How do I calculate the enthelpy of formation using combusion data

[Daniel2244]

The enthalpy of formation equation:
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?

 

[mjc123]

Your "enthalpy of formation" and "enthalpy of combustion" equations are wrong. There is no enthalpy of formation or combustion of the reaction, only the enthalpy of reaction. Your equations should read
ΔH_reaction = ΣΔHf_products - ΣΔHf_reactants
ΔH_reaction = ΣΔHc_reactants - ΣΔHc_products
The enthalpy of formation of elements in their standard states is zero, so ΔH_reaction = ΔHf(propan-1-ol), which is what you want to find.
Have you heard of a Hess's law cycle? Draw one involving the formation of propan-1-ol and the combustion of reactants and products. Then you should see how you get to the right answer, rather than using equations blindly.

 

[Chestermiller]

Please write out properly balanced chemical formulas for the combustion reactions of C, H2, and CH3CH2CH2OH.

 

[Daniel2244]

Your "enthalpy of formation" and "enthalpy of combustion" equations are wrong. There is no enthalpy of formation or combustion of the reaction, only the enthalpy of reaction. Your equations should read
ΔHreaction = ΣΔHfproducts - ΣΔHfreactants
ΔHreaction = ΣΔHcreactants - ΣΔHcproducts

My bad, I didn't realize I put ΔHƒ°reaction and ΔHc°reaction.

The enthalpy of formation of elements in their standard states is zero, so ΔHreaction = ΔHf(propan-1-ol), which is what you want to find.
Have you heard of a Hess's law cycle? Draw one involving the formation of propan-1-ol and the combustion of reactants and products. Then you should see how you get to the right answer, rather than using equations blindly.

Even when I draw the Hess cycle for formation I still get the equation ΔHreaction = ΣΔHfproducts - ΣΔHfreactants which gives the answer +316kjmol^-1 which is incorrect.

 

[Chestermiller]

I get $$4H^f_{H_2O}+3H^f_{CO_2}-\Delta H^c_{CH_3CH_2CH_2OH}=H^f_{CH_3CH_2CH_2OH}=-316\ kJ/mole$$

 

[mjc123]

Even when I draw the Hess cycle for formation I still get the equation ΔHreaction = ΣΔHfproducts - ΣΔHfreactants which gives the answer +316kjmol-1 which is incorrect.

That is because the numbers given are not enthalpies of formation. It is wrong to use this equation. You don't know ΔHf products and ΔHf reactants = 0.
See the attached Hess's law cycle. What are ΔH1, ΔH2 and ΔH3?

 

[Daniel2244]

Please write out properly balanced chemical formulas for the combustion reactions of C, H2, and CH3CH2CH2OH.

Yes, I get the same when I use the combustion cycle.

See the attached Hess's law cycle. What are ΔH1, ΔH2 and ΔH3?

ΔH1=-4336, ΔH2=-2326 and ΔH3=-2010 because that cycle gives ΔH2+ΔH3=ΔH1 and -4336. You draw the cycle differently to me, but I get the correct answer when I do it.

 

[Chestermiller]

I didn't use combustion cycles. I did it entirely algebraically:
$$4H^f_{H_2O}+3H^f_{CO_2}-H^f_{CH_3CH_2CH_2OH}=\Delta H^{c}_{{CH_3CH_2CH_2OH}}$$
with $$H^f_{CO_2}=H^c_{CO_2}$$
abd $$H^f_{H_2O}=H^c_{H_2O}$$
where the superscript f refers to heat of formation and the superscript c refers to heat of combustion.

 

[mjc123]

ΔH1=-4336, ΔH2=-2326 and ΔH3=-2010 because that cycle gives ΔH2+ΔH3=ΔH1 and -4336. You draw the cycle differently to me, but I get the correct answer when I do it.

ΔH3 is not -2010. Look at the direction of the arrow. How is it related to the heat of combustion of propanol?

 

[Daniel2244]

ΔH3 is not -2010. Look at the direction of the arrow. How is it related to the heat of combustion of propanol?

When I do it I get: