Calculating enthalpy change from temperature data for three acid-base reactions

[NickP717]

The problem says: Enthalpy is always reported with respect to a reactant. Calculate the enthalpy change, DeltaH, for each reaction in terms of kJ/mol for one of the reactants. (i.e either HCL or NaOH).

I had three reactions:

1. NaOH + HCL---->H20 +NaCl
2.NH4Cl +NaOH---->H20 +NH3 +NaCl
3. HCl +NH3---->NH4

The first reaction I measured out 25.3mL of 2.0M HCl into a foam cup, then I dumped in 25.4mL of NaOH solution into the cup... my lowest temperature was 22.6 degrees C (before two solutions re combined., and my highest was 35.7 degrees C (when the solutions were mixed).

The second reaction I measured 25mL of 2.0M NaOH into a foam cup and then dumped 25.4mL of 2.0M NH4Cl solution. My lowest temperature was 22.5 degrees C (before mixing), and highest was 23.8 degrees C (after mixing).

The third reaction I measured 25.3mL of 2.0M HCl into a foam cup and then dumped 25.5 mL of 2.0M NH3 solution. My lowest temperature was 22.4 degrees C (before mixing), and highest was 31. 7 degrees C (after mixing thr two solutions.

Can somebody help me and show me what to do possibly, I really have no clue. Thank you so much!

 

[Borek]

How much heat evolved in each case?

 

[Simon_Moore]

I would do it by working out the total amount of temperature rise over the whole reaction mixture (which you have been given), and then assume that the density of the two solutions is still 1.00 g/cc (which they will be, as near as makes no odds).

The delta H will then be:

Delta H = m.c.t

where m is the mass of the two solutions (technically the mass of the WATER in the two solutions but it's as near as makes no odds, as I said above), c is the specific heat of water, which is 4.2 J/g, and t is the temperature change.

First, work out the number of moles of HCl that you have used. This will be equal to 25.3/1000 x 2 = 0.0506 moles.

Now do the delta H calculation:

Delta H = (25.3+25.4) x 4.2 x 13.1 = 664.17 J

Therefore, 0.0506 M gives 664.17 J.

So, per mole (of HCl) it works out at (1/0.0506) x 664.17 = 13,125.89 J

So delta H wrt HCl is 13.13 kJ/mol.


Thats how I'd do it...but since I haven't done these for about 25 years I may be wrong !