A) 163 kJ \[mo{{l}^{-1}}\]
B) \[2.4\times {{10}^{2}}kJ\,mo{{l}^{-1}}\]
C) 1.63 kJ \[mo{{l}^{-1}}\]
D) \[2.38\times {{10}^{6}}kJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
Key Idea: Use the following formula to find\[\Delta {{G}^{o}}\]. \[\Delta {{G}^{o}}=-2.303\,RT\log {{K}_{p}}\] Given, \[{{K}_{p}}=2.47\times {{10}^{-29}}\] \[\therefore \] \[\Delta {{G}^{o}}=2.303\times 8.314\times 298\] \[\times \log 2.47\times {{10}^{-29}}\] \[=163000\,Jmo{{l}^{-1}}\] \[=163\,kJmo{{l}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
