Combustion of Hydrogen in a fuel cell at $300 K$ is represented as $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g) .$
If $∆ H$ and $∆ G$ are $- 241.60 k J m o l^{- 1}$ and $- 228.40 k J m o l^{- 1}$ are respectively for $H_{2} O$ then the value of $∆ S$ for the above process is

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∆ S = ∆ H - ∆ G T = - 241.6 - - 228.4 300 For 1 mole of H 2 O =- 44 J K -1 Hence, for 2 moles = − 88 J K -1

Chemistry - Thermodynamics - II Question with Solution | TestHub

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