as level enthalpy concept I don't understand

Say propane is burnt in oxygen to form carbon dioxide and water. Why is the oxygen "ignored" when calculating enthalpy change? Wouldn't the bonds in oxygen have to be broken so it react form carbon dioxide and water - so wouldn't it require energy?

Reply 1

KireLennox

6

I don’t think it is? I’m pretty sure you still need to add the bond energy for oxygen - is this from a specific question?

Reply 2

klafow

13

Original post by unsure-student

Say propane is burnt in oxygen to form carbon dioxide and water. Why is the oxygen "ignored" when calculating enthalpy change? Wouldn't the bonds in oxygen have to be broken so it react form carbon dioxide and water - so wouldn't it require energy?

is this for hess cycles? i doubt i'll explain it clearly but i'll try my best
(i'll be referencing example 7 on page 8 of this document btw. 3.2.1-enthalpy-changes.pdf - its propene so pretty similar)

in hess cycles, you calcuate the enthalpy change of reaction/whatever by using the enthalpy changes of combustion/formation of the products and reactants of an equation, which creates the cycle you see in the example. you don't ignore the oxygen when using enthalpy changes of combustion (as seen in example 9 actually) but you do for formation.
with enthalpy changes of formation, for any compounds in the equation you'll have the values for the energy needed to form each of those compounds from its elements and through some addition and subtraction you calculate what you need. for example, deltafH CO2 (g)= –394 kJ mol^-1
however, oxygen is already an element so you can't exactly form it from its elements if you get what i mean. so the enthalpy change of formation is 0, and that means it ends up getting ignored when calculating enthalpy changes.

i hope that made sense! reading it back it does sound a bit confusing, so if you have any qs let me know and i'll try my best to answer them 🙂

Reply 3

TypicalNerd

18

Original post by unsure-student

Say propane is burnt in oxygen to form carbon dioxide and water. Why is the oxygen "ignored" when calculating enthalpy change? Wouldn't the bonds in oxygen have to be broken so it react form carbon dioxide and water - so wouldn't it require energy?

It depends on the method you are using.

If you are working with bond enthalpies, you account for it.

If you are working with standard enthalpies of formation, you don’t need to account for the “formation” of oxygen from its standard state as the equation would be O2 (g) —> O2 (g). This of course must have an enthalpy change of 0, since no change occurs to the oxygen molecule in this process.

Reply 4

Nitrotoluene

8

Original post by unsure-student

Say propane is burnt in oxygen to form carbon dioxide and water. Why is the oxygen "ignored" when calculating enthalpy change? Wouldn't the bonds in oxygen have to be broken so it react form carbon dioxide and water - so wouldn't it require energy?

Hello unsure-student!

It is the result of Hess's Law.
Elements have 0 kJ/mol, so oxygen has 0 kJ/mol.

Have a look: Hess’s Law of Heat Summation

Bye,
Sandro

as level enthalpy concept I don't understand

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