HCJC Chem Prelim Energetics Qn (good practice for students!)

• UltimaOnline

  7 Jun 08, 13:39

  Arigato brought up a HCJC gas law qn from the prelim paper. I thought I'd share another interesting qn that many JC students struggle with.

   

  8.4g of NaF is dissolved in 250g of water. Given lattice enthalpy is -918 kJ/mol, hydration enthalpy of F- is -457 kJ/mol and hydration enthalpy of Na+ is -390kJ/mol, what is the initial temp of the water if final temp of soln is 20 deg C? Use 4.2 J g-1 K-1 for specific heat capacity of NaF(aq).

   

  Hint : Draw a hess law diagram to calculate solution enthalpy (note : lattice association enthalpy is exothermic, lattice dissociation enthalpy is endothermic; students often confuse the two). Next, use the formula : heat transferred = m c (delta)T. (note : capital T represents temperature in Kelvins, small t in deg C). Make (delta)T the subject, and hence find the initial temperature. (note : 'delta' means 'change in'). 

   

• Uncertain

  9 Jun 08, 12:41

  Initial Temperature = 80+ degree Celcius? <-- my ans a bit weird leh.

   

  The solution so hot at first? O.O

• UltimaOnline

  9 Jun 08, 14:36

  Yah, ur ans is a bit wierd. Keep going, continue your efforts.

• Uncertain

  9 Jun 08, 21:04

  Hi UltimaOnline,

  Your question is really great and I have managed to derive the answer after much thought.

  Solution (to those who are interested):

  H (solution of NaF) = -457 -390 + 918 = +71kJ/mol

  No. of mole of NaF dissolved = 8.4 / (23+19) = 0.2 mol

  Heat released = mc(delta)T

  -71000 * 0.2 (this is endothermic) = (250 + 8.4)(4.2)(20 - T)

  T = 33.08 C (initial temp)

   

  Ultima, is this correct?

   

• UltimaOnline

  9 Jun 08, 22:45

  Excellent!  *thumbs up!*