Thermodynamics and Thermochemistry 1 Question 15
15. The process with negative entropy change is
(2019 Main, 10 Jan II)
(a) synthesis of ammonia from $\mathrm{N}{2}$ and $\mathrm{H}{2}$
(b) dissociation of $\mathrm{CaSO}{4}(\mathrm{~s})$ to $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{SO}{3}(\mathrm{~g})$
(c) dissolution of iodine in water
(d) sublimation of dry ice
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Solution:
- The explanation of all the options are as follows :
(a) $\mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$,
$$ \Delta n_{g}=2-(1+3)=-2 $$
So, $\Delta S$ is also negative (entropy decreases)
(b) $\mathrm{CaSO}{4}(s) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(s)+\mathrm{SO}{3}(g)$,
$$ \begin{aligned} & \Delta n_{g}=(1+0)-0=+1 \ & \text { So, } \Delta S=+ \text { ve } \end{aligned} $$
(c) In dissolution, $\Delta S=+$ ve because molecules/ions of the solid solute (here, iodine) become free to move in solvated/dissolved state of the solution,
$$ \mathrm{I}{2}(s) \xrightarrow[(\mathrm{KI})]{\text { Water }} \mathrm{I}{2}(a q) $$
(d) In sublimation process, molecules of solid becomes quite free when they become gas,
$$ \begin{aligned} & \mathrm{CO}{2}(s) \longrightarrow \mathrm{CO}{2}(g) \ & \text { Dry ice } \end{aligned} $$
So, $\Delta S$ will be positive.
