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Hess' Law

Bond Average ...

Question

Bond Average Bond Energy (kJ/mol)
$C \equiv O$ 1075
$C = O$ 728
$C - C l$ 326
$C l - C l$ 243
From the above given data, calculate the heat of reaction for the following reaction :

$C O + {C l}_{2} \to C O {C l}_{2}$

+ 62

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- 62

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- 409

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+ 706

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Solution

The correct option is A $+ 62$ kJ
$C O + C l_{2} \to C O C l_{2}$

Heat of reaction = Sum of bond Energy of products - sum of bond energy of reactant.

Heat of reaction = $[728 + 2 \times 326] - [1075 + 243] = 1380 - 1318 = + 62 k J$

Hence, the correct option is A.

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Similar questions

m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

Δ H

2 C l_{2} + 2 H_{2} O \to 4 H C l + O_{2}

△ H_{r}^{o}

C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .

414 k J m o l^{- 1}

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

C - H = 414 k J m o l^{- 1} C - C l = 150 k J m o l^{- 1} C l - C l = 243 k J m o l^{- 1} H - C l = 432 k J m o l^{- 1}

C H_{4} (g) + 2 C l_{2} (g) ⟶ C H_{2} C l_{2} (g) + 2 H C l (g)

C - H = 414 K J / m o l

C - C l = 150 K J / m o l

C l - C l = 243 K J / m o l

H - C l = 432 K J / m o l

{C H}_{4 (g)} + {2 C l}_{2 (g)} \to {C H}_{2} {C l}_{2 (g)} + {2 H C l}_{(g)}

H_{2} + C l_{2} \to 2 H C l

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C l - C l

430

242

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