Answer to Question #346242 in General Chemistry for Ashter

Question

Calculate ΔG⁰ at 298 K for the reaction below.

Fe₂O₃(s) + 13CO(g) > 2Fe(CO)₅(g) + 3CO₂(g)

ΔH⁰_f (kJ/mol) -824.2 -110.5 -733.8 -393.5

S⁰ (J/mol•K) 87.4 197.6 445.2 213.6

Answer 1

∆H of the reaction = 3*∆H(CO₂) + 2*∆H(Fe(CO)₅) - 13*∆H(CO) - ∆H(Fe₂O₃):

∆H of the reaction = 3 *(-393,5)+ 2 *(-733,8) - (13 *(-110,5)) - (-824,2) = -387,4 kJ

∆S of the reaction = 3*∆S(CO₂) + 2*∆S(Fe(CO)₅) - 13*∆S(CO) - ∆S(Fe₂O₃):

∆S of the reaction = 3 * 213,6 + 2 * 445,2 - (13 * 197,6) - (87,4) = -1125 J = - 1,125 kJ

∆G = ∆H - T * ∆S

∆G = -387,4 - 298 * (-1,125) = -52,15 kJ

ANSWER: -52,15 kJ