Question

For the reaction,
4Fe(s) + 3O2(g) ---> 2Fe2O3(s) Entropy change and enthalpy change at 298K are -549.4 J/K/mol and -1648 * 10^3J/mol respectively. Calculate the free energy change for the reaction.

Answer 1

Dear Student,
GIven:
$$\begin{gathered} ∆S=-549.4J{K}^{-1}mo{l}^{-1},∆H=-1648\times {10}^{-3},T=298K \\ Accordingtogibbs-helmholtzequation \\ ∆G=∆H-T∆S \end{gathered}$$
Putting the values in equation we get
$∆G$= (-1648$\times$10³)- (298$\times$(-549.4))
Free energy $∆G$ = -1.48$\times$10⁶ Jmol^​-1 or -1.48$\times$10³ Jmol-1
Regards!