Calculate the enthalpy of vaporisation (in kJ/mol) for ethanol. Given, entropy change for the process is 109JK−1mol−1 and boiling point of ethanol is 78.5oC.
A
50.29kJmol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12.58kJmol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
63.52kJmol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
38.31kJmol−1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D38.31kJmol−1 We know that, △Svapourisation=△HvapourisationTb
Given: △Svapour=109JK−1mol−1 Tb=78.5+273=351.5K
Substituting these values in above equation, we get 109=△Hvap351.5 △Hvap=38313.5Jmol−1 =38.31kJmol−1