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Hess' Law

Following bon...

Question

Following bond energies are given: $C - H = 414 k J m o l^{- 1} C - C l = 150 k J m o l^{- 1} C l - C l = 243 k J m o l^{- 1} H - C l = 432 k J m o l^{- 1}$
How much energy would be required in the reaction :
$C H_{4} (g) + 2 C l_{2} (g) ⟶ C H_{2} C l_{2} (g) + 2 H C l (g)$

150 kJ

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130 kJ

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228 kJ

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571 kJ

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Solution

The correct option is A 150 kJ
$Δ H_{r}^{o} = Energy added to break the reactant bond +energy released in the making of product bonds So, Δ H_{r}^{0} = (4 \times 414) + (2 \times 243) - (414 \times 2 + 150 \times 2 + 2 \times 432) = 2142 - 1992 k J = 150 k J$

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Similar questions

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_{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 l - C l

C H_{4 (g)} + C l_{(g)} \to C H_{3} C l_{(g)} + H C l_{(g)}; Δ H = - 100.3 k J

C - H, C - C l, H - C l

413, 326

431

k J m o l^{- 1}

C H_{4} (g) + C l_{2} \to C H_{3} C l (g) + H C l (g)

δ H = - 25 k c a l .

C l - C l

x : y = 9 : 5

Δ H

- 25 k c a l

C H_{4} (g) + C l_{2} (g) \to C H_{3} C l (g) + H C l (g)

C l - C l

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