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### Topic: With regards to enthalpy change in the reaction of thermite  (Read 5380 times)

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#### Ace888

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##### With regards to enthalpy change in the reaction of thermite
« on: April 28, 2010, 08:18:20 PM »
Hi people,
I'm a little bit stuck in terms of getting the values to calculate for the change in Enthalpy

2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s)
Given,
Δ fH[Fe2O3(s)] = −826 kJ/mol and Δ fH[Al2O3(s)] = −1676 kJ/mol

Find delta H (change in enthalpy).. I know that you seperate the equation into 2, one the reactant and the other one is the product
Hp - Hr = Change in Enthalpy (Thats how I remember it)

Now, how do I find kJ/mol for Al and Fe if its not given? Can anyone help me out?

#### tamim83

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##### Re: With regards to enthalpy change in the reaction of thermite
« Reply #1 on: April 29, 2010, 08:32:58 AM »
What is the heat of formation for a pure element?

#### Ace888

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##### Re: With regards to enthalpy change in the reaction of thermite
« Reply #2 on: April 29, 2010, 09:14:02 AM »
I'm not sure about the heat of formation,
However, I found a website that also shows how to calculate the enthalpy change in the reaction of thermite!
BUT They did not consider Fe and Al alone? They just substracted the Product from the reactant..

Im really stuck with the ideology of this method.. please help me out in terms of the concept

#### tamim83

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##### Re: With regards to enthalpy change in the reaction of thermite
« Reply #3 on: April 29, 2010, 12:06:45 PM »
The heat of formation ( :delta:Hf) for any pure element in its standard state is 0.  Therefore, the heat of formation for Fe and Al (the metals themselves) is 0.  So, you do have everything you need for the calculation.  You are on the right track.