For the reaction,
$H_{2} O (l) ⇌ H_{2} O (g)$
at 373 K and 1 atm pressure

Δ H = 0

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Δ E = 0

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Δ H = T Δ S

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Δ H = Δ E

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Solution

The correct option is C $Δ H = T Δ S$
In the reaction, $H_{2} O (l) ⇌ H_{2} O (g)$
The system is at equilibrium.
Free energy equation is $Δ G = Δ H - T Δ S$

The free energy of the reaction at equilibrium will be zero.
$Δ G = 0$
$⟹ Δ H = T Δ S$

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