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

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

Step-by-step 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=(\Delta H)/(\Delta S)`

`T=(-70.0kJ)/(-0.300kJ/K)`

`T=233.3K`

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

Answer 2

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