Two kJ/mol values in Data booklet, what do they mean?

[IB`NOT`ez]

I'm really confused as to why there are two different kJ/mol values in Section 12 of the Data Booklet (see pic attached)

Usually I use the kJ/mol values from the first section and it comes out correct, but so what's the second one for? The 3rd section are entropy values right? 

Also, this is probably a really dumb question, but I still get confused from time to time with questions asking enthalpy changes. 

What I think is correct:

Bond enthalpy = Reactants - Products

Enthalpy change of formation = Products - Reactants

Combustion = Reactants - Products

Enthalpy change = Products - reactants (?)

Would really appreciate any clarification on this.

Thanks. 

[kw0573]

Your version of the Data Booklet is not up to date. You should use the third version: http://www3.sd71.bc.ca/School/highland/Programs/IB/Chemistry/Documents/May 2016 Chem Data Booklet 3rd Ed.pdf
There you can see that the second column of kJ/mol is for standard Gibb's free energy of formation. 

[kw0573]

33 minutes ago, IB`ez said:

Also, this is probably a really dumb question, but I still get confused from time to time with questions asking enthalpy changes. 

What I think is correct:

Bond enthalpy = Reactants - Products

Enthalpy change of formation = Products - Reactants

Combustion = Reactants - Products

Enthalpy change = Products - reactants (?)

Would really appreciate any clarification on this.

Thanks. 

The key concept is in 5.3: bond-forming releases energy. Hence bond enthalpies are negative. It is more intuitive if you take Physics HL, and is usually proven using calculus in a first-year university physics course. For the time being, you should equate stability with low energy. 

If using enthalpy of formation of species to calculate change in enthalpy, you always do 

(sum of enthalpies of formation of products) - (sum of enthalpies of formation of reactants).

But if using the absolute values of bond enthalpy instead of the actual negative values, you do 

(sum of given bond enthalpies for reactants) - (sum of given bond enthalpies for products)

to take the negative sign into consideration. It has nothing to do with if a reaction is combustion or not. 

I present a simplified derivation.
1) By definition of change in enthalpy: ΔH_reaction = H_products - H_{reactants =} ΔH_{f_products} - ΔH_{f_reactants}. "Change" in Chemistry/Physics always means final minus initial. The version with standard enthalpies of formation is commonly used, as we usually can't easily evaluate absolute enthalpy of species.

2) Because products/reactants are formed from the same elements, I state without detailed proof that the difference in the enthalpies of formation of the products/reactants are due to the different bonds. and that other energies, such as differences in vibrations or rotations, are negligible. You can learn more about this in a thermodynamics class. 
ΔH_{f_products} - ΔH_{f_reactants} = ΔH_{bond_products} - ΔH_{bond_reactants.}

3) We know that "bond-forming releases energy", (from syllabus 5.3), so bond enthalpies are negative. However the data booklet provides the absolute values, the positive values. 
ΔH_{bond_products} - ΔH_{bond_reactants}= -|ΔH_{bond_products}| - (-|ΔH_{bond_reactants}|) =  |ΔH_{bond_reactants}| - |ΔH_{bond_products}|. The required relationship is proven. 

[Msj Chem]

Simply put, bonding breaking is endothermic and bond making is exothermic. So to calculate the change in enthalpy, it's reactants - products. 

Using enthalpy of formation values, it's products - reactants. 

Using enthalpy of combustion values, it's reactants - products. 

For delta G and delta S, it's also products - reactants. 

[IB`NOT`ez]

9 hours ago, kw0573 said:

The key concept is in 5.3: bond-forming releases energy. Hence bond enthalpies are negative. It is more intuitive if you take Physics HL, and is usually proven using calculus in a first-year university physics course. For the time being, you should equate stability with low energy. 

If using enthalpy of formation of species to calculate change in enthalpy, you always do 

(sum of enthalpies of formation of products) - (sum of enthalpies of formation of reactants).

But if using the absolute values of bond enthalpy instead of the actual negative values, you do 

(sum of given bond enthalpies for reactants) - (sum of given bond enthalpies for products)

to take the negative sign into consideration. It has nothing to do with if a reaction is combustion or not. 

I present a simplified derivation.
1) By definition of change in enthalpy: ΔH_reaction = H_products - H_{reactants =} ΔH_{f_products} - ΔH_{f_reactants}. "Change" in Chemistry/Physics always means final minus initial. The version with standard enthalpies of formation is commonly used, as we usually can't easily evaluate absolute enthalpy of species.

2) Because products/reactants are formed from the same elements, I state without detailed proof that the difference in the enthalpies of formation of the products/reactants are due to the different bonds. and that other energies, such as differences in vibrations or rotations, are negligible. You can learn more about this in a thermodynamics class. 
ΔH_{f_products} - ΔH_{f_reactants} = ΔH_{bond_products} - ΔH_{bond_reactants.}

3) We know that "bond-forming releases energy", (from syllabus 5.3), so bond enthalpies are negative. However the data booklet provides the absolute values, the positive values. 
ΔH_{bond_products} - ΔH_{bond_reactants}= -|ΔH_{bond_products}| - (-|ΔH_{bond_reactants}|) =  |ΔH_{bond_reactants}| - |ΔH_{bond_products}|. The required relationship is proven. 

Thank you so much for that derivation, it helped clarify a lot of things. Also I didn't know there was a latest edition of the data booklet -- do you know if that one was also the one used in the May 2016 exam session? (not that important a question, just curious)

@MSJChem Thank you as well for the confirmation! 

[kw0573]

1 hour ago, IB`ez said:

Thank you so much for that derivation, it helped clarify a lot of things. Also I didn't know there was a latest edition of the data booklet -- do you know if that one was also the one used in the May 2016 exam session? (not that important a question, just curious)

@MSJChem Thank you as well for the confirmation! 

It was updated on May 2015, so yes.