# A2 Chemistry

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Thread starter 2 years ago
#1
Need help with (ciii) of this question, where you have to calculate standard enthalpy change of formation of NaN3.

The Na part in my born haber cycle calculation makes sense, but for Nitrogen, why dont we use its enthalpy change of atomisation as we always do for the anion as well as the cation?

And, is the equation "1 1/2 N2 (g) + 1e ---> N3- (g)" given in the table for the same part the equation for its electron affinity?
Last edited by Clark20; 2 years ago
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2 years ago
#2
because you need to use 3/2 N2 + e- -> N3 -
equation shows N2 gaining electrons, not N atoms
so don't atomise N2
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Thread starter 2 years ago
#3
Is there a way to know whether a gaseous ion gains electrons directly without first undergoing atomisation as this case, if suppose we arent given such an equation for a future question?

(Original post by BobbJo)
because you need to use 3/2 N2 + e- -> N3 -
equation shows N2 gaining electrons, not N atoms
so don't atomise N2
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2 years ago
#4
(Original post by Clark20)
Is there a way to know whether a gaseous ion gains electrons directly without first undergoing atomisation as this case, if suppose we arent given such an equation for a future question?
By looking at the ion formed

You should be given the equation as the standard enthalpy change for that reaction cannot be determined from the values in your data booklet
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Thread starter 2 years ago
#5
(Original post by BobbJo)
You should be given the equation as the standard enthalpy change for that reaction cannot be determined from the values in your data booklet
Standard enthalpy change of such an equation includes only the enthalpy change of electron affinity, right?

For e.g in the case above +142 kJmol-1 is the enthalpy change produced as a result of just adding up all 3 electron affinities of N and multiplying by 3 as there are 3 Nitrogen atoms in total?

Also, whats the reason N2 doesnt need atomisation in this case? Like, take Al2O3 as a similar example. 1 1/2 O2 is atomised there to 3O. Why cant that too accept electrons without atomisation first?
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2 years ago
#6
(Original post by Clark20)
Standard enthalpy change of such an equation includes only the enthalpy change of electron affinity, right?
No
For e.g in the case above +142 kJmol-1 is the enthalpy change produced as a result of just adding up all 3 electron affinities of N and multiplying by 3 as there are 3 Nitrogen atoms in total?
No
Also, whats the reason N2 doesnt need atomisation in this case? Like, take Al2O3 as a similar example. 1 1/2 O2 is atomised there to 3O. Why cant that too accept electrons without atomisation first?
N3- is formed, not N3-.
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Thread starter 2 years ago
#7
(Original post by Clark20)
For e.g in the case above +142 kJmol-1 is the enthalpy change produced as a result of just adding up all 3 electron affinities of N and multiplying by 3 as there are 3 Nitrogen atoms in total?
I meant, its the enthalpy change of 1st electron affinity of Nitrogen. And we multiply that by 3 as there are 3 atoms of Nitrogen* still I guess it includes more unknown enthalpy changes added up.

Can you please if possible explain by comparison why 1 1/2 N2 doesn't need atomisation in calculating the lattice enthalpy of NaN3 but 1 1/2 O2 is actually atomised in calculating the lattice enthalpy of Al2O3?
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Thread starter 2 years ago
#8
(Original post by BobbJo)
By looking at the ion formed
If it was 3N^-1 then it would've gone through atomisation?
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2 years ago
#9
(Original post by Clark20)
I meant, its the enthalpy change of 1st electron affinity of Nitrogen. And we multiply that by 3 as there are 3 atoms of Nitrogen* still I guess it includes more unknown enthalpy changes added up.

Can you please if possible explain by comparison why 1 1/2 N2 doesn't need atomisation in calculating the lattice enthalpy of NaN3 but 1 1/2 O2 is actually atomised in calculating the lattice enthalpy of Al2O3?
They gave you the structure of the azide ion in earlier parts. It is not 3 N- but N3-.

(Original post by Clark20)
If it was 3N^-1 then it would've gone through atomisation?
Yes
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Thread starter 2 years ago
#10
(Original post by BobbJo)
They gave you the structure of the azide ion in earlier parts. It is not 3 N- but N3-.

Yes
Alright. Thank you.
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Thread starter 2 years ago
#11
(Original post by BobbJo)
They gave you the structure of the azide ion in earlier parts. It is not 3 N- but N3-.

Yes
By the way, does N3- (3 is in subscript) indicate 3 Nitrogen atoms accept 1 electron each? Or only 1 of the 3 Nitrogens accepts an electron as shown in the dot and cross diagram?
Last edited by Clark20; 2 years ago
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2 years ago
#12
(Original post by Clark20)
By the way, does N3- (3 is in subscript) indicate 3 Nitrogen atoms accept 1 electron each? Or only 1 of the 3 Nitrogens accepts an electron as shown in the dot and cross diagram?
The latter. If it accepted 3 electrons, the charge would be -3. The charge is only -1
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Thread starter 2 years ago
#13
(Original post by BobbJo)
The latter. If it accepted 3 electrons, the charge would be -3. The charge is only -1
If the charge was -1 and its written as 3 N- that would have meant each Nitrogen accepts an electron, right?

The idea that usually calculating lattice enthalpy involves every single atom of the anions accepting an electron/electrons (For e.g Cl2 in NaCl, O2 in MgO, O2 in Al2O3) after going through atomisation, and how in this case, unusually, not all N atoms accept an electron each can be used to understand why it doesnt go through atomisation?

Like, does atomisation somehow ensure each atom (those atoms that become anions of course) in the compound accepts an electron? If suppose not all atoms of a compound gain an electron/electrons each (unlike in the example of calculating lattice enthalpies of MgO and NaCl, but as the example of NaN3-), can we assume atomisation hasnt simply happened?
Last edited by Clark20; 2 years ago
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