Combustion of Hydrogen in a fuel cell at 300 K is represented as 2 H 2 g + O 2 g → 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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ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

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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

Options:

Answer:

Solution:

$∆ S = \frac{∆ H - ∆ G}{T} = \frac{- 241.6 - (- 228.4)}{300}$

For 1 mole of $H_{2} O =- 44 J K^{-1}$

Hence, for 2 moles $= - 88 J K^{-1}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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