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Question: Answered & Verified by Expert
The pH of pure water at 298K and 308K are 7 and 6 respectively. Calculate the heat of formation of water from H+ and OH .
ChemistryChemical EquilibriumJEE Main
Options:
  • A ΔH=84.551 kcal/mol
  • B ΔH=-84.551 kcal/mol
  • C ΔH=44.981 kcal/mol
  • D ΔH=-44.981 kcal/mol
Solution:
2786 Upvotes Verified Answer
The correct answer is: ΔH=-84.551 kcal/mol
At 298K; [H+]=107
K w 1 = 10 14
At 308K; H+=106
K w 2 = 10 12
Now using
2.303 log 10 Kw 2 Kw 1 = ΔH R T 2 T 1 T 1 × T 2
2.303 log 10 10 12 10 14 = ΔH 2 308298 298×308
   ΔH=84551.4 kcal/mol 
=84.551 kcal/mol
Thus H2O H++OH-;   ΔH=84.551 kcal/mol
   H++OH-H2O;  ΔH-84.551 kcal/mol

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