10.
The Heat of vaporization of 1 mole of ethanol is 38.57646 Jk - mole - and its boiling
point is 351.5 k, the energy change is :
Answers
Answered by
0
Answer:
Answer
We know that
ΔS
vapour
=
T
bp
ΔH
vapour
Given, ΔS
vapour
=109.8JK
−1
mol
−1
T
bp
=78.5+273=351.5K
Substituting these values in above equation, we get
109.8=
351.5
ΔH
vapour
ΔH
vapour
=38594Jmol
−1
=38.594kJmol
−1
Similar questions