Question

For the reaction: $2Fe_{2}O_{3}(s)+3C(s) \longrightarrow 4Fe(s)+3CO_{2}(g)$, ΔH=467.9 kJ/mol and ΔS=0.56 kJ/mol K. Calculate the temperature at which the ΔG° becomes zero. What will be the direction of reaction above this temperature?

Answer 1

"For the reaction, $2Fe_{ 2 }O_{ 3 }(s)\quad +\quad 3C(s)\quad \longrightarrow \quad 4Fe(s)\quad +\quad 3CO_{ 2 }(g)$

Given,

$\Delta H\quad =\quad 467.9\quad kJ/mol\quad and\quad \Delta S\quad =\quad 0.56\quad kJ/molK$

We have to Calculate the temperature at which the ${\Delta G}^{ \circ}$becomes zero, so put \Delta G\quad =\quad 0

W.K.T. $\Delta G\quad =\quad \Delta H\quad -\quad T\Delta S\quad$

0 = 467.9 - T(0.56)

T = 835.5 (at which the ${\Delta G }^{ \circ}$becomes zero)

Above this temperature T = 835.5, ${\Delta G}$ becomes positive and so the direction of reaction is reversed."