Enthelpy of neutralization of HCl and NaOH

[whyonlyme]

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..

 

[Borek]

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]

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..

 

[Borek]

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.

 

[DeeNos]

I cannot provide a definitive answer without more information about the experiment and the conditions in which it was conducted. However, I can offer some insights and explanations that may help clarify the results.

Firstly, the enthalpy of neutralization is defined as the heat released or absorbed when an acid and a base react to form a neutral solution. This value is dependent on several factors, such as the strength of the acid and base, the concentration and volume of the solutions, and the temperature at which the reaction takes place.

In this experiment, you used 0.2 M solutions of HCl and NaOH, which are considered to be strong acids and bases. This means that they completely dissociate in water, releasing a large amount of energy when they react. The enthalpy change during neutralization is negative because energy is released in the form of heat.

The rise in temperature of the solution is a measure of the amount of heat released during the reaction. The equation you used to calculate the enthalpy change takes into account the water equivalent of the polythene bottle, which is the amount of water that would have absorbed the same amount of heat as the bottle. This value is subtracted from the total heat released to get the enthalpy change.

The value you obtained, -52.92 kJ, is a reasonable estimate for the enthalpy of neutralization of 0.2 M HCl and NaOH. However, it is important to note that this value may vary slightly depending on the specific conditions of the experiment, such as the accuracy of temperature measurements and the purity of the solutions.

The value of enthalpy of neutralization for a strong acid and a strong base is generally considered to be the same, regardless of the concentration of the solutions. This is because the strength of the acid and base does not change with concentration. However, if the solutions were not dilute, the volume of the solutions may affect the enthalpy change due to differences in heat capacity.

In conclusion, your answer of -52.92 kJ is a reasonable estimate for the enthalpy of neutralization of 0.2 M HCl and NaOH. It is important to note that the absolute value of the enthalpy of neutralization may differ slightly depending on the specific conditions of the experiment. Additionally, the enthalpy of neutralization is considered to be the same for dilute solutions of strong acids and bases.