On the basis of the following thermochemical data- - Hfo Haq-0 H2Ol-H-aq -OH-aq- -H-57-32 kJ H2g-12O2g-H2Ol- -H-286-20 kJThe value of enthalpy of formation of OH- ion at 25oC is-

The correct option is D 228.88 kJ In ionization of H_2O , H_2O(l)→H^+(aq) +OH^ (aq); ΔH=57.32 kJ ΔH=ΔH_f (OH (aq))+ΔH_f(H^+ (aq)) ΔH_f (H_ ...

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Heat of Formation

On the basis ...

Question

On the basis of the following thermochemical data: $(Δ H_{f}^{o} H_{a q}^{+} = 0)$
$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.20 k J$
The value of enthalpy of formation of $O H^{-}$ ion at $25^{o} C$ is:

- 22.88 k J

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- 228.88 k J

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+ 228.88 k J

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- 343.52 k J

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Solution

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Similar questions

H_{2} O_{(g)} ⟶ H_{(a q)}^{+} + {O H}_{(a q)}^{-}; Δ H = 57.32 k J

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

{O H}^{-}

25^{o} C

(Δ_{f} G^{0} (H)_{(a q)}^{+} = 0)

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

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286.20 k J

O H^{-}

25^{o} C

K O H

K_{(s)} + H_{2} O + a q \to {K O H}_{(a q)} + \frac{1}{2} H_{2}; Δ H = - 48.4 K . C a l

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}; Δ H = - 68.44 K . C a l

{K O H}_{(s)} + a q \to {K O H}_{(a q)}; Δ H = - 14.0 K . C a l

N H_{3}

- 46.0

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

H_{2}

- 436

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

N_{2}

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

N - H

N H_{3}

J

4.5 g

H_{2} O_{2}

1 a t m

25^{o} C

2 H_{2} O_{2} (l) ⟶ O_{2} (g) + H_{2} O (l)

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