"When Benzoic Acid is burned, the enthalpy of combustion is -3223.6 kJ/mol. Use this information and tabulated values of the standard enthalpies of formation of liquid water and carbon dioxide to calculate the standard enthalpy of formation of benzoic acid"
This is the question stated in my textbook
I started with:
C7H6O2 + 15/2O2 --> 7CO2 + 3H2O deltaH= -3223.6 kJ
2H2 + O2 --> 2H2O deltaH= -571.6 kJ
C + 1/2 O2 --> CO2 deltaH= -393.5 kJ
I don't understand how to get the target equation "standard enthalpy of formation of Benzoic Acid"
I thought it was 7CO2 + 3H2O --> C7H6O2 + 15/2O2
help..?
This is the question stated in my textbook
I started with:
C7H6O2 + 15/2O2 --> 7CO2 + 3H2O deltaH= -3223.6 kJ
2H2 + O2 --> 2H2O deltaH= -571.6 kJ
C + 1/2 O2 --> CO2 deltaH= -393.5 kJ
I don't understand how to get the target equation "standard enthalpy of formation of Benzoic Acid"
I thought it was 7CO2 + 3H2O --> C7H6O2 + 15/2O2
help..?
-
What is the answer for this question? I got -1972.7kJmol-1 .
If you say my answer is right, then I'll show you how I did it.
=)~
If you say my answer is right, then I'll show you how I did it.
=)~