Entropy changes for the process,
$H_{2} O (l) \to H_{2} O (s)$
at normal pressure and 274 K are given below.
$Δ S_{s y s t e m} = - 22.13, Δ S_{s u r r o u n d i n g} = + 22.05 J / m o l$
the process is non-spontaneous because :-

Δ S_{s y s t e m}

is negative

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Δ S_{s u r r o u n d i n g}

is positive

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Δ S_{u n i v e r s e}

is negative

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Δ S_{s y s t e m} \neq Δ S_{s u r r o u n d i n g}

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Solution

The correct option is C $Δ S_{u n i v e r s e}$ is negative
$Δ S_{u n i v e r s e} = Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g}$
$= - 22.13 + 22.05 = - 0.08$
for a spontaneous process, it must be positive.

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Entropy changes for the process, H2O(l)→H2O(s) at normal pressure and 274 K are given below. ΔSsystem=−22.13,ΔSsurrounding=+22.05 J/mol the process is non-spontaneous because :-

The correct option is C ΔSuniverse is negativeΔSuniverse=ΔSsystem+ΔSsurrounding =−22.13+22.05=−0.08 for a spontaneous process, it must be positive.

Entropy changes for the process,
$H_{2} O (l) \to H_{2} O (s)$
at normal pressure and 274 K are given below.
$Δ S_{s y s t e m} = - 22.13, Δ S_{s u r r o u n d i n g} = + 22.05 J / m o l$
the process is non-spontaneous because :-

The correct option is C $Δ S_{u n i v e r s e}$ is negative
$Δ S_{u n i v e r s e} = Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g}$
$= - 22.13 + 22.05 = - 0.08$
for a spontaneous process, it must be positive.