# Enthalpy of Reaction (Hess's Law)

## Homework Statement

I have been provided with the results of a lab:

Reaction 1:
NaOH(s) --> Na+(aq) + OH-(aq)
Reaction 2:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
OH-(aq) + H+(aq) --> H2O(l)
Reaction 3:
NaOH(s) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
NaOH(s) + H+(aq) --> H2O(l) + Na+(aq)

The enthalpies for each reaction:
∆H°r: -5.80 kJ (Reaction 1)
∆H°r: -5.06 kJ (Reaction 2)
∆H°r: -16.7 kJ (Reaction 3)

I have been asked to add the equations of 1 and 2 (Hess's Law), and to compare the sum with equation 3.

Hess's Law.

## The Attempt at a Solution

NaOH(s) --> Na+(aq) + OH-(aq) ∆H°r: -5.80 kJ (Reaction 1)
OH-(aq) + H+(aq) --> H2O(l) ∆H°r: -5.06 kJ (Reaction 2)

NaOH(s) + H+(aq) + OH-(aq) --> Na+(aq) + H2O(l) + OH-(aq) ∆H°r: -10.9 kJ (Reaction 1+2 intermediate)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -10.9 kJ (Reaction 1+2)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -16.7 kJ (Reaction 3)

So, as can be seen, the equations for reactions 1&2 are added and are identical to that of reaction 3. However, the enthalpies of reaction are way different. I do not know how I can account for this steep a difference, as all the lab results were provided with the question. The only difference is that, in the addition of reactions 1&2, there was a hydroxide ion on both sides of the equation that was cancelled out (since I was asked for the net ionic equations for each reaction). Could this account for the difference, or does it have no role in the enthalpy of reaction, given that it is a spectator ion?

If I did nothing wrong, I suspect I am supposed to identify this discrepancy between reaction 1+2 with reaction 3.

I'm looking into possible sources of error; the only thing I can think of is that the experimenters assumed the specific heat capacity of each solution to be same as that of water? As with density.

Could this account for the discrepancy?

chemisttree
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Gold Member

## Homework Statement

I have been provided with the results of a lab:

Reaction 1:
NaOH(s) --> Na+(aq) + OH-(aq)
Reaction 2:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
OH-(aq) + H+(aq) --> H2O(l)
Reaction 3:
NaOH(s) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
NaOH(s) + H+(aq) --> H2O(l) + Na+(aq)

The enthalpies for each reaction:
∆H°r: -5.80 kJ (Reaction 1)
∆H°r: -5.06 kJ (Reaction 2)
∆H°r: -16.7 kJ (Reaction 3)

I have been asked to add the equations of 1 and 2 (Hess's Law), and to compare the sum with equation 3.

Hess's Law.

## The Attempt at a Solution

NaOH(s) --> Na+(aq) + OH-(aq) ∆H°r: -5.80 kJ (Reaction 1)
OH-(aq) + H+(aq) --> H2O(l) ∆H°r: -5.06 kJ (Reaction 2)

NaOH(s) + H+(aq) + OH-(aq) --> Na+(aq) + H2O(l) + OH-(aq) ∆H°r: -10.9 kJ (Reaction 1+2 intermediate)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -10.9 kJ (Reaction 1+2)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -16.7 kJ (Reaction 3)

So, as can be seen, the equations for reactions 1&2 are added and are identical to that of reaction 3. However, the enthalpies of reaction are way different. I do not know how I can account for this steep a difference, as all the lab results were provided with the question. The only difference is that, in the addition of reactions 1&2, there was a hydroxide ion on both sides of the equation that was cancelled out (since I was asked for the net ionic equations for each reaction). Could this account for the difference, or does it have no role in the enthalpy of reaction, given that it is a spectator ion?

If I did nothing wrong, I suspect I am supposed to identify this discrepancy between reaction 1+2 with reaction 3.

How did you get H+aq? From HAaq?

How did you get H+aq? From HAaq?

Not sure what you mean by HAaq, but the hydrogen ions arise from the dissolution of hydrochloric acid.

chemisttree