Need serious help with these enthelpy of neutralisation calculations

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The discussion centers on calculating the enthalpy of neutralization for reactions involving NaOH and various acids. The user is obtaining results significantly lower than the expected value of around 58 kJ, with calculations yielding -25.4 kJ for HCl and 27.08 kJ for nitric acid, while sulfuric acid gives a closer result of 52.416 kJ. The user is using the correct formulas but questions whether their approach to mass and specific heat is accurate. The response clarifies that enthalpy values are generally similar due to the net ionic equation for water formation, with variations possibly arising from hydration interactions of ions. Understanding these factors is essential for accurate enthalpy calculations in neutralization reactions.
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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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