$O (g) + e^{-} ⟶ O^{-} (g), △ H_{e g} = - v e$

True

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False

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Solution

The correct option is A True

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Similar questions

O (g) + e^{-} ⟶ O^{-} (g), △ H_{e g} = - v e

O_{(g)} + e^{-} \to O_{(g)}^{-} Δ H_{1}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{- 2} Δ H_{2}

O_{2 (g)} \to O_{2 (g)}^{+} + e^{-} Δ H_{3}

O_{(g)} \to O_{(g)}^{+} + e^{-} Δ H_{4}

H_{2 (g)} \to H_{2 (g)}^{+} + e^{-} Δ H_{5}

H_{(g)} \to H_{(g)}^{+} + e^{-} Δ H_{6}

O_{(g)} + 2 e^{-} \to O^{- 2} Δ H_{7}

O_{(g)} + e^{-} \to O_{(g)}^{-}, Δ H_{1}

F_{(g)} + e^{-} \to F_{(g)}^{-}, Δ H_{2}

C l_{(g)} + e^{-} \to C l_{(g)}^{-}, Δ H_{3}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}, Δ H_{4}

O_{(g)} + e^{-} \to O_{(g)}^{-} Δ H_{1}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{- 2} Δ H_{2}

O_{2 (g)} \to O_{2 (g)}^{+} + e^{-} Δ H_{3}

O_{(g)} \to O_{(g)}^{+} + e^{-} Δ H_{4}

H_{2 (g)} \to H_{2 (g)}^{+} + e^{-} Δ H_{5}

H_{(g)} \to H_{(g)}^{+} + e^{-} Δ H_{6}

O_{(g)} + 2 e^{-} \to O^{- 2} Δ H_{7}

O^{2 -} (g)

O (g) + e^{-} = O^{-} (g); Δ H^{0} = - 142 k J m o 1^{- 1}

O (g)^{-} + e^{-} = O^{2 -} (g); Δ H^{0} = 844 k J m o 1^{- 1}

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