Calculate C-Cl bond enthalpy from following reaction - C H 3 Cl g - Cl 2 g - C H 2 Cl 2 g -HCl g - H o -104kJ- If C-H- Cl-Cl and H-Cl bond enthalpies are 414- 243 and 431kJ mol -1 respectively-

Δ H=∑ H _ reactant bond ∑ H _ product bond 104=( 3× H _ C H + H _ C Cl + H _ Cl Cl ) ( 2× H _ C H + H _ C Cl + H _ H Cl ) 104=( 3× 414+X ...

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Enthalpy

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Question

Calculate $C - C l$ bond enthalpy from following reaction :
$C H_{3} C l (g) + {C l}_{2} (g) ⟶ C H_{2} {C l}_{2} (g) + H C l (g)$ $Δ H^{o} = - 104 k J$ . If $C - H$ , $C l - C l$ and $H - C l$ bond enthalpies are $414$ , $243$ and $431 k J {m o l}^{- 1}$ respectively.

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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}

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

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

H - H, C l - C l

H - C l

436, 243

432

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

Δ H

C_{2} H_{4} + C l_{2} \to C l H_{2} C - C H_{2} C l

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

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