Question

In which of the processes, the entropy will decrease

• A. $H _{2}( g ) \rightarrow 2 H ( g )$
• B. Liquid water crystallizes to ice
• C. Temperature of ice is raised from $10\, K$ to $115\, K$
• D. $2 NaHCO _{3}( s ) \rightarrow Na _{2} CO _{3}( s )+ CO _{2}( g )+ H _{2} O ( s )$

Answer 1

Answer: (B)

(a) $H_{2}(g) \rightarrow 2 H(g) $
$\Rightarrow \Delta n_{g}>0 $
$\Rightarrow \Delta S >0$
(b) $\underset{\text{water}}{H _{2} O ( l )} \rightarrow \underset{\text{Ice}}{H _{2} O ( s )} $
$\Rightarrow $ degree of randomness of $H _{2} O$ molecules decreases,
when freezing of water into ice takes place,
$\Rightarrow \Delta S<0$
(c) With increase in temperature, $KE$ of $H _{2} O$ molecules of ice also increase. $So , \Delta S>0$.
(d) $2 NaHCO _{3}(s) \rightarrow Na _{2} CO _{3}(s)+ CO _{2}(g)+ H _{2} O (s)$
Here, $\Delta n_{g}>0$
$ \Rightarrow \Delta S>0$
So, entropy decreases when liquid water crystallises into ice.