[tg22542]

The enthalpy of combustion (ΔH°c) of 2-pentanol (C5H12O) is -3315.40 kJ/mol. Using the appropriate information given below, calculate the enthalpy of formation (ΔH°f), in kJ/mol, for 2-pentanol


My attempt :

2C5H12O + 15O2 -> 10CO2 + 12H2O

ΔHrx = [(10* -393.51) + (12*-285.83)] - (2*ΔHf)

ΔHf = [[(10* -393.51) + (12*-285.83)] - (-3315.4)] /2

= -2023.0 kJ/mol


Why is this incorrect? I have no idea what I'm doing wrong..

[Corribus]

I'm not sure what you're off by, but enthalpy of combustion is defined as the amount of energy to burn 1 mole of your starting material to completion. Are you basing your calculation off of 1 mole of 2-pentanol?

[tg22542]

Yup.. and that would be my problem haha.. Whoops

Thank you!