final temp of NH4ClO4? chemistry equation?

Question

The dissolution of NH4ClO4 in H2O is endothermic.
DeltaH for the solution is 33.5 KJ/mol.
If you prepare a 1.00m solution of NH4ClO4, with the water beginning at 25C, what is the final temperature of the solution?
Note: Assume that the specific heats of pure water and the solution are the same, 4.18 j (k*g).

Answer 1

didn't you already ask this question. my previous answer is...

here's some hints..

1) do it on a 1 liter basis for ease. doesn't really matter though

2) how many moles of ammonium perchlorate are present in 1 liter if the concentration is 1 molar?

3) how much energy is absorbed as the ammonium perchlorate is dissolved

4) how many grams of water are present in 1 liter of pure water.

5) assume 1 liter of solution contains 1 liter of pure water. It will actually be slightly less. save that for another class.

6) energy absorbed by ammonium perchlorate = energy lost by water = m x cp x dT



answer ... ( A = ammonium perchlorate, W = water)

1 molar A x (1 mole A/liter water) x (33.5 KJ/mole A) = Energy gained by A = Energy lost by W = 33.5 KJ

so

-33.5 KJ = 1 liter W x (1 Kg W/ liter W) x (4.18 KJ/(Kg x °K) x dT

so dT = -33.5 / 4.18 in °K = 8 °K

so the answer is 17°C

Answer 2

You don't give the volume of water or solution, so I'll assume 1 L. Let NH4ClO4 be called AC. Let the solution be called S.

1molAC/1000gS x 33.5kJ/1molAC x 1000J/1kJ x 1K-g/4.18J = (33.5)(1000)/(1000)(4.18) = 8.00degK = 8degC temperature change. 25C - 8C = 17C.

The first factor comes from the solution makeup, assuming 1L or 1000g. The second factor comes from heat of solution. The moles AC cancel, leaving kJ per 1000gS. The third factor comes from conversion of kJ to J. The fourth factor comes from the heat capacity of water. Both grams of S (back at the beginning) and J cancel, leaving degK of temperature change.