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Heat of Formation

Given that ;...

Question

Given that $;$
$H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32$ kJ
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02$ kJ
Then calculate the enthalpy of formation of $O H^{-}$ at $25^{o}$ C.

A

- 228.8

kJ

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B

- 343.52

kJ

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C

+ 228.8

kJ

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D

+ 343.52

kJ

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Solution

The correct option is A $- 228.8$ kJ

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Similar questions

Q. Consider the following reactions.

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

m o l^{- 1}

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

m o l^{- 1}

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

m o l^{- 1}

Enthalpy of formation of

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

Then, calculate the enthalpy of formation of

O H^{-}

25^{o} C

Q. Given that :

2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g)

Δ H = - 787

H_{2} (g) + 1_{/ 2} O_{2} (g) \to H_{2} O (l)

Δ

= - 286

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)

Δ

= - 1310

Heat of formation of acetylene is:

Q. Consider the following reactions :

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}

Enthalpy of formation of

H_{2} O (l)

Q. On the basis of the following thermochemical data:

(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

The value of enthalpy of formation of

O H^{-}

25^{o} C

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