November 14, 2024, 01:51:35 AM
Forum Rules: Read This Before Posting


Topic: lattice enthalpy and enthalpy of hydration (chem II) (cant sleep)  (Read 4119 times)

0 Members and 1 Guest are viewing this topic.

Offline noapologies01

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
The lattice enthalpy of CaCl2(s) is -2260 kJ/mol and the enthalpy of hydration is –2340
kJ/mol.  How much heat will be released when 0.153 mol of CaCl2(s) dissolves in
water?
 
A. 12.2 kJ
B. 38.1 kJ
C. 28.7 kJ
D. 80.0 kJ
E. 17.4 kJ

the answer is A.

i did Q=ndeltaH twice, with the same mol each time but with the two different heats. i substracted the values and got 12.2 kJ. did i do this right. please explain it, if i did it right too. i thought i had to do something with water and i just got the answer by luck

Offline Sam (NG)

  • Chemist
  • Full Member
  • *
  • Posts: 223
  • Mole Snacks: +12/-3
  • Gender: Male
  • Surface Modification
Re: lattice enthalpy and enthalpy of hydration (chem II) (cant sleep)
« Reply #1 on: June 19, 2008, 04:45:42 AM »
Construct a Born-Haber cycle, then use Hess' Law to calculate the molar enthalpy change on dissolution of the calcium chloride.  Multiply this answer by 0.153 and you're there.

Sponsored Links