# Born Haber Cycle

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#1

Thanks!
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7 years ago
#2
Na is solid because you don't find sodium gas naturally. Sodium is naturally a solid. The definition of enthalpy of formation is the standard enthalpy change 1 one mole of a compound is formed from its elements in their standard states
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#3
(Original post by Munrot07)
Na is solid because you don't find sodium gas naturally. Sodium is naturally a solid. The definition of enthalpy of formation is the standard enthalpy change 1 one mole of a compound is formed from its elements in their standard states
I dont understand why its in 2 steps rather than just the 1: NaCl(s) ==> Na(g) + Cl(g)
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7 years ago
#4
(Original post by Zenarthra)
I dont understand why its in 2 steps rather than just the 1: NaCl(s) ==> Na(g) + Cl(g)
You have to do it in small steps that we know the enthalpy for and can measure. We know the enthalpy change of Na(s) to Na(g) so can use it. We know the enthalpy change of 1/2Cl2(g) to Cl(g) and we know the enthalpy change of NaCl(s) to Na(s) + Cl(g). You have to do every small step cause we know the enthalpy changes of these small steps. The enthalpy change of NaCl(s) to Na(g) + Cl(g) is hard to measure.
0
7 years ago
#5
(Original post by Zenarthra)
I dont understand why its in 2 steps rather than just the 1: NaCl(s) ==> Na(g) + Cl(g)
You need to split it into two steps because you can only find an accurate enthalpy change of small sections of the whole process.

1/2cl2 (g) + Na(s) are the elements in their standard states.
They then need to be ionisised into a gas hence 1/2cl2(g) + Na(g).
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#6
(Original post by Munrot07)
You have to do it in small steps that we know the enthalpy for and can measure. We know the enthalpy change of Na(s) to Na(g) so can use it. We know the enthalpy change of 1/2Cl2(g) to Cl(g) and we know the enthalpy change of NaCl(s) to Na(s) + Cl(g). You have to do every small step cause we know the enthalpy changes of these small steps. The enthalpy change of NaCl(s) to Na(g) + Cl(g) is hard to measure.
So is the atomisation enthalpy for Na:
NaCl(s) ==> Na(g) + 1/2Cl2(g)

And Atomisation enthalpy for Cl:
NaCl(s) ==> 1/2Cl2(g) + Na(s)
0
7 years ago
#7
(Original post by Zenarthra)
So is the atomisation enthalpy for Na:
NaCl(s) ==> Na(g) + 1/2Cl2(g)

And Atomisation enthalpy for Cl:
NaCl(s) ==> 1/2Cl2(g) + Na(s)
Na(s) --> Na(g) is the atomisation of Na

1/2Cl2(g) --> Cl(g) is the atomisation of Cl2

You don't start at NaCl(s) that is the final product after the born haber cycle (or from formation). You start with Na(s) + 1/2Cl2(g)
0
#8
(Original post by Munrot07)
Na(s) --> Na(g) is the atomisation of Na

1/2Cl2(g) --> Cl(g) is the atomisation of Cl2

You don't start at NaCl(s) that is the final product after the born haber cycle (or from formation). You start with Na(s) + 1/2Cl2(g)
thnaks!
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