Question

In which of the processes, the entropy will decrease

• A. $H_2(g)\to 2H(g)$
• B. Liquid water crystallizes to ice
• C. Temperature of ice is raised from $10K$ to $115K$
• D. $2NaHC{O}_3(s)\to N{a}_{2}C{O}_3(s)+C{O}_2(g)+H_{2}O(s)$

Answer 1

Answer: (B)

(a) $H_2(g)\to 2H(g)$
$\Rightarrow \Delta n_g>0$
$\Rightarrow \Delta S>0$
(b) $\underset{\text{water}}{H_{2}O(l)}\to \underset{\text{Ice}}{H_{2}O(s)}$
$\Rightarrow$ 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) $2NaHC{O}_3(s)\to N{a}_{2}C{O}_3(s)+C{O}_2(g)+H_{2}O(s)$
Here, $\Delta n_g>0$
$\Rightarrow \Delta S>0$
So, entropy decreases when liquid water crystallises into ice.