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Heat of Formation

Calculate the...

Question

Calculate the enthalpy of formation of water, given that the bond energies of $H - - H$ , $O - - O$ , and $O - - H$ bond are $433 k J m o l^{- 1}$ , $492 k J m o l^{- 1}$ , and $464 k J m o l^{- 1}$ , respectively.

Δ H = - 249 k J

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Δ H = + 249 k J

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Δ H = - 649 k J

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None of these

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Solution

The correct option is A $Δ H = - 249 k J$
The balanced chemical reaction for the formation of water is as shown.
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (g)$
$= (\begin{matrix} BE of H_{2} + \frac{1}{2} BE of O_{2} \end{matrix}) - (2BE of O-H)$
$= 433 + (\frac{1}{2} \times 492) - (2 \times 464)$
$= 433 + 246 - 928$
$= - 249 k J$

Hence, the enthalpy of formation of water is -249 kJ.

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Similar questions

H_{2} (g)

H_{2} O (g)

^{- 1}

^{- 1}

^{- 1}

H_{2} (g)

H_{2} O (g)

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

433 k J m o l^{- 1}

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

298 K

O - H

H_{2} O (g) \to H (g) + O H (g); Δ H = 498 k J

{m o l}^{- 1}

O H (g) \to H (g) + O (g); Δ H = 428 k J

{m o l}^{- 1}

O - H

(δ H)

2 C_{2} H_{2 (g)} + 5 O_{2 (g)} \to 4 C O_{2 (g)} + 2 H_{2} O_{(g)}

C - H, C \equiv C, O = O, C = O, O - H

414, 814, 499, 724

640 k J m o l^{- 1}

298

O - H

H O H

H_{2} O (g) \to H (g) + O H (g), Δ H = 498

m o l^{- 1}

O H (g) \to H (g) + O (g), Δ H = 428

m o l^{- 1}

O - H

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