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
Show Answer
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,
$$ \begin{aligned} & CO _2(s) \longrightarrow CO _2(g) \ & \text { Dry ice } \end{aligned} $$
So, $\Delta S$ will be positive.
