# What is the standard Gibbs free energy for this reaction? In kJ/mol At what temperature T_eq do the forward and reverse rusting reactions occur in equilibrium? In Kelvin?

## The chemical reaction that causes iron to rust in air is given by 4Fe + 3O₂ → 2Fe₂O₃ in which ΔHrxn⁰ =-1648.4 kJ/mol ΔSrxn⁰ = -543.7 J/mol*K

May 2, 2016

$\Delta {G}^{\circ} = - 1486 \text{ ""kJ/mol}$

$\text{T"=3032" ""K}$

#### Explanation:

$\Delta {\text{G"^@=Delta"H"^@-"T"Delta"S}}^{\circ}$

Under standard conditions $\text{T" =298"K}$

$\therefore \Delta {\text{G}}^{\circ} = - 1648.4 \times {10}^{3} - \left(298 \times - 543.7\right)$

$\Delta \text{G"^@=-1486" ""kJ/mol}$

At equilibrium $\Delta \text{G} = 0$

$\therefore 0 = \Delta {\text{H"^@-"T"Delta"S}}^{\circ}$

• assuming the enthalpy and entropy changes don't change much with temperature we can use the standard values.

:."T"=(Delta"H"^@)/(Delta"S"^@)

$\text{T"=(-1648.4xx10^3)/(-543.7)" ""K}$

$\text{T"=3032" ""K}$

Strictly speaking, that is not the equation for rusting. This requires both air and water and the product, rust, is hydrated iron(III) oxide $\textsf{F {e}_{2} {O}_{3.} x {H}_{2} O}$.