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Given information:

  • The boiling point of benzene at atmospheric pressure is $353~\mathrm{K}$
  • the enthalpy of vaporization of benzene is $30.8~\mathrm{kJ~mol^{−1}}$ at this temperature.
  • The molar heat capacities of the liquid and vapour are $136.1~\mathrm{J~K^{−1}~mol^{-1}}$ and $81.7~\mathrm{J~K^{−1}~mol^{-1}}$, respectively, and may be assumed temperature independent.

Calculate the entropy change of the system, the surroundings and hence the universe when $1~\mathrm{mol}$ of benzene vapour at $343~\mathrm{K}$ and atmospheric pressure becomes liquid benzene at $343~\mathrm{K}$. Also, will this process occur spontaneously?

I know that $\mathrm{d}S = \frac{\mathrm{d}H}{T}$ therefore,
$\displaystyle\mathrm{d}S = \frac{-30.8 \times 10^3~ \mathrm{J}}{343~\mathrm{K}}=-89.8~\mathrm{J~K^{-1}}$ which is the entropy of the system.

How do I continue from there, utilising the molar heat capacities given?

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  • $\begingroup$ I would use the clausius clapeyron relation to find the equilibrium constant at the lower temperature and then convert to gibbs free energy and then solve from delta S. There probably are many different ways to solve though. $\endgroup$
    – Andy
    Commented Apr 18, 2015 at 22:53
  • $\begingroup$ @pikachugee -89.8 is the change in entropy of the surroundings. If that were the system, that would mean that there is a decrease in entropy if a liquid transitioned into a gas. $\endgroup$
    – user3735
    Commented Apr 19, 2015 at 3:36
  • $\begingroup$ I can solve it if, you clear whether, heat capacity is for constant volume or constant pressure, $C_v$ or $C_p$ $\endgroup$
    – RE60K
    Commented Apr 19, 2015 at 10:23

2 Answers 2

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You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this: $$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$

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You already determined the $\Delta S$ for benzene at 353 K by dividing the $\Delta H$ at 353 K by the temperature 353 K. Now all you need to do is determine the $\Delta H$ at 343 K using Hess' law? Once you do that, you can get the $\Delta S$ at 343 K.

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