For combustion of hydrogen H 2 (g) + 1/2O 2 (g) -> H 2 O(l) the reaction enthalp

Question

For combustion of hydrogen H2(g) + 1/2O2(g) -> H2O(l) the reaction enthalpy is-286 kJ/mol and the reaction Gibbs free energy is -237 kJ/mol.Calculate the maximum amount of electrical work that may begenerated by oxidizing 1 mole of hydrogen to water in a fuelcell Please include final answer in kJ For combustion of hydrogen H2(g) + 1/2O2(g) -> H2O(l) the reaction enthalpy is-286 kJ/mol and the reaction Gibbs free energy is -237 kJ/mol.Calculate the maximum amount of electrical work that may begenerated by oxidizing 1 mole of hydrogen to water in a fuelcell Please include final answer in kJ

Explanation / Answer

H2(g) + 1/2 O2(g) -> H2O(l) here reaction enthalpy i.e H = -286 KJ/mol and gibbs free energy i.e G = -237 KJ/mol now we know from the laws of thermodyanmics that G = H - TS from the second law of thermodyanmics we know that entropy is given by: S = U/T so U = ST now putting the value we get, G = H - U so U = H - G          = -286 + 237          = 49 KJ/mol now for one mole the elctrical work genereated is 49 KJ(ans)

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