Question

For vaporization of water at 1 bar,
AH = 40.63 kJ mol- and AS = 108.8
J K-mol"!. At what temperature,
AG= 0 ?
a. 273.4 K
b. 393.4K
c. 373.4 K
d. 293.4 K​

Answer 1

Answer:

373.4 K

Explanation:

∆G=∆H-T∆S

∆G=0

∆H=T∆S

T=40.63 × 10³/108.8

=373.4 K