature and pressure are controlled so that they are held constant as the reaction mixture is approaching equilibrium. That tells you that, at equilibrium, the G of the mixture is minimized. $\endgroup$ – Chet Miller Jul 17 '17 at 12:15
  • $\begingroup$ Yes, but it doesn't tell me what the $\xi$ of equilibrium is. I'd like to know how fast the reaction would progress. I guess I'm assuming there's some point when the reaction is no longer favorable. $\endgroup$ – ignorance Jul 17 '17 at 12:34
  • $\begingroup$ You have $\sum{\nu_i \mu_i}=0 $, and $\mu_i$ is a function of $\xi$ $\endgroup$ – Chet Miller Jul 24 '17 at 14:58
  • $\begingroup$ @ChesterMiller, how do you get the dependance? Do you have to look it up in a table or use Stat Mech to get the partition function? The tables for Gibbs free energy of formation en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation (which I assume are the $\mu$'s) don't include a dependance on any of the $ T, P, \mu $ like I would expect physics would depend on. $\endgroup$ – ignorance Jul 24 '17 at 16:57
  • $\begingroup$ Would you be willing to settle for the chemical potential equation for an ideal gas species in terms of its partial pressure? $\endgroup$ – Chet Miller Jul 24 '17 at 18:31

I think I was over thinking it. It doesn't seem that difficult, but let me illustrate with an example.

Given we are interested in 4 species of particles, we would expect a total Hamiltonian like, \begin{align} \mathcal{H}\big(\mathbb{Q}, \mathbb{P}; \vec{N}\big) &= \sum_{i=1}^4 \mathcal{H}_i\big(\mathbb{Q}_i \subset \mathbb{Q}, \mathbb{P}_i \subset \mathbb{P} ; \vec{N}_i\big) + U\big(\mathbb{Q}, \mathbb{P}\big) \end{align} where $ | \mathbb{Q}_i | = | \mathbb{P}_i | = N_i $.

Thinking about a single species system, computing the Isothermal-Isobaric ensemble gets you the Gibbs free energy, \begin{aligned} G(T, P, N) &= -k T\log \Delta (N,P,T) \\ &= -k T \log \left[ {\frac {C}{h^{3}}}\int {\mathrm {e}}^{{-\beta (H(q,p) + PV)}}~d^{{3N}}q~d^{{3N}}p~dV \right] \end{aligned}

If you compute this for a multi-species particle, then you can get $ \sum_i \mu_i \nu_i = \sum_i \frac{dG}{dN_i}\frac{dN_i}{d\xi} = 0 $ as the chemical equilibrium point as a function of $ P, T $.

Ideal Gas Example

For an ideal gas (see Section 2.4 these lecture notes), \begin{align} G(T, P, N) = -N k_B T \log \left[ \frac{k_B T}{P} \cdot \frac{(2\pi m k_B T)^{3N/2}}{h^{3N}} \right] \end{align} in chemistry, its written like $ G(T, P, N) = G(P_0) + nRT \ln\left(\frac{P}{P_0}\right) $. I'm ignoring the fact that this differs from the Stat Mech version and just using the stat mech version.

Going back to the multi-species problem, if we drop the interaction term $ U\big(\mathbb{Q}, \mathbb{P}\big) $ and assume an ideal gas, then \begin{align} G(T, P, N) &= -\sum_i N_i k_B T \log \left[ \frac{k_B T}{P} \cdot \frac{(2\pi m_i k_B T)^{3N_i/2}}{h^{3N_i}} \right] \\ &= \sum_i G_i \end{align} I am just taking $ P, T $ to be the same for each species because they are Lagrange multipliers and Laplace transform variables, so having multiple doesn't make much sense to me

Then, \begin{align} \sum_i \mu_i \nu_i = \sum_i \frac{dG_i}{dN_i}\frac{dN_i}{d\xi} = - \sum_i \nu_i k_B T \log \left[ \frac{k_B T}{P} \right] + 2 \big( N^{(0)}_i + \nu_i \xi \big) k_B T \log \left[\frac{(2\pi m_i k_B T)^{3/2}}{h^{3}} \right] \end{align}

Setting this equal to zero, taking $ m_i = m $, and the translation partition function $ {\displaystyle \Lambda ={\frac {h}{\sqrt {2\pi mk_{B}T}}}} $, \begin{align} \xi = \frac{\sum_i \log \left[ \left(\frac{k_B T}{P}\right)^{\nu_i} \Lambda^{6 N^{(0)}_i} \right]}{\sum_i \log \left[\Lambda^{-6 \nu_i } \right]} \end{align}

  • $\begingroup$ "I am just taking P,T to be the same for each species because they are Lagrange multipliers and Laplace transform variables, so having multiple doesn't make much sense to me" Love that sentence. Just shows the difference between thinking like a experimentalist and a theoretician. $\endgroup$ – user224659 Jan 10 '20 at 6:52

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