Chem help?

Question

The value of ΔHvaporizaAon of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.

Calculate ΔS, ΔSsurr, and ΔG for the vaporiza$on of one mole of this substance at 72.5°C and 1 atm.

ΔS = 132 J/K·∙mol ΔSsurr = -132 J/K·∙mol ΔG = 0 kJ/mol 

 

i I was given the answers in green but need to know how to calculate it  I also need to explain why the values are negative, positive., and 0  

 

Answer 1

Step 1:

The value of of substance X is 45.7 kJ/mol.

The boiling point is .

.

Here vaporization is an endothermic process.

For endothermic process .

.

From the conversation of energy, .

Step 2:

is called as Gibbs free energy.

From Gibbs free energy .

Solution :

.

.

.