Chemical Thermodynamics Question 407

Question: For the auto-ionization of water at $ 25{}^\circ C, $

$ H_2O(l)\leftrightharpoons {{H}^{+}}(aq)+O{{H}^{-}}(aq) $ equilibrium constant is $ {{10}^{-14}} $ . What is $ \Delta G{}^\circ $ for the process-

Options:

A) $ \simeq 8\times 10^{4}Jmo{{l}^{-1}} $

B) $ \simeq 3.5\times 10^{4}Jmo{{l}^{-1}} $

C) $ \simeq 2\times 10^{4}Jmo{{l}^{-1}} $

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ \Delta {{G}^{{}^\circ }}=-RT $ ln K $ =-8.314J{{K}^{-1}},mo{{l}^{-1}}\times 298K\times 2.303log{{10}^{-14}} $

$ =7.98\times 10^{4}\simeq 8\times 10^{4}J,mo{{l}^{-1}} $



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