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  4. If EA1 and EA2 for oxygen atom -142kJmol- 1 and +844kJmol- 1 . The energy released form 2O+2e- arrow 2O- will be

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Q. If $EA_{1}$ and $EA_{2}$ for oxygen atom $-142kJmol^{- 1}$ and $+844kJmol^{- 1}$ . The energy released form $2O+2e^{-} \rightarrow 2O^{-}$ will be

NTA AbhyasNTA Abhyas 2020Thermodynamics

Solution:

$O+e^{-} \rightarrow O^{-}$

$\Delta H=-142kJmol^{- 1}$

So $2O+2e^{-} \rightarrow 2O^{-}$

$\Delta H=2\times \left(\right.-142\left.\right)$

$=-284kJmol^{- 1}$