Question

For a particular reaction, Δ H ∘ = − 46.7 kJ and Δ S ∘ = − 123.8 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer 1

Answer: 377 K

Explanation:

According to Gibbs equation

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change = -46.7 kJ=   -46700 J

$\Delta S$ = entropy change  = -123.8 J/K

T = temperature in Kelvin  = ?

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

$\Delta H=T\Delta S$

$T=\frac{\Delta H}{\Delta S}=\frac{ -46700J}{-123.8J/K}=377K$

Thus at 377 K the reaction change from nonspontaneous to spontaneous