- #1
mrxtothaz
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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.
Homework Equations
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 canceled 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.