From Castellan's Physical Chemistry, chapter 12 [1, p. 276]:

12.18 At $\pu{25 °C}$ we have for rhombic sulfur: $ΔG_\mathrm{f}^\circ = 0,$ $ΔS^\circ = \pu{31.88 ± 0.17 J K-1 mol-1};$ and for monoclinic sulfur: $ΔG_\mathrm{f}^\circ = \pu{63 J mol-1},$ $ΔS^\circ = \pu{32.55 ± 0.25 J K-1 mol-1}.$ Assuming that the entropies do not vary with temperature, sketch the value of $μ$ versus $T$ for the two forms of sulfur. From the data determine the equilibrium temperature for the transformation of rhombic sulfur to monoclinic sulfur. Compare this temperature with the experimental value, $\pu{95.4 °C},$ noting the uncertainties in the values of $ΔS^\circ.$

I have plotted the graph of $μ$ vs $T$ successfully, but unable to find the equilibrium temperature. The answer is showing 119 °C and possible range from 83 °C to 277 °C.

How do I find it mathematically? I have used the equation $\mathrm d\mu = -S\,\mathrm dT + V\,\mathrm dp,$ but lot of the things are unknown. Can anyone solve it completely with calculation?

my worked out graph


  1. Castellan, G. W. Physical Chemistry, 3rd ed.; Addison-Wesley: Reading, Mass, 1983. ISBN 978-0-201-10386-1.
  • $\begingroup$ Ask yourself: Is there anything here that is constant, which would allow you to simplify the equation? $\endgroup$ – theorist Sep 22 '19 at 20:06
  • $\begingroup$ I already tried to solve by making chemical potential and Equilibrium temperature as constants but still failed, so I request you please solve the problem. $\endgroup$ – Sayantan Bhanja Sep 23 '19 at 0:09
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    $\begingroup$ That's not how it works for homework problems on this site. We're supposed to provide guidance so that you can solve it, not solve it for you. In this problem, 𝝻 and T are not constants, they're variables. However, p is constant, which gives you a nice simplification of the equations you're working with. $\endgroup$ – theorist Sep 23 '19 at 1:49
  • $\begingroup$ I already considered p to be constant I have used the equation du = -SdT, on integrating both sides from limits 298K to T I got u-u° = -S (T-298) now for both Rhombic and monoclinic sulphur u must be same in equilibrium, and also T is same , since T is equilibrium temperature, now Gf for rhombic sulphur is zero and Gf for monoclinic sulphur is 63, both values are given so and for rhombic S =31.88 and for monoclinic S=32.55 so using the equation u =-Sdt for both the sulphur we get u -0= -31.88(T-298) and u - 63= -32.55(T-298) these two equations u is same so we find the equil temp $\endgroup$ – Sayantan Bhanja Sep 23 '19 at 2:05
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    $\begingroup$ You've got it set up correctly, so your problem seems to be just with the algebra: When I set -31.88 (T- 298) = 63 - 32.55 (T - 298), and solve for T, I get T = 392 K = 119 C. $\endgroup$ – theorist Sep 23 '19 at 3:24

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