- #1

Daniel2244

- 125

- 5

Thread moved from the technical forums, so no Homework Template is shown

The enthalpy of formation equation:

ΔHƒ°

When using the enthalpy of formation equation you get:

-2010-((-394x3)+(-286x4))= 316Kjmol

However, this answer is wrong. On the mark scheme, the answer is -316Kjmol

So I used the enthalpy of combustion equation:

ΔHc°

((-394x3)+(-286x4))--2010-= -316Kjmol

When using the enthalpy of combustion equation I got the correct answer. This is where I get confused as the question is asking for the enthalpy of formation of propan-1-ol, so shouldn't I use the enthalpy of formation equation? Or do I have to use the enthalpy of combustion equation as I am using combustion data?

ΔHƒ°

_{reaction}=∑Δhƒ°_{(products)}-ΣΔHƒ°_{(reactants)}When using the enthalpy of formation equation you get:

-2010-((-394x3)+(-286x4))= 316Kjmol

^{-1}However, this answer is wrong. On the mark scheme, the answer is -316Kjmol

^{-1}.So I used the enthalpy of combustion equation:

ΔHc°

_{reaction}=∑Δhc°_{(reactants)}-ΣΔHc°_{(Products)}((-394x3)+(-286x4))--2010-= -316Kjmol

^{-1}When using the enthalpy of combustion equation I got the correct answer. This is where I get confused as the question is asking for the enthalpy of formation of propan-1-ol, so shouldn't I use the enthalpy of formation equation? Or do I have to use the enthalpy of combustion equation as I am using combustion data?