# $\Delta \bar{h}_{vap}$ and $\Delta \bar{s}_{vap}$ from vapor pressure vs. temperature data

I'm trying to work on the following two problems: The answers are given below: 4.2: 57.7 kJ/mol, 122.6 J/(K-mol), 12 kJ/mol 4.3: -58.8 kJ per mole of dimer

For 4.2, I used the equation

$$\frac{\mathrm{d} lnP}{\mathrm{d} T}=\frac{\Delta \bar{h}_{vap} }{\bar{R}T^2}$$

and made a least squares line using linear algebra, taken the natural log of the pressure and tabulating it versus the temperature, and taking the slope of that least squares line to be the slope $$\frac{\mathrm{d} lnP}{\mathrm{d} T}$$. After doing that, in the equation

$$\frac{\mathrm{d} lnP}{\mathrm{d} T}=\frac{\Delta \bar{h}_{vap} }{\bar{R}T^2}$$,

I took $$T$$ to be the average temperature over the range of data we were given, and that is 386.425 K.

So for $$\Delta \bar{h}_{vap}$$, I got 59.132 kJ/mol, which is not far off the mark from what the textbook answers are. The problem is that the problem never gives a $$T_{vap}$$ or an identity of the liquid. When I tried to find $$\Delta \bar{s}_{vap}$$ from the equation

$$\Delta \bar{s}_{vap} = \frac{\Delta \bar{h}_{vap} }{T_{vap}}$$,

I got 153.02 J/(K-mol), which is very far off the book's answer of 122.6 J/(K-mol). Finding $$\Delta \bar{g}_{vap}$$ will just be a consequence of finding the molar enthalpy and entropy of vaporization. I'm confused as to how to solve this problem.

For 4.3, I honestly don't know which equations to use or how to figure out the equilibrium constant.

• Note that plain text titles are preferred on Chemistry SE site.// The symbols for enthalpy and entropy are $H$ and $S$ // You can find useful upright vs italic Jul 8 at 14:03
• It would have been better to fit ln(p) vs 1/T. What do you get when you do it that way? The change in G should be zero at all temperatures. Jul 8 at 18:29
• Using their data for the changes in H and S, the corresponding T would be 470 K, which doesn’t make sense. Jul 8 at 18:34
• I would say it is poorly posed. It should not be asking for a value of delta G, since the starting equation is based on delta G being zero. Jul 8 at 23:22
• $$\frac{d\ln{p}}{d(1/T)}=-\frac{\Delta H}{R}$$ is the same as your starting equation. Jul 8 at 23:25 Here is a plot of p on a logarithmic scale vs 1/T on a linear scale. It is almost a perfect straight line. The slope is $$-\Delta H/R$$