A) \[1.63 kJ mo{{l}^{-1}}\]
B) \[2.4\,\,\times \,\,{{10}^{2}}\,kJ\,mo{{l}^{\text{-}1}}\]
C) \[1.63\,\,kJ\,mo{{l}^{\text{-}1}}\]
D) \[2.8\,\,\times \,\,{{106}^{2}}\,kJ\,mo{{l}^{\text{-}1}}\]
Correct Answer: A
Solution :
As we know that, \[\Delta G{}^\circ = -2.303 \,RT \,log\,\,{{K}_{p}}\] Therefore, \[\Delta G{}^\circ -2.303 \times \left( 8.314 \right) \times \left( 298 \right) (log 247\times 1{{0}^{-}}^{29})\] \[\Delta G{}^\circ = 16,3000 J mo{{l}^{-}}^{1}\,\,=\,\,163 kJ mo{{l}^{-}}^{1}\]You need to login to perform this action.
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