[jennielynn_1980]

So I did an experiment where I had to calculate heat of reaction.
For the first one I got
ΔH = m x ΔT x Q
m = 205.5g
ΔT = 22.3˚C -22.6˚C (trial 1) 22.6˚C ? 26˚C (trial 2)
Q= 4.18 x 10-3 kJ/g˚C

ΔH= 205.5g x (-4.3˚C) x (4.18 x 10-3 kJ/g˚C)
     = -3.7 kJ (trial 1)
ΔH = 205.5g x (-3.4˚C) x (4.18 x 10-3 kJ/g˚C)
      = -2.9 kJ (trial 2)

Now I have to give the answer in kJ/mol of NaOH.  So in this particular reaction I used 5.5g of NaOH which turns out to be:
mol = g        
          g/mol
 = 5.5g/40.0g per mol
= 0.14 mol
 
So do I divide my answer from part one (-3.7 kJ and -2.9kJ) by the number of moles to get kJ per mole?  I know this is a bit of a dumb question but the thing is that the answer I get is pretty far off from what I think it should be so I dont' know if I am doing it wrong.
Thanks

[Mitch]

So if you used 5.5g of NaOH then why did you plug in m = 205.5g?

[enahs]

So if you used 5.5g of NaOH then why did you plug in m = 205.5g?

Because I would guess it was in 200 mL of water.



You are calculating delta H for everything, the water and the NaOH...and?
What else is in the solution besides the water and NaOH? What exactly is the experiment? Is it neutralizing an acid?



Also next time look at your results when doing them, those are quite different, you should have done another one and gotten two sets of data more consistent.


I am guessing this was "coffee-cup calorimeter". Remember when you are assuming no heat is lost or gained by the solution then:
q_rxn + q_solution = 0.

What you are doing is calculating q_sol, not q_rxn.
q_rxn = -q_sol in this case (if that is what you are doing?).

I use q (heat) and not H because you are making them the same with your assumptions, but they are not.


I have some other problems with your data. Your change in temperatures you typed do not math what you said the initial and final where for trial one.
If it was a typical acid-base neutralization then the temperature would increase, so I am guessing it was not that?

Note, your change in temperature says it was an endothermic reaction, yet your H is for exothermic (see q+q=0).

[jennielynn_1980]

In this experiment I did three different reactions where I mixed HCl and NaoH and I was asked to do two trials.
In this particular reaction I mixed 5.5g of NaOH with 200mL of 1.0mol/L of HCl solution.
The initial temperature was 22.3C and 22.6C for trials one and two respectively.  The  highest temp. reached was 26.6C and 26.0C respectively for trials one and two.  This is the data I was asked to record.

So then I was asked to calculate the heat lost or gained by each reaction per mole of NaOH using this equation:
?H = m x ?T x Q
where
m  = total mass of reaction solution (205.5g)
?T = temperature change (22.3?C -22.6?C (trial 1) 22.6?C - 26?C (trial 2))
Q = specific heat (I was told to assume that it was 4.18 x 10^-3kJ/gC)

SO that is where I got my numbers from.  I am also fairly sure my numbers are supposed to be negative becasue the 2nd question I have to answer is "explain why deltaH is negative in all the reactions"

SO assuming I have done it correctly, how do I find the kJ per mole of NaOH?  Is it how I suggested in the first post or have I still got something wrong?

Thanks

[Mitch]

you have that 5.5gNaOH caused 4.18X10^-3KJ/gC.

1molNaOH * ( MW NAOH/ 1mol NaOH) * (4.18X10^-3KJ/gC  /  5.5gNaOH)

[enahs]

SO that is where I got my numbers from.  I am also fairly sure my numbers are supposed to be negative becasue the 2nd question I have to answer is "explain why deltaH is negative in all the reactions"

I miss understound your first post, it read as 22.3-22.6 for trial 1 and 22.6-26 for trial 2.