Need serious help with these enthelpy of neutralisation calculations

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The discussion focuses on the calculations of enthalpy of neutralization for reactions involving NaOH and various acids, specifically HCl, nitric acid, and sulfuric acid. The user calculated the moles of HCl as 0.025 and applied the formula for enthalpy, yielding results of -25.4 kJ for HCl and 27.08 kJ for nitric acid, while sulfuric acid provided a closer value of 52.416 kJ. The discrepancies in results are attributed to hydration interactions of ions in solution, affecting the enthalpy values. The ideal enthalpy of neutralization is approximately 58 kJ, which is consistent across strong acid-base reactions.

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jackal22
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ok i keep reading that ideally they (neutralisation enthalpies) are meant to be around 58KJ

here is my problem:

im reacting 50cm3 of NaOH with 25cm3 of HCl

the HCl is 1 mole NaOH is also 1 mole.

my moles calculation is: (1x25)/1000 which gives me 0.025

the i do :

((mass x specific heat x change in temp)/moles)/1000

i get -25.4 which is no where near 58 ish...

same thing when i use nitric acid instead, i end up with 27.08. However when i use sulphuric acid i get a result of 52.416 (sulphuric acid is 0.5 molar)

so my question is, am i getting something wrong? I am using the mass of the substance needed to neutralise the acid for mass (reading it off a graph) & I am using 4.2 for specific heat

also is there any reason for why some are better at neutralising is it like to do with more H+ ions ?

i would really appreciate any help
 
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Neutralization enthalpies are based on the net ionic equation for water

H+ (aq) + OH- (aq) ----> H2O (l)

that's why the enthalpies are fairly the same. The slight differences may pertain to the hydration interactions of the respective components (for instance, Na + and Cl-) with water.
 
thanks, =)
 

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