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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics2 Marks
Question
Calculate the total entropy change, $\Delta \mathrm{S}_{\text {total }}$ and state if the process is spontaneous, when one mole of liquid mercury $\mathrm{Hg}(\mathrm{l})$ changes to mercury vapour, $\mathrm{Hg}(\mathrm{g})$ at 298 K . The molar entropy of vapourisation of Hg is $99 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ and molar enthalpy of vapourisation is $59.1 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$.
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Answer

$\Delta\text{S}_\text{surr}=\frac{-\Delta\text{H}_\text{sys}}{\text{T}}=\frac{-59100\text{J mol}^{-1}}{298\text{K}}$$=-198\text{JK}^{-1}\text{mol}^{-1}$
$\Delta\text{S}_\text{total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{surr}$
$=99\text{J}-198\text{J}=-99\text{JK}^{-1}\text{mol}^{-1}$
Since $\Delta\text{S}_\text{total}=-\text{ve},$ therefore, process is non-spontaneous at 298K.

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