Thermodynamics and Thermochemistry - Result Question 15
15. The process with negative entropy change is
(2019 Main, 10 Jan II)
(a) synthesis of ammonia from $N _2$ and $H _2$
(b) dissociation of $CaSO _4(s)$ to $CaO(s)$ and $SO _3(g)$
(c) dissolution of iodine in water
(d) sublimation of dry ice
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Answer:
Correct Answer: 15. (a)
Solution:
The explanation of all the options are as follows :
(a) $N _2(g)+3 H _2(g) \longrightarrow 2 NH _3(g)$,
$\Delta n _g=2-(1+3)=-2$
So, $\Delta S$ is also negative (entropy decreases)
(b) $CaSO _4(s) \stackrel{\Delta}{\longrightarrow} CaO(s)+SO _3(g)$,
$\begin{aligned} & \Delta n _g=(1+0)-0=+1 \\ & \text { So, } \Delta S=+ \text { ve } \end{aligned}$
(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,
$ I _2(s) \xrightarrow[(KI)]{\text { Water }} I _2(a q) $
(d) In sublimation process, molecules of solid becomes quite free when they become gas,
$ \underset{\text { Dry ice }}{CO _2(s)} \longrightarrow CO _2(g) $
So, $\Delta S$ will be positive.
BETA
