# Heat of Formation: Solve Ethylene C2H4 Combustion

• moxy
In summary: I multiplied by 2 mol...so I think all of my numbers are the same as yours, I just never used the value for O2 (a huge duh on my part). I guess I thought that the "zero enthalpy" applied because O2 (g) is oxygen's natural state at room temperature.This was a HUGE help, this problem has been eating up my time all night, and I'm so glad to finally understand where I was going wrong! Thanks!In summary, the question is about finding the amount of heat released in the complete combustion of ethylene (H2C=CH2) using bond energies. The correct approach is to calculate the enthalpy of the reactants and products by adding

## Homework Statement

This is a question from a sample exam that I can't figure out. The answer is 1297 kJ/mol, but I don't know how to solve it. Any help is appreciated.

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The following are average bond energies (kJ/mol):
C−H 413 O−H 463 C=C 614
C−C 348 O−O 146 C=O 799
C−O 358 O2 495

What is the amount of heat released in the complete combustion of ethylene, H2C=CH2?

## Homework Equations

H(reaction) = H(products) - H(reactants)

## The Attempt at a Solution

I've written the balanced equation...

C2H4 + 3O2 -> 2H2O + 2CO2

Now, I think I need to break down the reaction and add all of the energies from broken bonds (reactants) together, and then subtract that sum from the energy released from bonds made (products).

Hf of reactants :
C2H4 -> one C=C bond, four C-H bonds = (614 + 4*413) = 2266 kJ/mol * 1mol = 2266 kJ
O2 -> this is an element in its normal state, so the enthalpy is zero

Hf of products :
H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJI continued like this, but the answer I got was wrong. Am I completely off base? I have a feeling I'm misunderstanding the question and approaching it in the wrong way.

This is the raw equation I got [4(413)+614+3(495)]-[4(463)+4(799)]=-1297 kJ/mol

I'm pretty sure its going to be negative because its Change in H formation=D(reactant bonds)-D(product bonds).

"Hf of products :
H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJ"

In your original equation there are two O-H bonds "per H2O molecule." But you have 2 H2O molecules, so you'll have 4 O-H bonds. Also, drawing Lewis diagrams help.

C2H4 + 3O2 -> 2H2O + 2CO2
So [(4(C−H)+C=C+3(O=O))-(4(O−H)+4(C=O))]

kuahji said:
This is the raw equation I got [4(413)+614+3(495)]-[4(463)+4(799)]=-1297 kJ/mol

I'm pretty sure its going to be negative because its Change in H formation=D(reactant bonds)-D(product bonds).

"Hf of products :
H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJ"

In your original equation there are two O-H bonds "per H2O molecule." But you have 2 H2O molecules, so you'll have 4 O-H bonds. Also, drawing Lewis diagrams help.

C2H4 + 3O2 -> 2H2O + 2CO2
So [(4(C−H)+C=C+3(O=O))-(4(O−H)+4(C=O))]
I multiplied by 2 mol...so I think all of my numbers are the same as yours, I just never used the value for O2 (a huge duh on my part). I guess I thought that the "zero enthalpy" applied because O2 (g) is oxygen's natural state at room temperature.

This was a HUGE help, this problem has been eating up my time all night, and I'm so glad to finally understand where I was going wrong! Thanks!

kuahji said:
This is the raw equation I got [4(413)+614+3(495)]-[4(463)+4(799)]=-1297 kJ/mol

I'm pretty sure its going to be negative because its Change in H formation=D(reactant bonds)-D(product bonds).

"Hf of products :
H2O -> two O-H bonds = (2*463) = 926 kJ/mol * 2mol = 1852 kJ
CO2 -> two C=O bonds = (2*799) = 1598 kJ/mol * 2mol = 3196 kJ"

In your original equation there are two O-H bonds "per H2O molecule." But you have 2 H2O molecules, so you'll have 4 O-H bonds. Also, drawing Lewis diagrams help.

C2H4 + 3O2 -> 2H2O + 2CO2
So [(4(C−H)+C=C+3(O=O))-(4(O−H)+4(C=O))][/QUOTE]

This helped becuase I didnt get the same answer yours but I figured my mistake though.

## 1. What is heat of formation?

Heat of formation is the amount of heat energy released or absorbed during the formation of a chemical compound from its constituent elements, at a constant pressure and temperature.

## 2. How is heat of formation calculated?

Heat of formation is calculated by subtracting the sum of the heat of formation of the products from the sum of the heat of formation of the reactants.

## 3. What is ethylene (C2H4)?

Ethylene (C2H4) is a colorless and flammable gas that is commonly used as a chemical intermediate in the production of various plastics, solvents, and other industrial products.

## 4. What is the heat of combustion of ethylene (C2H4)?

The heat of combustion of ethylene (C2H4) is the amount of heat energy released when one mole of ethylene reacts with oxygen to form carbon dioxide and water, under standard conditions. It is approximately -1411 kJ/mol.

## 5. How does ethylene (C2H4) combustion contribute to climate change?

Ethylene (C2H4) combustion contributes to climate change by releasing carbon dioxide, a potent greenhouse gas, into the atmosphere. This can contribute to the Earth's overall increase in temperature and lead to negative impacts on the environment and human health.