Use the bond energies to estimate - H for this reaction- H2g - O2g- H2O2g

The correct option is B 127 kJ As we know, Δ H = ∑ BE_R ∑ BE_P So, for H_2(g) + O_2(g)⟶ H_2O_2(g) $Δ H = 436+499 (2×460 + 142) = 127 kJ/mol ...

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Hess' Law

Use the bond ...

Question

Use the bond energies to estimate $Δ H$ for this reaction:
$H_{2} (g) + O_{2} (g) ⟶ H_{2} O_{2} (g)$

- 127 k J

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- 109 k J

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- 400 k J

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- 800 k J

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Solution

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Similar questions

Δ H

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

H - H

436 k J

{m o l}^{- 1}

O - O

142 k J

{m o l}^{- 1}

O = O

499 k J

{m o l}^{- 1}

H - O

460 k J

{m o l}^{- 1}

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

Δ H

Δ U

25^{o} C

\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to O H (g); Δ H = 10.06 k c a l

H_{2} (g) \to 2 H (g); Δ H = 104.18 k c a l

O_{2} (g) \to 2 O (g); Δ H = 118.32 k c a l

O - H

O H

Δ H

k J

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

H_{2} O (g) + C (g) \to C O (g) + H_{2} (g)

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

For the reaction,

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$ ; $Δ H = - 571 k J$ .

Bond energies: $(H - H) = 435 k J$ , $(O = O) = 498 k J$ .

Calculate the average bond energy of $(O - H)$ using the above data.

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