DOFPro Team

React But Maintain Your Equilibrium Part 1 & Part 2
You are not expected to fully understand the derivation at this point. It’s just worth being exposed to what advanced-level work in chemical engineering is like.
From Thermodynamics, equilibrium is reached when the total Gibbs Energy is minimized. Mathematically the criterion is:
\[ \sum \nu_i G^\circ_i + RT \sum \ln \hat{a}_i^{\nu_i} = 0 \]
or rearranged
\[ \prod \hat{a}_i^{\nu_i} = \exp \left( - \frac{\sum \nu_i G^\circ_i}{RT}\right) \equiv K \]
The \(\hat{a}_i\) are the activities of the various species where
\[ \hat{a}_i \equiv \frac{\hat{f}_i}{f^\circ_i} \]
\(\hat{f}_i\) is the fugacity of species \(i\) in solution.
\(f^\circ_i\) is the reference fugacity of species \(i\).
\(\sum \nu_i G^\circ_i\) is often written as \(\Delta G^\circ\) and is similar to \(\Delta H^\circ\).
\(K\) is the equilibrium constant. It varies only with temperature.
What are the units of \(K\)?
For ideal gases (see upcoming videos) the fugacity in solution is numerically equal to the partial pressure, and the reference fugacity is the ideal gas state at the system \(T\) and a pressure of 1 bar.
With these substitutions:
\[ K = \prod \hat{a}_i^{\nu_i} = \prod \left( \frac{p_i}{\mathrm{1\ bar}}\right)^{\nu_i} \]
or
\[ K\ (\mathrm{1\ bar})^\nu = \prod p_i^{\nu_i} \]
\(K\) is still dimensionless but may not appear to be.
Often problems are posed in mole fractions instead of partial pressures. Since for an ideal gas (see Ideal Gases EXPOSED: Are They Even Real?!)
\[ y_i = \frac{p_i}{P} \]
then
\[ p_i = y_i P \]
Substituting
\[ K\ (\mathrm{1\ bar})^\nu = \prod (y_i P)^{\nu_i} =P^\nu \prod (y_i)^{\nu_i} \]
or
\[ K\ (\mathrm{1\ bar})^\nu P^{-\nu} =\prod (y_i)^{\nu_i} \]
which is where we will begin. There are later videos that go into more detail about the earlier steps and explain when you’d need them. For the present we’ll only deal with equilibrium reactions involving ideal gases and pure solids.
As a reminder, \(K\) depends only on temperature but the equilibrium composition depends on total pressure and the presence of inerts.
With a fair amount of calculus one can show that
\[ \frac{d \ln K}{dT} = -\frac{\Delta \hat{H}^\circ_r}{RT^2} \]
which, if the enthalpy of reaction is constant with temperature integrates to
\[ \ln\frac{K}{K_0} = -\frac{\Delta \hat{H}^\circ_r}{R} \left(\frac{1}{T} - \frac{1}{T_0} \right) \]
Thanks for watching!
The Full Story companion video is in the link in the upper left. The companion video, React But Maintain Your Equilibrium Part 2, in the series, is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.

