On the basis of the following thermochemical data H 2 O g - H aq - - OH aq - - H-57-32kJ H2g-12O2g- H 2 O l - H-286-2kJThe value of enthalpy of formation of OH- ion at 25 o C is-

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Calorimetry

On the basis ...

Question

On the basis of the following thermochemical data
$H_{2} O_{(g)} ⟶ H_{(a q)}^{+} + {O H}_{(a q)}^{-}; Δ H = 57.32 k J$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} ⟶ H_{2} O_{(l)}; Δ H = - 286.2 k J$
The value of enthalpy of formation of ${O H}^{-}$ ion at $25^{o} C$ is:

+ 288.88 k J

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- 343.52 k J

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- 22.88 k J

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- 228.88 k J

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Solution

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(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

(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

O H^{-}

25^{o} C

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

O H^{-}

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

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

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