# How do I calculate the enthelpy of formation using combusion data

• Daniel2244
In summary: H^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}$$This is how you would get the correct answer. Daniel2244 Thread moved from the technical forums, so no Homework Template is shown 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? #### Attachments • 20190108_191530.jpg 38.6 KB · Views: 3,338 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 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. Please write out properly balanced chemical formulas for the combustion reactions of C, H2, and CH3CH2CH2OH. mjc123 said: 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. mjc123 said: 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. 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$$Daniel2244 said: Even when I draw the Hess cycle for formation I still get the equation ΔHreaction = ΣΔHfproducts - ΣΔHfreactants which gives the answer +316kjmol-1 which is incorrect. That is because the numbers given are not enthalpies of formation. It is wrong to use this equation. You don't know ΔHf products and ΔHf reactants = 0. See the attached Hess's law cycle. What are ΔH1, ΔH2 and ΔH3? #### Attachments • Hess's law cycle.png 3.8 KB · Views: 649 Chestermiller said: Please write out properly balanced chemical formulas for the combustion reactions of C, H2, and CH3CH2CH2OH. Yes, I get the same when I use the combustion cycle. mjc123 said: See the attached Hess's law cycle. What are ΔH1, ΔH2 and ΔH3? ΔH1=-4336, ΔH2=-2326 and ΔH3=-2010 because that cycle gives ΔH2+ΔH3=ΔH1 and -4336. You draw the cycle differently to me, but I get the correct answer when I do it. I didn't use combustion cycles. I did it entirely algebraically:$$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.

Daniel2244 said:
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?

mjc123 said:
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:

#### Attachments

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## 1. What is enthalpy of formation?

Enthalpy of formation is the change in enthalpy that occurs when one mole of a compound is formed from its constituent elements in their standard states.

## 2. How is enthalpy of formation calculated?

Enthalpy of formation can be calculated using the formula: ΔH°f = ΣnΔH°f(products) - ΣmΔH°f(reactants), where n and m are the coefficients of the products and reactants, respectively.

## 3. What is combustion data?

Combustion data refers to the heat released during the complete combustion of a substance. This data is often used to calculate enthalpy of formation.

## 4. How do I find the combustion data for a substance?

The combustion data for a substance can be found in reference tables or online databases. It is typically listed in units of kilojoules per mole (kJ/mol).

## 5. Can enthalpy of formation be negative?

Yes, enthalpy of formation can be negative if more energy is released during the formation of the compound than is required to break the bonds of the constituent elements. This indicates that the compound is energetically stable.

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