Enthelpy of neutralization of HCl and NaOH

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SUMMARY

The enthalpy of neutralization for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) was experimentally determined using 0.2 M solutions. The calculated enthalpy change was approximately -52.92 kJ for 1 M HCl, while online sources reported values around -57.1 kJ. The discrepancy is attributed to measurement errors and the specific heat capacity used in calculations, which should be 4.184 J/g°C instead of 1.184 J. Accurate temperature measurement and proper unit usage are critical for reliable results.

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There is the experiment of Enthalpy of neutralization of strong HCl and strong base NaOH

Homework Statement

Chemicals: 0.2 M HCl and 0.2 M NaOH
Polythene bottle. Water equivalent of polythene bottle e=900
t1=29.7 c
temp of mixture= 32 c
Rise in temperature = 2.3 c

The attempt at a solution
enthalpy change during neutralization of 100 mL 0.2 M HCl=-(200+e)(rise in temp)1.184 J

For 1 M HCl = -(200+e)(rise in temp)1.184/0.2×1000
=-52.92 KJ

This was my answer. But on the internet, I found the answer of same around -57.1 KJ

Some of my frnd's answer was near to it. I tried many times, but my answer was nearly -53 KJ..

Actually, I performed this experiment alone, as I was absent during this test. So, please tell me am I right or wrong?

What is the absolute value of enthalpy of neutralization of strong acid HCl and strong base NaOH.
Is it same for dilute solution and for strong or different? what are the exact values..


Please answer soon..
 
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Hard to say what you did, as the data is not complete and partially wrong - you have not listed volumes of NaOH nor HCl, specific heat of water is not 1.184 (and it is not in J) and so on. But assuming these are just typos and problems with the description, and your approach was correct, you are off by about 7% - which is not bad, assuming you measure the temperature change with 0.1°C accuracy. 0.1°C/2.3°C means an intrinsic error of about 4%.
 
whyonlyme said:
The attempt at a solution
enthalpy change during neutralization of 100 mL 0.2 M HCl=-(200+e)(rise in temp)1.184 J

Here it was.. Sorry, I didn't mention it in data.

But actually, in my practical book, there is no need of volume in calculation. It shows some calculation, initial line is

For 1 M HCl = -(200+e)(rise in temp)4.184/0.2×1000

We just have to find rise in temp. by thermometer in polythene bottle. and put it in the calculation, and u get ur answer.

And sorry, that is 4.184 not 1.184. A conversion factor..
 
Last edited:
OK< I have missed that 100 mL part.

Isn't 200 in the book formula the total volume? Each time someone writes something like that, ignoring units, it is a possible source of confusion.

Please don't use txtspeak at the forum.
 

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