- #1
zenterix
- 774
- 84
- Homework Statement
- I've been studying chemical potential in a chemistry course and in a thermodynamics course.
- Relevant Equations
- There is a derivation that I always come back to but I still don't fully understand it.
That derivation is of the chemical equilibrium constant.
Suppose we have a reaction where reactants ##A## and ##B## react to form products ##C## and ##D## according to the balanced equation
$$\nu_AA+\nu_BB\rightarrow \nu_CC+\nu_DD$$
Suppose the reaction has occurred to some extent and we have ##a,b,c,## and ##d## moles of reactants and products.
Let the reaction proceed infinitesimally such that the number of moles changes as follows
$$(aA,bB,cC,dD)\rightarrow ((a-\epsilon\nu_A)A, (b-\epsilon\nu_B)B, (c+\epsilon\nu_C)C, (d+\epsilon\nu_D)D)$$
Then, by definition of chemical potential we have
$$\Delta G=G_{\text{after}}-G_{\text{before}}$$
$$=((a-\epsilon\nu_A)\mu_A+(b-\epsilon\nu_B)\mu_B+(c-\epsilon\nu_C)\mu_C+(d-\epsilon\nu_D)\mu_D)-(a\mu_A+b\mu_B+c\mu_C+d\mu_D)$$
$$=\epsilon((\mu_C\nu_C+\mu_D\nu_D)-(\mu_A\nu_A+\mu_B\nu_B))$$
We can now sub in for the chemical potentials using the equation
$$\mu_i(T,P)=\mu_i^\circ+RT\ln{\left (\frac{P_i}{1\ \text{bar}}\right )}$$
We reach
$$\Delta G=\epsilon \left [ (\mu_C^\circ\nu_C+\mu_D^\circ\nu_D)-(\mu_A^\circ\nu_A+\mu_B^\circ\nu_B)+RT\ln{\left ( \frac{P_C^{\nu_C}P_D^{\nu_D}}{P_A^{\nu_A}P_B^{\nu_B}} \right )} \right ]$$
Okay, my question concerns the next step.
In my notes, I have the following
$$\Delta G_{rxn}^\circ=\nu_C\mu_C^\circ(\text{pure})+\nu_D\mu_D^\circ(\text{pure})-\nu_A\mu_A^\circ(\text{pure})-\nu_B\mu_B^\circ(\text{pure})$$
Therefore
$$\Delta G=\epsilon (\Delta G_{rxn}^\circ+RT\ln{Q})$$
Now, these notes are based on lectures from MIT OCW's Thermodynamics course.
The lectures in that course are quite confusing and the lecturers unfortunately routinely make many uncorrected mistakes on the blackboard every lecture. It's difficult sometimes to know what is what.
Here is what I have in my notes exactly
So, I think "pure" has to do with a gas being by itself or in a mixture. The chemical potential of the gas in the mixture is smaller than the chemical potential of the pure gas.
This is a result that was derived previously
$$\mu_i(\text{mix},T,P_T)=\mu_i(\text{pure},T,P_T)+RT\ln{(x_i)}\tag{1}$$
It appears that ##\Delta G^\circ_{rxn}## is the Gibbs energy of reaction that we commonly see in tables and does not include entropy of mixing.
In the calculations in this post, we are trying to reason about a mixture (of reactants and products in the reaction in question). I can't really identify where (1) is used in all of this derivation.
$$\nu_AA+\nu_BB\rightarrow \nu_CC+\nu_DD$$
Suppose the reaction has occurred to some extent and we have ##a,b,c,## and ##d## moles of reactants and products.
Let the reaction proceed infinitesimally such that the number of moles changes as follows
$$(aA,bB,cC,dD)\rightarrow ((a-\epsilon\nu_A)A, (b-\epsilon\nu_B)B, (c+\epsilon\nu_C)C, (d+\epsilon\nu_D)D)$$
Then, by definition of chemical potential we have
$$\Delta G=G_{\text{after}}-G_{\text{before}}$$
$$=((a-\epsilon\nu_A)\mu_A+(b-\epsilon\nu_B)\mu_B+(c-\epsilon\nu_C)\mu_C+(d-\epsilon\nu_D)\mu_D)-(a\mu_A+b\mu_B+c\mu_C+d\mu_D)$$
$$=\epsilon((\mu_C\nu_C+\mu_D\nu_D)-(\mu_A\nu_A+\mu_B\nu_B))$$
We can now sub in for the chemical potentials using the equation
$$\mu_i(T,P)=\mu_i^\circ+RT\ln{\left (\frac{P_i}{1\ \text{bar}}\right )}$$
We reach
$$\Delta G=\epsilon \left [ (\mu_C^\circ\nu_C+\mu_D^\circ\nu_D)-(\mu_A^\circ\nu_A+\mu_B^\circ\nu_B)+RT\ln{\left ( \frac{P_C^{\nu_C}P_D^{\nu_D}}{P_A^{\nu_A}P_B^{\nu_B}} \right )} \right ]$$
Okay, my question concerns the next step.
In my notes, I have the following
$$\Delta G_{rxn}^\circ=\nu_C\mu_C^\circ(\text{pure})+\nu_D\mu_D^\circ(\text{pure})-\nu_A\mu_A^\circ(\text{pure})-\nu_B\mu_B^\circ(\text{pure})$$
Therefore
$$\Delta G=\epsilon (\Delta G_{rxn}^\circ+RT\ln{Q})$$
Now, these notes are based on lectures from MIT OCW's Thermodynamics course.
The lectures in that course are quite confusing and the lecturers unfortunately routinely make many uncorrected mistakes on the blackboard every lecture. It's difficult sometimes to know what is what.
Here is what I have in my notes exactly
So, I think "pure" has to do with a gas being by itself or in a mixture. The chemical potential of the gas in the mixture is smaller than the chemical potential of the pure gas.
This is a result that was derived previously
$$\mu_i(\text{mix},T,P_T)=\mu_i(\text{pure},T,P_T)+RT\ln{(x_i)}\tag{1}$$
It appears that ##\Delta G^\circ_{rxn}## is the Gibbs energy of reaction that we commonly see in tables and does not include entropy of mixing.
In the calculations in this post, we are trying to reason about a mixture (of reactants and products in the reaction in question). I can't really identify where (1) is used in all of this derivation.