When water in a beaker freezes to form ice, then which of the following will be a correct statement.

S_{s y s t e m}

decreases whereas

S_{s u r r o u n d i n g}

increases

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S_{s y s t e m}

and

S_{s u r r o u n d i n g}

both increases

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S_{s y s t e m}

increases whereas

S_{s u r r o u n d i n g}

decreases

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S_{s y s t e m}

and

S_{s u r r o u n d i n g}

both decreases

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Solution

The correct option is A $S_{s y s t e m}$ decreases whereas $S_{s u r r o u n d i n g}$ increases
$H_{2} O (l) ⇌ H_{2} O (s)$
For this process $Δ H$ is negative, and it is a reversible process.
$S o, Δ G = 0$
$∴ Δ S_{s y s} = \frac{Δ H}{T_{M P}} = n e g a t i v e$
For reversible process
$Δ S_{s u r r o u n d i n g} = - (Δ S_{s y s})$
$Therefore, here Δ S_{s u r r o u n d i n g} = p o s i t i v e$
So as water freezes, randomness decreases i.e $(Δ S)_{s y s t e m}$ decreases and oppositely $(Δ S)_{s u r r o u n d i n g}$ increases.

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