[John623]

Let's say that when mixing some H₂SO₄ and NaOH together we produced 2 kJ of heat.

The thermochemical equation for the reaction is: 0.5H₂SO₄ + NaOH -> 0.5Na₂SO₄ + H₂O

I believe the formula for enthalpy of neutalisation for these two compounds will be 2kJ/(the number of moles of the limiting factor)

Let's say that we had 1.1 moles of H₂SO₄ and 2 moles of NaOH, this would make NaOH the limiting factor

So would we divide by 2? Or would we divide by 1, as 1 mol of H₂SO₄ is used when 2 moles of NaOH are used?

Then let's say we had 1 moles of H₂SO₄ and 2.1 moles of NaOH, this would make H₂SO₄ the limiting factor

So would we divide by 1?

I can't tell if you're supposed to divide by the number of moles of the acid that are actually used, or the number of moles of the base that are actually used.

[thetada]

"The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. It is a special case of the enthalpy of reaction. It is defined as the energy released with the formation of 1 mole of water(my emphasis)." (Wikipedia)

work out the number of moles of water produced and divide by that figure. In this case, it will be the same as the number of moles of NaOH that reacted.

[John623]

I see, thank you.

I was confused because I keep seeing this everywhere: http://www.chm.davidson.edu/vce/Calorimetry/HeatOFNeutralization.html


ΔHneut =   (heat of neutralisation)/n

where n is the moles of the limiting reactant. I don't know why it didn't say the number of moles of water produced instead.

[thetada]

Well of course the limiting reagent is absolutely relevant to the discussion, but the page you linked is talking about a neutralisation reaction with a 1:1 stoichiometry, which simplifies the discussion in one sense, but complicates it for the context you were considering.