Respuesta :
i) Entropy change for the surroundings when 2.13 moles of Fe(s) react at standard conditions is 52.89 K/J.
According to Hess law, “at constant temperature, heat energy changes (enthalpy – ΔHrec) accompanying a chemical reaction will remain constant, irrespective of the way the reactants react to form product”.
Heat of reaction = sum of heats of reaction of products - sum of heats of reaction of reactants or
ΔH0rxn = ∑n×ΔHf0(products)−∑n×ΔHf0(reactants)
= ΔHf0 FeCl2(s) + ΔHf0 H2(g) - (ΔHf0 Fe(s) + 2 *ΔHf0 HCl(aq))
= (-341.8 KJ + 0 KJ)-(2*-167.2 KJ+1*0 KJ)
= -7.4 KJ
the enthalpy change when 2.13 moles of Fe(s) react at standard conditions = 2.13(-7.4) KJ = -15.762 KJ
ΔS°surroundings = -ΔH0rxn /T
= -(-15.762 KJ)/298 K
= 52.89 J/K
ii) Entropy change for the surroundings when 2.23 moles of C2H4(g) react at standard conditions is 341.2 J/K.
According to Hess law,
ΔH0rxn = ∑n×ΔHf0(products)−∑n×ΔHf0(reactants)
= ΔHf0 CH3CH2OH(g) - (ΔHf0 C2H4(g) + ΔHf0 H2O(g)))
= (-235.1 KJ)-(52.3 KJ+(-241.8 KJ)
= -45.6 KJ
the enthalpy change when 2.23 moles of C2H4(g) react at standard conditions = 2.23(-45.6) KJ = -101.688 KJ
ΔS°surroundings = -ΔH0rxn /T
= -(-101.688 KJ)/298 K
= 341.2 J/K
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