Enthalpy change for the reaction,
${4 H}_{g} ⟶ {2 H}_{2 (g)}$ is -869.6 kJ.
The dissociation energy of H-H bond is-

-434.8 kJ

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-869.6 kJ

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+434.8 kJ

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+217.4 kJ

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Solution

The correct option is C +434.8 kJ
$H_{2} \to 2 H = 869.6 = 434.8$

$\frac{869.6}{2} = 434.8$

Option C

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Q. Calculate the enthalpy change for the reaction,

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C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction.

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	336.81 kJ / mol	606.68 kJ / mol	410.87 kJ / mol	431.79 kJ / mol

If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ $m o l^{- 1}$ , the value of enthalpy change for the reaction

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Enthalpy change for the reaction,4Hg⟶2H2(g) is -869.6 kJ.The dissociation energy of H-H bond is-

The correct option is C +434.8 kJH2→2H=869.6=434.8869.62=434.8Option C

Enthalpy change for the reaction,
${4 H}_{g} ⟶ {2 H}_{2 (g)}$ is -869.6 kJ.
The dissociation energy of H-H bond is-

The correct option is C +434.8 kJ
$H_{2} \to 2 H = 869.6 = 434.8$

$\frac{869.6}{2} = 434.8$

Option C