What is the Correct Molar Enthalpy of Reaction for NaOH + NH4Cl?

[gstullo]

I'm having some trouble with getting the correct entahlpy for the rxn:

NaOH + NH4Cl --> NaCl + NH3 + H2O

I've gotten the net ionic to be:

OH + NH4 --> NH3 +H2O

individuals to be sum of products (-) sum of reactants:

(-80.83+-285.84)-(-229.94 (for OH) and NH4 doesn't show anyvalue in my book.

This comes out to -136.73 which according to the little answer checker I've got here, is wrong. Where am I amiss?

Can anyone shed some light?

 

[t!m]

The NH4+ ion has a standard heat of formation of -132.5 kJ/mol. Assuming everything else is right, that should do it.

 

[GCT]

The net ionic equation should simply be the formation of water

$$H^+_{(aq)} + OH^-_{(aq)} \xrightarrow ~H_2O_{(l)}$$
This should be around -55kJ/mol

although it actually depends on the solubility of ammonium chloride, I may be wrong.