The following thermodynamic data was obtained for an unknown compound. Delta Hvap = 31.3 kj/mol and Delta S Vap= 79.7 kj/mol .Calculate the normal boiling point of this compound in Celcius..

Question

Thank you!

Chemistry - DrBob222, Tuesday, April 30, 2013 at 12:31am
dG = dHvap - TdSvap
dG at boiling point = 0; therefore,
dHvap = TdSvap
You know dHvap and dSvap, solve for T. Check to make sure you typed the problem correctly. Check that dSvap is really in kJ/mol (or is it J/mol instead).
T will be in kelvin, convert to C.

- DrBob222,
The dSvap is J/(k*mol). I subtract the dHvap and dSvap, I got 44 Kelvin then I convert it to celsius 229.15 C. My final answer doesn't seems right. Help me please!

Answer 1

What did you subtract? I don't see a subtraction anywhere in the problem except for conversion of K to C.
dG = dH - TdS
0 = dH - TdS
dH = TdS
31300 = T*79.7
T = 31300/79.7
T = ? K
C = ?K-273.15 = ?

Answer 2

I realized where I made a mistake for this problem. This time, I got this one right.

Thank you so much!