A) \[{{E}_{f}}=60\,kJ/mol;{{E}_{b}}=100kJ/mol\]
B) \[{{E}_{f}}=30\,kJ/mol;{{E}_{b}}=70kJ/mol\]
C) \[{{E}_{f}}=80\,kJ/mol;{{E}_{b}}=120kJ/mol\]
D) \[{{E}_{f}}=70\,kJ/mol;{{E}_{b}}=30kJ/mol\]
Correct Answer: C
Solution :
\[Ef=80kj/mole;\,Eb=120kj/mole\] given, \[A\left( g \right)B\left( g \right)\] \[\Delta H=-40\,kg/mole\] \[\frac{Ef}{Eb}=\frac{2}{3}\]we know that \[Ef-Eb=\Delta H\] \[Ef-Eb=-40\] \[Eb\frac{2}{3}-Eb=-40\] Eb = 120 kg/mole Ef = 80 kg/moleYou need to login to perform this action.
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