Chemical Thermodynamics Question 272
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}} $
BETA
