Question

If δh = -70.0 kj and δs = -0.300 kj/k , the reaction is spontaneous below a certain temperature. calculate that temperature.

Answer 1

The  temperature  for  the  reaction  is  calculated  as  follows

delta  H/ delta S


delta  H=-70 Kj
delta  S  =-0.300  kj/k
temperature  is  therefore = -70kj/-0.300kj/k  =233.33 K

Answer 2

Answer: The reaction is spontaneous below a temperature of 233.3 K.

Explanation:

Using Gibbs Helmholtz equation:

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

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change  = -70 kJ

$\Delta S$ = entropy change  = -0.300 kJ/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

At equilibrium : $T\Delta S=\Delta H$

$T=\frac{\Delta H}{\Delta S}$

$T=\frac{-70.0kJ}{-0.300kJ/K}$

$T=233.3K$

Thus the reaction is spontaneous below a temperature of 233.3 K.