Question

Given that for the vaporization of benzene ΔHvap= 30.7 kJ/mol and ΔSvap = 87.0 J/(K⋅mol), calculate ΔG for the vaporization of benzene at the following temperatures.

part A-73, part B-80, part C-90 degrees C

Answer 1

1)

ΔHvap = 30.7 KJ/mol

ΔSvap = 87 J/mol.K

= 0.087 KJ/mol.K

T= 73.0 oC

= (73.0+273) K

= 346 K

use:

ΔGvap = ΔHvap - T*ΔSvap

ΔGvap = 30.7 - 346.0 * 0.087

ΔGvap = 0.598 KJ/mol

Answer: 0.598 KJ/mol

B)

ΔHvap = 30.7 KJ/mol

ΔSvap = 87 J/mol.K

= 0.087 KJ/mol.K

T= 80.0 oC

= (80.0+273) K

= 353 K

use:

ΔGvap = ΔHvap - T*ΔSvap

ΔGvap = 30.7 - 353.0 * 0.087

ΔGvap = -0.011 KJ/mol

Answer: -0.011 KJ/mol

C)

Now we have:

ΔHvap = 30.7 KJ/mol

ΔSvap = 87 J/mol.K

= 0.087 KJ/mol.K

T= 90.0 oC

= (90.0+273) K

= 363 K

use:

ΔGvap = ΔHvap - T*ΔSvap

ΔGvap = 30.7 - 363.0 * 0.087

ΔGvap = -0.881 KJ/mol

Answer: -0.881 KJ/mol