[rozannz]

Hey

I would really appreciate help on the calculations, because I am just not quiet getting it.

I had 3 different reactions I had to do.

First, some things to note:

Since all the solutions are dilute, then the density of each soltuion can be assumed to be 1.00g/mL. So, 100mL has a mass of 100g.

Reaction 1 was the dissolution of NaOH. So it was 200mL of water. The initial temp was taken, then 5.5g of NaOH was placed in, and a second temp was taken.

Reaction 2 was a heat of reaction between aqueous NaOH and aqueous HCL. 100mL of 1.0mol/L of HCl in a vessel, and the ame of NaOH. Initial temps are taken of each substance. Then, they are mixed together, and the new temp. is recorded.

Finally, Reaction 3 is the heat of reaction between solid NaOH and aqueous HCl. We have 200mL of 1.0mol/L of HCl in a vessel, we record the temp. Then we take 5.5g of NaOH and we pour it into the acid. We then take that new temp.

Of course the equation that is used for Heat or Rection is deltaH = m * deltaT * C.

The calculations are difficult because there is more then that calculation...

Not, for Reaction 1, dont think, this is how I did it:


deltaH = m * c * deltaT
deltaH = 200g * 1cal/moldegcel * 0.00418kj/cal * (13degcel)
deltaH = 10.868kJ

since its 5.5g though:

-10.868kj/0.1375 = -79.04kj/mol

However, Reaction 2

2) deltaH = m * c * deltaT
deltaH = 200g * 1cal/moldegcel * 0.00418kJ/cal * (10degcel)
deltaH = -8.36kJ

-.836kJ/0.1=-83.6kJ/mol


and I dont know how to do with 3...

Any help would be appreciated.

[[[[Now i need to write the net eqaution for each reaction and note the value of delta H for each reaction.]]]]]

plzzzzzzzzzz i needd *delete me*!

[Sev]

For reaction 3, I would solve it by calculating the heat change of the water (m.C.deltaT).  Then, use the number of moles of the yield limiting reagent (in this case it appears to be NaOH) to work out delta H (per mole).

The products are salt and water.