Calculate the enthalpy - - H- for the given reaction using the the given bond energies - C2H4 - Cl2- ClH2C-CH2ClBond energieskJ- mol C-C 347 C - C 612 C-Cl 341 C-H 414 Cl-Cl 243

The correct option is C Δ H = 174 kJ Δ H^o _rxn = ΣΔ H^o _reactant bonds ΣΔ H^o _product bonds Δ H^o _rxn= #160; [4Δ H^o _C H+ Δ H^o _C=C+Cl ...

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Standard XII

Chemistry

Heat of Reaction

Calculate the...

Question

Calculate the enthalpy , $Δ H$ , for the given reaction using the the given bond energies ?
$C_{2} H_{4} + C l_{2} \to C l H_{2} C - C H_{2} C l$
Bond energies kJ/ mol
$C - C$ $347$
$C = C$ $612$
$C - C l$ $341$
$C - H$ $414$
$C l - C l$ $243$

Δ H = - 800 k J

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Δ H = - 680 k J

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Δ H = - 174 k J

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Δ H = + 174 k J

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Δ H = + 200 k J

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Solution

The correct option is C $Δ H = - 174 k J$
$Δ H_{r x n}^{o} = Σ Δ H_{reactant bonds}^{o} - Σ Δ H_{product bonds}^{o} Δ H_{r x n}^{o} = [4 Δ H_{C - H}^{o} + Δ H_{C = C}^{o} + C l - C l] - [Δ H_{C - C}^{o} + 4 Δ H_{C - H}^{o} + 2 Δ H_{C - C l}^{o}]$
$Δ H_{r x n}^{o} = [Δ H_{C = C}^{o} + C l - C l] - [Δ H_{C - C}^{o} + 2 Δ H_{C - C l}^{o}]$
$Δ H_{r x n}^{o} = [612 + 243] - [347 + 2 (341)]$
$Δ H_{r x n}^{o} = 855 - 1029$
$Δ H_{r x n}^{o} = - 174 k J$

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Similar questions

Q. Calculate the

Δ H

of the reaction given (figure).
Bond energy for

C - H

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415 k J

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326 k J

respectively.

Q. The reaction

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

has

δ H = - 25 k c a l .

From the data given below, what is the bond enthalpy of

C l - C l

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Bond	Bond Energy (kcal)
$C - C$	$84$
$C - C l$	$81$
$C - H$	$x$
$C l - C l$	$y$
$H - C l$	$103$

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x : y = 9 : 5

Q. Calculate the bond energy of

C l - C l

bond from the following data:

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

. Also the bond enthalpies of

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

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Q. What is the

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Q. Consider the reactions:

C (s) + 2 H_{2} (g) ⟶ C H_{4} (g); Δ H = - X k c a l

C (g) + 4 H (g) ⟶ C H_{4} (g); Δ H = - X_{1} k c a l

C H_{4} (g) ⟶ C H_{3} (g) + H (g); Δ H = + Y k c a l

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Bond energies	kJ/ mol
$C - C$	$347$
$C = C$	$612$
$C - C l$	$341$
$C - H$	$414$
$C l - C l$	$243$

Calculate the enthalpy , ΔH, for the given reaction using the the given bond energies ?C2H4+Cl2→ClH2C−CH2ClBond energieskJ/ molC−C347C=C612C−Cl341C−H414Cl−Cl243

The correct option is C ΔH=−174kJΔHorxn=ΣΔHoreactant bonds−ΣΔHoproduct bondsΔHorxn=[4ΔHoC−H+ΔHoC=C+Cl−Cl]−[ΔHoC−C+4ΔHoC−H+2ΔHoC−Cl] ΔHorxn=[ΔHoC=C+Cl−Cl]−[ΔHoC−C+2ΔHoC−Cl]ΔHorxn=[612+243]−[347+2(341)]ΔHorxn=855−1029ΔHorxn=−174kJ

Calculate the enthalpy , $Δ H$ , for the given reaction using the the given bond energies ?
$C_{2} H_{4} + C l_{2} \to C l H_{2} C - C H_{2} C l$
Bond energies kJ/ mol
$C - C$ $347$
$C = C$ $612$
$C - C l$ $341$
$C - H$ $414$
$C l - C l$ $243$