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Consider the reaction Fe(s) + 2HCl(aq)FeCl2(s) + H2(g) for which H° = -7.400 kJ and S°...

Consider the reaction Fe(s) + 2HCl(aq)FeCl2(s) + H2(g) for which H° = -7.400 kJ and S° = 107.9 J/K at 298.15 K.

(1) Calculate the entropy change of the UNIVERSE when 1.552 moles of Fe(s) react under standard conditions at 298.15 K. Suniverse = J/K

(2) Is this reaction reactant or product favored under standard conditions?

(3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'. Submit Answer

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Answer #1

Ferst 2 H Clar e Cla(s) + H₂(g) Ho=-7.400 kJ so - 107.9 JlL at 298-15k (1) for 1.552 moles. SH° = 1.5524 (-7.400) kJ = - 11-4(2) (3) Asuniverse > 0 reaction product favored. AH =-ve A Suniverse = tve product favored by both enthalpy and entropy.

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