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Physical Chemistry

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Joined:	14 Apr 2008
Post:	55

5 May 2008 19:50:41 IST

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400

Thermochemistry

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150 ml of 0.5 N nitric acid solution at 25.35 degree celcius was mixed with 150 ml of 0.5 N NaOH at the same temperature.Final temperature was28.77 degree celcius. Calculate the heat of neutralization.

Ans: -13.68 kcal

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Shikhar Shrivastav

Cool goIITian

Joined: 6 May 2007

Posts: 87

5 May 2008 23:14:16 IST

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Since the soln is dilute so for final soln -300ml volume implies 300gms of water. as 1ml =1gm of water.

Heat of neutralisation is the heat liberated per gram equivalent -defn.

Eqnts=.15*.5=.075

change in temp =3.42

So heat =ms*(change in temp)

300*1*3.42=1026

1026/.075=13680cal=13.68 kcal.

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