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LoginHCJC Chem Prelim Energetics Qn (good practice for students!)
5 posts
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Moderator'A' & 'O' Levels Chemistry & Biology Ex-MOE Teacher Tutor
840 posts since May '05-
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').
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Moderator'A' & 'O' Levels Chemistry & Biology Ex-MOE Teacher Tutor
840 posts since May '05
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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?
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Moderator'A' & 'O' Levels Chemistry & Biology Ex-MOE Teacher Tutor
840 posts since May '05
*thumbs up!*
sorry noahie
