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Topic: Enthalpy of neutralization  (Read 5619 times)

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Offline blern

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Enthalpy of neutralization
« on: October 18, 2011, 04:07:10 PM »
Hi.  I was wondering if I did this problem correctly

A dilute solution of HCl with a mass of 610.29g and containing 0.33183 mol of HCl was exactly neutralized in a calorimeter by the NaOH in a 615.31g of a comparably dilute solution.  The temperature increased from 16.784 to 20.610 degrees Celsius.  The specific heat of the HCl was 4.301J/g degrees Celsius and that of the NaOH was 4.045J/g degrees Celsius.  The heat capacity of the calorimeter was 77.99J/degrees Celsius.  Assume that the original solutions made independent contributions to the total heat capacity of the system following their mixing.  Calculate the ΔH of the neutralization in Joules/mole of H+.

So I know:
ΔT=3.829 degrees C
Ccal=77.99J/degrees C
CHCl=4.031J/g degrees C
CNaOH=4.045J/g degrees C

I'm not positive in the next step.  I assumed that since the solutions independently contributed to the total heat capacity that I do a calculation for each so:

ΔH surroundings=(77.99C/J x 3.826C) + (4.031J/gC x 610.29g x 3.826C) + (4.046J/gC x 615.31g x 3.826C)= 294.4J + 9412J + 9525J = 19235.4J
so ΔH reaction= -19234.4J

To find the J/mol H+ I did -19235.4J/.33183 mol H+ since I was given .33183 mol HCl which = -57968J/mol H+.

Is this the correct way to do the problem?

Offline Borek

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Re: Enthalpy of neutralization
« Reply #1 on: October 18, 2011, 05:48:26 PM »
I just skimmed - and I don't see anything blatantly wrong.
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