# Working out energies of hydrogen bonds by comparing chemical potential of vapour and condensed phase - equation derivation

I want to ask a question about working out the energy of a hydrogen bond between two water molecules, $$w_{AA}$$ using the chemical potentials of vapour and condensed phases.

I was reading K. Dill, Molecular Driving Forces, 2nd Ed, 2011, p. 257 which lines out the framework for the Clapeyron equation and I was confused on the following:

The book states that you can calculate $$w_{AA}$$ using the following formula, which is derived by calculating the chemical potentials of the vapour $$\mu_v$$ and $$\mu_c$$ respectively:

$$\mu_v = kT\ln\left(\frac{p}{p_{int}^{*}}\right)$$

where $$\mu_v$$ is the chemical potential of a a pure substance in the liquid phase, $$p$$ is the partial pressure of the vapour phase and $$p_{int}^*$$ is the standard vapour pressure

The chemical potential of the condensed phase, $$\mu_c$$ is achieved by differentiation of the Helmholtz equation, with the knowledge that the $$TS$$ term of $$U-TS$$ is set to $$0$$ since a pure liquid's molecules are indistinguishable upon interchange.

$$\mu_c = \left(\frac{\partial F}{\partial N}\right)_{T,V} = \frac{zw_{AA}}{2}$$

where $$z$$ is the number of neighbours of each molecule in the liquid phase, $$w_{AA}$$ is the bonding interaction between two molecules, and divided by 2 to account for any double counting.

Using $$\mu_A = \mu_B$$, both terms can be equated:

$$kT\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2}$$

and more simply written as

$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$

and for those interested, this can be expressed as:

$$p=p_{int}^*e^{\frac{zw_{AA}}{2kT}}$$

Now, I was presented with the following question from the book:

Given that water’s vapor pressure is $$1 \ atm$$ at $$T = 100^{\circ}$$ and $$0.03 \ atm$$ at $$T = 25^{\circ}C$$, find the value of $$zw_{AA}$$.

This all makes sense.

However, Dill states the following

Take the logarithm of the equation to get the boiling pressure $$p_1$$ at temperature $$T_1$$ as $$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$ and the boiling pressure $$p_2$$ at temperature $$T_2$$ as $$\ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$
Subtract the second equation from the first to get $$\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ neglecting the small temperature dependence on $$p_{int}^*$$.

However, I'm not sure whether this derivation is correct. Below shows my working out from the second equation stated from the start of the body:

First of all, they transformed

$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$

to

$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$

and somehow stated the relationship between the gas constant $$R$$ and $$k$$ is $$R = 2k$$ which doesn't make sense.

Also, the denominator term containing the $$2$$ disappears in both of their initial statements for $$T_1$$ and $$T_2$$ e.g. $$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$, yet it returns in the final equation $$\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ and I can't see that from the following working that I did to prove this equation does not work (taking $$R = 2k$$ to be true):

$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}, \ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$

Subtracting the second equation from the first:

$$\ln\frac{P_1}{P_2} = \frac{zw_{AA}}{RT_1} - \frac{zw_{AA}}{RT_2} = \frac{zw_{AA}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

which does not compare with the derived result.

My question is therefore two-fold:

1. How has Dill interchanged $$R$$ and $$k$$ in his solution?
2. Is there an error in his derivation for his final expression.

The solution (as transcribed above) is also attached as a screenshot for illustration (for illustrative purposes only):

• Switching between k and R is just a matter of changing units from J to J/mol (or energy to molar energy units). The factor of two should be retained and its disappearance from some of the expressions is was probably a typo. – Buck Thorn Oct 9 '20 at 16:33
• Going between R and k is more than just a dimensional conversion. You also need to divide or multiply by $N_A$, since R is J/K/mol, while k is J/K/particle: $R = N_A k$. – theorist Oct 10 '20 at 2:02
• @theorist you are strictly right of course. You encounter this type of manipulation so often that you ignore the finer details. Forgot to note: upon conversion from k to R, $w_{AA}$ must be expressed as a molar quantity. – Buck Thorn Oct 10 '20 at 7:49
• @BuckThorn yeah that's what I thought - all quantites relevant would then have to be expressed explicitly as molar quantites - shame that this book doesn't choose to mention it properly. Many thanks for your suggestions! I think I've solved the matter! – vik1245 Oct 12 '20 at 1:12