1. I found it always helpful to find an answer to a question in many different ways. This allows us to check the self-consistency of different methods. With that in mind, I'm constructing this question, the answer to which is not clear to me. I would hope that I would learn something from your efforts. Consider the case of Langmuir adsorption with a surface exposed to a solution of a species of concentration C. We are interested to find the adsorbed amount. C. This equation can potentially be derived using the method described on page 48 of the notes. See: G. Application of Chemical Potential in Chemical Reaction. D This equation is also derivable using the method described on page on page 51 40 48 of 145 F3D70 Ni log i Ni As we have done above, we can obtain other state functions from dE=-PdV+TdS + H8 by adding d(PV) and/or subtracting d (TS) on the left and right sides of the above equations. So, the Helmholtz free energy becomes d(E-TS) = -PdV + TdS - d(TS) + iid Ni Or, dF = -PdV-SdT+di H9 H10 And Gibbs free energy becomes: dG = VdP-SdT + H11 Chemical potential can also be obtained from the Gibbs free energy as: Mi= T,P H12 Which is the partial molar Gibbs free energy at constant temperature and constant pressure. And from Helmholtz free energy as: Mi = ON TV G. Application of Chemical Potential in Chemical Reaction 10 Let us consider the following reaction CH4+202 = CO2 + 2H2O The differential change of the Helmholtz free energy is: dF (or AF) = -pdV - S&T + d At constant temperature and volume, dF=HdN =-(HCH)dN - Ho (2dN) + co (dN) + 0 (2dN) or dF = (-HCH4-2o+ co +2o)d G2 H13 G1