H 2g - 1 - 2 O 2g - H 2 Ol- - H-286-2kJ- H 2 O l - H aq - -O H aq - - H-57-3kJ- Enthalpy of ionisation of OH- in aqueous solution is-

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Heat of Reaction

H 2g + 1 / 2 ...

Question

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}$ ; $Δ H = - 286.2 k J$ ; $H_{2} O_{(l)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}$ ; $Δ H = + 57.3 k J$ . Enthalpy of ionisation of $O H^{-}$ in aqueous solution is:

- 228.5 k J

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+ 228.5 k J

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- 343.5 k J

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zero

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Solution

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Similar questions

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

H_{2} O (l)

H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02 k J

O H^{-}

25^{o} C

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.77 k J / m o l e

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 241.84 k J / m o l e

;

H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02

O H^{-}

25^{o}

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.2 k J

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l); Δ H = - 1410.8 k J

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