A) \[CaC{{O}_{3}}(s)\xrightarrow{\Delta }CaO(s)+C{{O}_{2}}(g)\]
B) \[2l(g)\xrightarrow{{}}{{I}_{2}}(g)\]
C) \[N{{a}^{+}}(g)+C{{l}^{-}}(g)\xrightarrow{{}}NaCl(s)\]
D) \[{{H}_{2}}O(l)\xrightarrow{{}}{{H}_{2}}O(s)\]
Correct Answer: A
Solution :
The solid \[\text{CaC}{{\text{O}}_{\text{3}}}\text{(s)}\]with low entropy is transformed into a solid \[\text{CaO(s)}\]with low entropy and a gas \[\text{C}{{\text{O}}_{\text{2}}}\]with high entropy. Hence \[\Delta S,\]entropy change is positive.
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