Calculating Enthalpy Change: HNO3 + KOH

[Blade]

Well up to this point, the problems have been calorimeter problems, however, at the end of the assignment has different "looking" type of problems.

i.e.
"What would the change in enthalpy, in kJ/mol, be for the reaction of 50.0 mL of 1.00 M HNO(3) with 50.0 mL of 1.00 M KOH?"
These don't give a specific heat, change in temperature, or density. Which I have been using for the previous problems. (change in enthalpy = (SH)(mass)(change in temp)) Is there something I'm overlooking? It seems simple...
[As far as I know, I'm not suppose to use anything else besides a periodic table]

 

[chem_tr]

As far as I know, formation of inorganic salts require heat to dissolve in water; so this will be an endothermic reaction. You can confirm this reasoning by considering some cooling mixtures; in some of them, sodium nitrate is used to provide a lower freezing temperature.

 

[Bystander]

Let's nudge this back in the correct direction --- you've done some other "neutralization" problems? Right? Sodium hydroxide plus hydrochloric acid? Right? A neutralization reaction.

Words to the wise from Bystander: "I have no qualms about letting you be dragged off into the woods to be mugged and raped by the forces of darkness, ignorance, and superstition who prowl this forum. When I "type" (speak), you "read" (listen) and heed. When I'm silent, don't think for a minute you've got the straight dope.

chem_tr's okay, but prone to step in when he should sit back --- this is a good case in point --- no help on this question would have been better than what was offered. See the "chem guru" thread.

 

[Blade]

Jeez, took me a while, but finally realized that these questions were actually connected to some previous questions *hits self on head*

So, I'm guessing this is what they stemmed out of:

1. Which I solved the rxn (47.3 mL of 2.00M nitric acid, 47.1 mL of 2.00M sodium hydroxide) change in enthalpy as -209 kJ/mol*deg

These are paired with I assume with nitric, and NaOH in portions --^
So... since enthalpy is an extensive property... it would seem to suggest that the solutions would just be proportions of the answer I determined correct?

["What would the change in enthalpy, in kJ/mol, be for the reaction of 50.0 mL of 1.00 M nitric acid with 50.0 mL of 1.00 M KOH?"

"What would the change in enthalpy, in kJ/mol, be for the reaction of 25.0 mL of 1.00 M nitric acid with 25.0 mL of 1.00 M KOH?"]

 

[Bystander]

Jeez, took me a while, but finally realized that these questions were actually connected to some previous questions *hits self on head*

So, I'm guessing this is what they stemmed out of:

1. Which I solved the rxn (47.3 mL of 2.00M nitric acid, 47.1 mL of 2.00M sodium hydroxide) change in enthalpy as -209 kJ/mol*deg

These are paired with I assume with nitric, and NaOH in portions --^
So... since enthalpy is an extensive property... it would seem to suggest that the solutions would just be proportions of the answer I determined correct?

["What would the change in enthalpy, in kJ/mol, be for the reaction of 50.0 mL of 1.00 M nitric acid with 50.0 mL of 1.00 M KOH?"

"What would the change in enthalpy, in kJ/mol, be for the reaction of 25.0 mL of 1.00 M nitric acid with 25.0 mL of 1.00 M KOH?"]

"1" gives you the enthalpy of neutralization (formation of water from H⁺ and OH^- ) --- this is the major heat for the reaction. Generally, it is necessary to consider enthalpy of dilution for those cases in which a different final concentration occurs (1m NaNO₃ in "1" differs from .5m KNO₃ in two ways, concentration and cation species). Do you have dilution experiments that allow you to make such a correction? If not, you are looking at an approximation that enthalpy of dilution is negligible compared to enthalpy of neutralization.