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

Question

If EA1 and EA2 for oxygen atom -142kJmol- 1 and +844kJmol- 1 . The energy released form 2O+2e- arrow 2O- will be (A) -986kJmol- 1 (B) -702kJmol- 1 (C) -284kJmol- 1 (D) -1688kJmol- 1

Tardigrade Admin · Accepted Answer

O+e- arrow O- Δ H=-142kJmol- 1 So 2O+2e- arrow 2O- Δ H=2× (.-142.) =-284kJmol- 1

Q. If $E A_{1}$ and $E A_{2}$ for oxygen atom $- 142 k J m o l^{- 1}$ and $+ 844 k J m o l^{- 1}$ . The energy released form $2 O + 2 e^{-} \to 2 O^{-}$ will be

$- 986 k J m o l^{- 1}$

$- 702 k J m o l^{- 1}$

$- 284 k J m o l^{- 1}$

$- 1688 k J m o l^{- 1}$

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

$Δ H = - 142 k J m o l^{- 1}$

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

$Δ H = 2 \times (- 142)$

$= - 284 k J m o l^{- 1}$

Q. If EA1​ and EA2​ for oxygen atom −142kJmol−1 and +844kJmol−1 . The energy released form 2O+2e−→2O− will be

−986kJmol−1

−702kJmol−1

−284kJmol−1

−1688kJmol−1