Kaplan 45 - Chemistry Discrete Hess Law

[lamborghiniMD]

Ok so I'm a little confused about how Kaplan decided to solve this problem:

Calculate the change in enthalpy of the formation of Fe2O3
1. 6Fe2O3 --> 4Fe3O4 + O2 (H = 472 kJ/mol)
2. 3Fe + 2O2 --> Fe3O4 (H = -1118.4 kJ/mol)


Here's what I know:
- You need to reverse equation 1 so that 6Fe2O3 is formed, thus H = -472
- Hrxn = Hproducts - Hreactants
- substitute the value of formation in the second equation into the first
- thus, 4Fe3O4 + O2 --> 6Fe2O3 (H = -472 kJ/mol)
so you have 4(-1118.4) + O --> 6(Fe2O3) = - 472

then Hrxn (-472) = Hproducts (6Fe2O3) - Hreactants (4 x -1118.4),
so then you just solve for Fe2O3 right?


Here's how Kaplan says it should be:

Hrxn (+ 472) = Hproducts (4 x - 1118.4) -Hreactants (6Fe2O3)

which is completely the opposite of what I did. But which way is correct?

The only reason why I reversed equation 1 was because the question asked for the enthaply formation of Fe2O3, so you had to reverse equation 1 right?

Can some help clarify this? Thank you!

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[indianjatt]

haha. I misread the question

 

[V5RED]

Ok so I'm a little confused about how Kaplan decided to solve this problem:

Calculate the change in enthalpy of the formation of Fe2O3
1. 6Fe2O3 --> 4Fe3O4 + O2 (H = 472 kJ/mol)
2. 3Fe + 2O2 --> Fe3O4 (H = -1118.4 kJ/mol)


Here's what I know:
- You need to reverse equation 1 so that 6Fe2O3 is formed, thus H = -472
- Hrxn = Hproducts - Hreactants
- substitute the value of formation in the second equation into the first
- thus, 4Fe3O4 + O2 --> 6Fe2O3 (H = -472 kJ/mol)
so you have 4(-1118.4) + O --> 6(Fe2O3) = - 472

then Hrxn (-472) = Hproducts (6Fe2O3) - Hreactants (4 x -1118.4),
so then you just solve for Fe2O3 right?


Here's how Kaplan says it should be:

Hrxn (+ 472) = Hproducts (4 x - 1118.4) -Hreactants (6Fe2O3)

which is completely the opposite of what I did. But which way is correct?

The only reason why I reversed equation 1 was because the question asked for the enthaply formation of Fe2O3, so you had to reverse equation 1 right?

Can some help clarify this? Thank you!

We are given Hrxn. We are given Hformation for Fe3O4(in a rather odd way since usually it is just presented as Hformation, not as a whole reaction). We already know that Hformation of O2 is zero. We want to find Hformation of Fe2O3.

Fe2O3 is the reactant. Fe3O4+O2 are the products.

Hrxn=4*H(Fe3O4)+4*H(O2)-6*H(Fe2O3)
Thus.
472kJ=(-4*1118kJ)-(6*HFe2O3)
Thus.
(472kJ+4*1118kJ)/(-6)=HFe2O3=-824kJ which is almost exactly the value for Hformation of Fe2O3 in the table in my chemistry text.

 

[lamborghiniMD]

Ok, but why must 6Fe2O3 be the reactant? Shouldn't you reverse the first equation so that 6Fe2O3 is the products for the formation?

There was a similar problem in AAMC FL #3 where this was the case.

I guess I don't understand why you wouldn't reverse the first equation when the question asked for the formation of Fe2O3.

 

[Singhp03]

We are given Hrxn. We are given Hformation for Fe3O4(in a rather odd way since usually it is just presented as Hformation, not as a whole reaction). We already know that Hformation of O2 is zero. We want to find Hformation of Fe2O3.

Fe2O3 is the reactant. Fe3O4+O2 are the products.

Hrxn=4*H(Fe3O4)+4*H(O2)-6*H(Fe2O3)
Thus.
472kJ=(-4*1118kJ)-(6*HFe2O3)
Thus.
(472kJ+4*1118kJ)/(-6)=HFe2O3=-824kJ which is almost exactly the value for Hformation of Fe2O3 in the table in my chemistry text.

thats the answer I got👍

 

[V5RED]

Ok, but why must 6Fe2O3 be the reactant? Shouldn't you reverse the first equation so that 6Fe2O3 is the products for the formation?

There was a similar problem in AAMC FL #3 where this was the case.

I guess I don't understand why you wouldn't reverse the first equation when the question asked for the formation of Fe2O3.

You are thinking about Hess law problems where you set the reactions so that they produce a net reaction. In that case you sum the enthalpies of each reaction and that leads to a net enthalpy of reaction.

This can be done for this problem, but it takes a few more steps than doing it the way Kaplan showed.

If we wanted to do that for this problem we would need to flip reaction 1 and multiply reaction 2 by 4.

1. 4Fe3O4 + O2 --> 6Fe2O3 (H = -472 kJ))
2. 4*(3Fe + 2O2 --> Fe3O4 (H = -1118.4 kJ/mol))
Net. 12Fe + 8O2 --> 6Fe2O3 (H=-4*1118.4+-472)

In this case Hreaction is -4945.6kJ/6mol Fe2O3. Thus Hformation of Fe2O3 is -824kJ/mol.

 

[lamborghiniMD]

Yes you're right. I came up with the same answer too. Thanks !

 

[circulus vitios]

Ok, but why must 6Fe2O3 be the reactant? Shouldn't you reverse the first equation so that 6Fe2O3 is the products for the formation?

There was a similar problem in AAMC FL #3 where this was the case.

I guess I don't understand why you wouldn't reverse the first equation when the question asked for the formation of Fe2O3.

Could someone explain this since it was never answered?

What is the difference because Hrxn and Hformation? Isn't a formation a reaction? Man, TBR's treatment of this section is embarrassingly bad.

 

[circulus vitios]

Could someone explain this since it was never answered?

What is the difference because Hrxn and Hformation? Isn't a formation a reaction? Man, TBR's treatment of this section is embarrassingly bad.

Anyone?