Question

For the decomposition of calcium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 178.5kJ/mol ΔS∘rxn 161.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous.

Answer 1

Answer : The temperature in kelvins is, $T>1108.695K$

Explanation : Given,

$\Delta H$ = 178.5 KJ/mole = 178500 J/mole

$\Delta S$ = 161.0 J/mole.K

Gibbs–Helmholtz equation is :

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

As per question the reaction is spontaneous that means the value of $\Delta G$ is negative or we can say that the value of $\Delta G$ is less than zero.

$\Delta$

The above expression will be:

$0>\Delta H-T\Delta S$

$T\Delta S>\Delta H$

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

Now put all the given values in this expression, we get :

$T>\frac{178500J/mole}{161.0J/mole.K}$

$T>1108.695K$

Therefore, the temperature in kelvins is, $T>1108.695K$