From the following bond energies:
H—H bond energy: 431.37 kJ mol $^{- 1}$
C=C bond energy: 606.10 kJ mol $^{- 1}$
C—C bond energy: 336.49 kJ mol $^{- 1}$
C—H bond energy: 410.50 kJ mol $^{- 1}$
Enthalpy for the reaction, $C H_{2} = C H_{2} + H_{2} \to C H_{3} - C H_{3}$ will be

Question

From the following bond energies:
H—H bond energy: 431.37 kJ mol

^{- 1}

C=C bond energy: 606.10 kJ mol

^{- 1}

C—C bond energy: 336.49 kJ mol

^{- 1}

C—H bond energy: 410.50 kJ mol

^{- 1}

Enthalpy for the reaction,

C H_{2} = C H_{2} + H_{2} \to C H_{3} - C H_{3}

will be

Accepted Answer

Correct answer: B. (b) For the given reaction, enthalpy of reaction can be calculated as =B .E . (reactant) —B.E. (product) = [B .E_ (C=C )+B .E_ (H-H )+4 B .E_ (C-H ) ] - [B .E_ (c-C )+6 B .E_ (c-H ) ] = [606.10+431.37+4 410.50 ] - [336.49+6 410.50 ] =2679.47-2799.49=-120.02 kJmo1^-1

$From the following bond energies: H—H bond energy: 431.37 kJ mol^{- 1} C=C bond energy: 606.10 kJ mol^{- 1} C—C bond energy: 336.49 kJ mol^{- 1} C—H bond energy: 410.50 kJ mol^{- 1} Enthalpy for the reaction, C H_{2} = C H_{2} + H_{2} \to C H_{3} - C H_{3} will be$

Solution

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