The standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous athanol is -168.6 kj/mol. calculate the vapour pressure of ehtanol at

P=133.71mmHg

Explanation:

the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous ethanol is -168.6 kj/mol. calculate the vapour pressure of ethanol at

assumption:

is that temperature is at 25C, at standard pressure of 1bar(750mmHg)

ethanol is an ideal gas

The free energy of ethanol liquid does not vary with pressure,

C2H5OH(l)⟶C2H5OH(g)

free energy of formation on the reactant side is  -174.9 kj/mol

fro the product side is  -168.6 kj

∅Gvap-∅G(l)=-168.6kj/mol-(-174.9kj/mol)

+6.3kj/mol

∅G=∅Gvap+RTlnK-∅Gliq

∅G=0

0=+6.3kj/mol+8.314Jk/mol/k(298)InK

-6.3/(RT)=Lnk

taking the exponential of both sides

0.178=k

k=p/

P^0=refers to the pressure of ethanol vapour at its standard state

partial pressure , which is 750 mmHg

P=0.178*750

P=133.71mmHg

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr