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..
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!
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 = ?
I realized where I made a mistake for this problem. This time, I got this one right.
Thank you so much!
To calculate the normal boiling point of the compound, you can use the equation:
ΔG = ΔHvap - TΔSvap
At the boiling point, ΔG is equal to zero. Therefore, the equation becomes:
0 = ΔHvap - TΔSvap
Rearranging the equation, we get:
T = ΔHvap / ΔSvap
Given that ΔHvap = 31.3 kJ/mol and ΔSvap = 79.7 J/(K*mol), let's first convert ΔHvap into J/mol:
ΔHvap = 31.3 kJ/mol x 1000 J/1 kJ = 31,300 J/mol
Now we can substitute the values into the equation:
T = 31,300 J/mol / 79.7 J/(K*mol)
Calculating this, we find:
T ≈ 392.98 K
To convert this to Celsius, subtract 273.15 from the temperature in Kelvin:
T = 392.98 K - 273.15 K ≈ 119.83°C
Therefore, the normal boiling point of the compound is approximately 119.83°C.