Enthalpy Change in Dissolving Ammonium Nitrate in Water

[louise82]

Homework Statement

A coffee cup calorimeter can be used to investigate the process that occurs when solid ammonium nitrate dissolves in water:

NH4NO3 (s) --> NH4+ (aq) + NO3 – (aq)

25.0 g of solid NH4NO3 at 23.0 degrees C is added to 250.0 mL of H2O at the same temperature, and after the solid is all dissolved the temperature is measured to be 15.6 degrees C. Calculate the enthalpy change (in kJ/ mol NH4NO3) for this process. Assume that the density and specific heat capacity of the solution are the same as for water. Is the process endothermic or exothermic?

Homework Equations

(I think): q=(c) x (m) x (change in temp)

qsol + qrxn = 0 (or) -qsol = qrxn

The Attempt at a Solution

? I do know that the specific heat for water is 4.184J/g degrees C and its density is 1.000g/mL

 

[louise82]

And I think its endothermic since the temp change is positive.

 

[JerusalemLuver]

Alright let's think this through, I'm studying enthalopy changes right now so I'm going to give this a try​

You're right about the equation:
"H = mass*C (or specific heat capacity)*Change in Temp"

I honestly haven't heard of the second equation you mentioned but I know I can solve this problem with just the first equation

So in this case:​

Mass=Total volume of the solution= 25g(NH4NO3) + 250g(H2O)= 275g (Solution)
**You know that 1g of water is equal to 1mL of water**

C = 4.184 J/g*C

Change in Temp = 15.6 deg - 23 deg = - 7.4

Now multiply the mass*C*Change in Temp=
275g*4.184(J/g*C) *-7.4 = -8 514.44 J = -8.51444KJ

That is -8.51444 KJ per 25g of NH4NO3. But the question is asking for the # of J for every mole of NH4NO3. So we have to find how many moles of NH4NO3 there are in 25g usinf the molar mass of NH4NO3:

25g*(1mol/80g) = 0.3125 mols

Now we know that it's -8.51444 KJ for 0.3125 mols ,divide and you get -27.246208 kJ/mol of NH4NO3

As for it being exo or endo -thermic, i'd say it's exothermic, or it released heat energy (think of it this way it had 23deg and it gave away a bunch of heat till it reached 15.6 deg) Another way you know this is b/c the final enthalapy change is negative, and another rule is that if the enthalapy change is neg then it's exothermic, positive and it would be endothermic

**Remember Change in Temp= Temp Final(or 15.6 deg) - Temp Initial (23 deg) = - 7.4

Hope that helps!

 

[louise82]

Thanks, that helped a lot!

 

[chemisttree]

It's endothermic...

 

[louise82]

Yeah, it is endothermic. I realized I was solving the problem for the solution instead of the the reaction. So, it ended up being 27kJ/mol. Thanks so much for the help though. I have only seen one problem worked in class.

Just for clarification: since the enthalpy change for the solution was negative, does that mean it IS an exothermic reaction for the solution, but an endothermic reaction for the reactants?

 

[chemisttree]

Endothermic means that heat is absorbed by the reaction. Exothermic means heat is produced. What will the sign of enthalpy be for heat absorption? For heat produced?

 

[louise82]

Heat absorbed, enthalpy change should be positive. Heat released, enthalpy change should be negative?

What is giving me problems, is that when I work the problem, (as done in the previous post), it comes out to be -27kJ/mol. But, when I looked up the enthalpy in my textbook, it is listed as positive 27kJ/mol. How do I account for the sign change?

 

[chemisttree]

I think the temperature change definition should be

Temperature (initial) - Temperature (final) = Temperature change

 

[Bystander]

T_f - T_i is the definition. Exothermic means that heat is given off, or if not allowed to escape from the system, raises the system temperature; endothermic means that the system temperature decreases, and the system can absorb heat from the surroundings. The sign of the heat is that for heat absorbed by the system to return it to its initial temperature.