Understanding Heat of Neutralization in Acid-Base Reactions

[Kqwert]

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?

 

[scottdave]

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?

 

[Kqwert]

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?

 

[scottdave]

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.

 

[BvU]

What about my first question, that relating to the deltaH calculations?

Did you google 'heat of neutralization' ?

 

[Kqwert]

I did it now, and it seems like we simply ignore the cations in our deltaH calculations if they do not react after dissociation..?

 

[BvU]

What dissociation ? (and Cl- is not a cation)

 

[Kqwert]

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?

 

[BvU]

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 ?)

 

[Chestermiller]

The heat of neutralization of HCl with NaOH is -57.9 kJ/mole. This is mainly due to the reaction of H+ with OH-.