Question

Calculate the total entropy change for the transition at $368$ K of $1$ mol of sulphur from the monoclinic to the rhombic solid state. Given $\Delta H=-401.7Jmo{l}^{-1}$ for the transition. Assume the surroundings to be an ice-water. Both at $0^{o}C$?

• A. $-1.09J{K}^{-1}$
• B. $1.47J{K}^{-1}$
• C. $0.38J{K}^{-1}$
• D. $0.76J{K}^{-1}$

Answer 1

Answer: (C)

$0.38J{K}^{-1}$

$\Delta S_{\left(system\right)}=\frac{1\times -401.7}{368}=-1.09J{K}^{-1}$

The ice-water both absorbs the 401.7 J $mo{l}^{-1}$ at temperature 273 K.

$\therefore \Delta S_{surrounding}=\frac{1\times 401.7}{273}=1.47J{K}^{-1}$ and $\Delta S_{\left(universe\right)}$$=-1.09+1.47=0.38J{K}^{-1}$

$\because △S_{universe}=△S_{system}+△S_{surroundings}$