Chemistry Understanding Heat of Neutralization in Acid-Base Reactions

AI Thread Summary
The discussion centers on the heat of neutralization in acid-base reactions, specifically between NaOH and HCl. It clarifies that the net reaction can be simplified to H+ + OH- → H2O, ignoring the Na+ and Cl- ions, which do not participate in further reactions. The deltaH values for the proposed reaction and the simplified reaction are considered equivalent, as the ions remain unchanged in solution. The confusion about the production of 55.6 moles of water from 1 mole of H+ and OH- is resolved by noting that the additional volume comes from the total solution, not from the reaction itself. The heat of neutralization is confirmed to be approximately -57.9 kJ/mole, primarily attributed to the reaction of H+ with OH-.
Kqwert
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Homework Statement
What will the temperature be if 1 mole NaOH (aq) and 1 mole HCl (aq) are mixed when the total volume of the solutions are 1 L and both solutions were at 25 °C prior to the mixing?
Relevant Equations
.
I have a couple of questions related to this task.

The reaction that I proposed was this:

NaOH(aq) + HCl(aq) <--> Na+ + Cl- + H2O (l)

where as the solution manual have this net reaction, as nothing will happen with the Na+ and Cl- ions:

H+ + OH- <--> H2O.

I assume these reactions will yield the same deltaH value? If that is true, Is deltaH(for example) for deltaH(NaOH(aq)) = deltaH(Na+(aq))+deltaH(Cl-(aq)) ?

Secondly, if we go from the reaction proposed by the solution manual. How can 1 mole of H+ and OH- yield 55.6 moles of water (1 L)? (I thought it would just yield 1 mole H2O). Is it due to the rest of the water in the solution?
 
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Kqwert said:
Problem Statement: What will the temperature be if 1 mole NaOH (aq) and 1 mole HCl (aq) are mixed when the total volume of the solutions are 1 L and both solutions were at 25 °C prior to the mixing?
Relevant Equations: .

I have a couple of questions related to this task.

The reaction that I proposed was this:

NaOH(aq) + HCl(aq) <--> Na+ + Cl- + H2O (l)

where as the solution manual have this net reaction, as nothing will happen with the Na+ and Cl- ions:

H+ + OH- <--> H2O.

I assume these reactions will yield the same deltaH value? If that is true, Is deltaH(for example) for deltaH(NaOH(aq)) = deltaH(Na+(aq))+deltaH(Cl-(aq)) ?

Secondly, if we go from the reaction proposed by the solution manual. How can 1 mole of H+ and OH- yield 55.6 moles of water (1 L)? (I thought it would just yield 1 mole H2O). Is it due to the rest of the water in the solution?
You got the answer in your last paragraph. How many moles of water will produce from the NaOH and HCl ? Where must the rest of it come from?
 
1 mole of water, correct? And then the rest of it must come from the fact that it's a 1L solution?

What about my first question, that relating to the deltaH calculations?
 
It's been several years since I've done these calculations, but HCl in solution will be H+ ions and Cl- ions. Same with NaOH in solution. It will already be Na+ ions and OH- ions.
 
Kqwert said:
What about my first question, that relating to the deltaH calculations?

Did you google 'heat of neutralization' ?
 
I did it now, and it seems like we simply ignore the cations in our deltaH calculations if they do not react after dissociation..?
 
What dissociation ? (and Cl- is not a cation)
 
Sorry - cation and anion!

I mean if we have the following equation:

NaOH(aq) + HCl(aq) <--> Na+(aq) + Cl-(aq) + H2O (l)

we can simply write it as

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

and calculate deltaH for that equation, as no further reaction happens with the Na+ and Cl- cation/anion?
 
That's right. They were ions in water and they remain ions in water.

I expect the small change in enthalpy from the decrease in concentration can be ignored (@Chestermiller ?)
 
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The heat of neutralization of HCl with NaOH is -57.9 kJ/mole. This is mainly due to the reaction of H+ with OH-.
 
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