S0H2(g)=130.6JK−1 mole e−1S0H2O(ℓ)=69.9JK−1 mole S0O2(g)=205JK−1 mole then absolute entropy change of H2O(ℓ)→H2(g)+12O2(g)

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$\begin{aligned} S^{0} H_{2} (g) = 130.6 J K^{- 1} mole e^{- 1} \\ S^{0} H_{2} O (ℓ) = 69.9 J K^{- 1} mole \\ S^{0} O_{2} (g) = 205 J K^{- 1} mole \end{aligned}$
then absolute entropy change of
$H_{2} O (ℓ) \to H_{2} (g) + \frac{1}{2} O_{2} (g)$

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${ΔS}^{0} = S_{products} - S_{Reactants}$

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