Question

For a reaction that has ∆H of -30 kJ and ∆S of -62 J/K, estimate at what temperature will the reaction change from spontaneous to non-spontaneous reaction?

Answer 1

Now We know that

$∆H=-30KJ\\∆S=-62J/K$

Now

$∆G=∆H-T∆S$

$∆G>0$ Non spontaneous

$∆G<0$ spontaneous

$∆G=0$ equilibrium

$∆H-T∆S>0$

$T<\frac{∆H}{∆S}\\T<483.87K$

equilibrium temperature

$T=\frac{-30000}{-62}=483.87K$

Spontaneous temperature

$T>483.87K$