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### Topic: Calculating Delta H of the reaction  (Read 37912 times)

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#### sabah

• Regular Member
•   • Posts: 9
• Mole Snacks: +0/-2 ##### Calculating Delta H of the reaction
« on: December 06, 2007, 12:42:06 AM »
When sodium Chloride is strongly heated in a flame, the flame takes on the yellow color associated with the emission spectrum of sodium atoms. the reaction that occurs in the gaseous state is

Na(g) + Cl- (g) -----> Na (g) + Cl (g)

Calculate delta H for this reaction.

i really need help with the question. even though my teacher solved it a little i still do not understand it.

This is how she did the question.

Na + e ----> Na  delta H = - Ii
Cl- --------> Cl + e  delta H = - EA
----------------------------------------
delta H =  delta H1 + Delta H 2

pleaseeeee can anyone explain it to me... #### champ

• Regular Member
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• Mole Snacks: +0/-7 ##### Re: Calculating Delta H of the reaction
« Reply #1 on: December 08, 2007, 07:05:54 PM »
here is the solution for this question....

Na + e ----> Na  delta H = - Ii
Cl- --------> Cl + e  delta H = - EA
----------------------------------------
delta H =  delta H1 + Delta H 2
Ii is stand for Ionization energy (losing of electron)

EA is stnad for Electron Affinity..(gaining of electron)

U just have to look for the First Ionization Energy of Sodium and first Electron Affinity for Cl.Then cancle out the electrons. U should have the following equation..
Na(g) + Cl- (g) -----> Na (g) + Cl (g)
After getting this equation then just simply use the Hess's law to calculate the Delta H...UNits are in Kj/mole...