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Which of the following reactions show a decrease in entropy? a. CaO(s) + H2O(l) Ca(OH)2 (s)

B2H2O (l) 2 H2 (g) + O2 (g) c. H2CO3 (aq) CO2 (g) + H2O (l) d. 2KClO3 (s) 3O2 (g) + 2KCl (s)

Verified Answer :

antonsandiego

The correct answer is a. CaO(s) + H2O(l) Ca(OH)2 (s)
Decreasing in entropy means that you have to get a solid from the liquid
therefore : CaO(s) + H2O(l) Ca(OH)2 (s)
This formula will make it absolutely clear 
\((g) -\ \textgreater \ (l) -\ \textgreater \ (s) more molecules -\ \textgreater \ less molecules\)

Answer: The correct answer is Option a.

Explanation:

Entropy is defined as the measurement of randomness in a system.

Entropy increases as we move from solid state to liquid state to gaseous state and it decreases as we move from gaseous state to liquid state to solid state.

From the options above:

Option a: \(CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)\)

Entropy is getting decreased because liquid reactants are changing its state to solid state.

Option b: \(2H_2O(l)\rightarrow 2H_2(g)+O_2(g)\)

In this reaction, entropy is increasing because the state is changing from liquid to gaseous state.

Option c: \(H_2CO_3(aq.)\rightarrow CO_2(g)+H_2O(l)\)

In this reaction, entropy is increasing because the state is changing from liquid to gaseous state.

Option d: \(2KClO_3(s)\rightarrow 3O_2(g)+2KCl(s)\)

In this reaction, entropy is increasing because the state is changing from solid to gaseous state.

Hence, the correct answer is Option a.

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