Calculate the S(univ) for the following phase change at 25 deg celcius. The boiling point of heptane is 98.0 Celcius and has a Delta Hvap= +21.6kj/mol.

C7H16(l)--> C7H16(g) Delta S= +79.0j/k

I know that the equation is
Delta Suniv= DSsys + DSsurr

but for DSurr= -0.216J/371k= 5.82*10^-4

But idk how to get Dsys...

Please explain it to me thoroughly!!!

8 answers

dSsys = dH/T
I think you should look at your dH value. Isn't 21.6 kJ/mol = 21600 J/mol
So since

Delta S= +79.0J/K

is Delta S equal to Delta Ssys?
I don't get the relationship of Delta S and sys and surr with delta S (entropy)

that would make it become +79.0J/K DSsys

and then DSurrs= (-21600J/mol*1mol)/298K= -72.5J/K

Suniv= +79.0J/K+-72.5J/K= +6.5
DrBob222
Yes, dS for the system is 79.0 J/K*mol but I would use T at boiling point of 98C (371K) to solve for this because you're using dHvap and it makes sense to use T at the vapor point.

I note the problem asks for dSuniv at 298. I think you can get that by
dG = -dH/T = (dSuniv) = -RT*lnK using 371 for T and solve for K @ 371 kelvin; then
dG = dGo + RTlnK using T = 298 and K from above and solve for dGo.
I did as you said and I got the wrong answer. the correct asnwer is supposed to be +6.5, and I am getting -25.8..

Can you just show me how to do it with numbers and steps. I am just more confused now.
Could I just use the formula

DSuniv= DSsys- (DHsys/T)
79.0J/K- 21600/298k= +6.5
Calculate the S(univ) for the following phase change at 25 deg celcius. The boiling point of heptane is 98.0 Celcius and has a Delta Hvap= +21.6kj/mol.

C7H16(l)--> C7H16(g) Delta S= +79.0j/k

I know that the equation is
Delta Suniv= DSsys + DSsurr

PLEASE SOMEONE HELP ME WITH THIS PROBLEM, I CANT FIGURE IT OUT!!!

All I know is Delta Ssys= +79.0