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# Born-Haber Cycles watch

1. I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:

/\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
=-1708 kJ mol-1

But the answer is supposed to be -288 kj mol-1. Can anyone help please?
2. (Original post by Ridium)
I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:

/\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
=-1708 kJ mol-1

But the answer is supposed to be -288 kj mol-1. Can anyone help please?
Eformation = everything else added together
= -288 KJmol^-1

It's not that difficult O_O
3. (Original post by Ridium)
I missed a couple of lessons on Born-Haber cycles so i don't understand it fully but I've been doing some reading up and i can answer most questions. I've just had a bit of trouble with this working out the enthalpy of formation in this question:

Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy is -648 kJmol-1. This is the cycle i made:

/\Hf = -(+109)-(+107)-(+494)+(-314)+(-684)
=-1708 kJ mol-1

But the answer is supposed to be -288 kj mol-1. Can anyone help please?
just at a glance you have started off with I2(g) when it's a solid. You have to add in the enthalpy of vaporisation of iodine (x1/2).
4. Iodine is meant to start off solid at the start, just notice you already had enthalpy of atomisation for Iodine lol. Anyway just add all the values and you get -288.
5. All clockwise arrows=anti clockwise arrows
So, enthalpy formation=109+107+494+(-314}+(-684)
= -288 kj mol-1
6. The following are endothermic

Step 1: Na(s) + &#189;I2(s) --> Na(s) + &#189;I2(g) this is &#189; the enthalpy of vaporisation of iodine

Step 2: Na(s) + ½I2(g) --> Na(s) + I(g) this is &#189; the bond enthalpy of iodine
Step 3: Na(s) + I(g) --> Na(g) + I(g) this is the enthalpy of atomisation of sodium
Step 4: Na(g) + I(g) --> Na+(g) + I(g) this is the ionisation of sodium
---------------------------------------------

Then the exothermic steps:

Step 5: Na+(g) + I(g) --> Na+(g) + I-(g) this is the electron affinity of iodine

Step 6: Na+(g) + I-(g) --> NaI(s) this is the lattice enthalpy

----------------------------------------------

add up the endothermic steps and the exothermic steps keeping the signs correct and you will get the enthapy of formation.
7. (Original post by charco)
just at a glance you have started off with I2(g) when it's a solid. You have to add in the enthalpy of vaporisation of iodine (x1/2).
At A-Level, I think we presume atomisation assumes vaporisation as well.
8. (Original post by Hippysnake)
At A-Level, I think we presume atomisation assumes vaporisation as well.
That's not the issue here. Iodine is a solid under standard conditions so the ΔHf equation MUST read:

Na(s) + &#189;I2(s) --> NaI(s)

Atomisation of iodine would include &#189;(vaporisation + bond dissociation) but you have to start from solid iodine.
9. Yeah i just realised that i put I2 as gas, i was meant to put it as solid and why didn't i get -288 in the start it was so obvious . I mistakenly went anti-clockwise Thanks for the help guys.

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