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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Solution:

  1. 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.



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