I encountered following reaction:

$$\ce{CO + 1/2 O2 -> CO2}$$

I am provided with the composition of the various gases at $\pu{1000K}$ and am asked the direction of the reaction by finding the affinity of the reaction. I know the way to find affinity is:

$$A= \mu_\ce{CO2}-\mu_\ce{CO} -\frac{1}{2}\mu_\ce{O2}$$

I know the change in chemical potential considering ideal mixing. But how do I go about finding the absolute value of chemical potential, rather than the change, because the expression requires the chemical potential itself rather than the change? I thought of writing $A$ as $-\Delta G$, but for that also I need some reference value of $G$ and from there I can integrate $\Delta H-T\Delta S$, but again I can't figure out how reference state is chosen?

Please suggest something.

  • $\begingroup$ Potentials are standard not "absolute"... $\endgroup$ – Mithoron Mar 28 '18 at 14:58
  • $\begingroup$ I meant the "actual value of chemical potential". Suggest something. $\endgroup$ – Ankur Singh Mar 28 '18 at 15:00

Firstly, since $\overline{\mu_\alpha} = \overline{g_\alpha}$, the two approaches that you propose are basically the same: $\overline{\mu_\ce{CO2}}-\overline{\mu_\ce{CO}} -\frac{1}{2}\overline{\mu_\ce{O2}}$ is the same as the change in $G$ per mol of forward reactions.

Let's suppose that we can calculate the true, absolute value of the molar Gibbs energy at some reference physical and chemical state, labelled $^*$. The molar Gibbs energy at any arbitrary state is then

$$ \overline{g} = \overline{g}^* + \Delta \overline{g}^\text{chem} + \Delta \overline{g}^\text{phys} $$

Where the to $\Delta g$ terms account for the fact that the state of interest may have a different chemical and/or physical state than the reference state. If we substitute this back into your expression, we find

$$\begin{aligned} A &=\overline{\mu}_\ce{CO2}-\overline{\mu}_\ce{CO} -\frac{1}{2}\overline{\mu}_\ce{O2} \\ &=\overline{g}_\ce{CO2}-\overline{g}_\ce{CO} -\frac{1}{2}\overline{g}_\ce{O2} \\ &=(\overline{g}^*_\ce{CO2}+\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys}) -(\overline{g}^*_\ce{CO}+\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys}) -\frac{1}{2}(\overline{g}^*_\ce{O2}+\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys}) \end{aligned} $$

If we collect the $g^*$ components, their sum is

$$ \sum \overline{g}^* = \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^* $$

But then, by definition of the reference chemical state,

  • $\overline{g}^*_\ce{CO2} = \overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO2}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)
  • $\overline{g}^*_\ce{CO} = \overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)

Substituting this back:

$$\begin{aligned} \sum \overline{g}^* &= \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^* \\ &= (\overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}) - (\overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}) - \frac{1}{2}\overline{g}_\ce{O2}^* \\ &= 0 \end{aligned}$$

(The term cancels to zero because the reaction is balanced)

Thus $A$ has no dependence at all on the absolute value of $g$. We can cancel the sum of the $g^*$ components and write

$$ A = (\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys}) -(\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys}) -\frac{1}{2}(\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys}) $$

In practice, we typically define state $^*$ by:

  • Physical State: $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$
  • Chemical State: pure elements in their most stable form at $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$

and then re-label/re-name the chemical component of $\overline{g}$:

  • $\Delta \overline{g}^\text{chem} \rightarrow \overline{g}^\circ_f$ the "Gibbs energy of formation" (change in Gibbs energy associated with forming the substance from pure elements at the reference physical state)

To actually evaluate this, you'd

  • Look up $\sum \overline{g}^\circ_f$ for this particular reaction (which might be labelled as $\Delta G$) in a table, if one is available. If one isn't, you could look up the individual species' $\overline{g}^\circ_f$ (if you have tables which present them), or calculate them from the species' $\overline{h}^\circ_f$ and $\overline{s}^\circ$ (again, from tables).
  • For each species, evaluate $\Delta \overline{g}^\text{phys}$ (the difference in $\overline{g}$ associated with changing the species' physical state from $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$ to the $T$ and $P$ of interest) using some property model (equation or table) for each substance in turn

My Notation: For some extensive quantity $X$

  • $x$ represents $X$ per mass
  • $\overline{x}$ represents $X$ per amount
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