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From the following bond energies:

H-H bond energy : 431.37 k J  mol^{-1}

C=C bond energy : 606.10 k J  mol^{-1}

C-C bond energy : 336.49 k J  mol^{-1}

C-H bond energy : 410.50 k J  mol^{-1}

Enthalpy for the reaction,

 

 

Option: 1

-243.6 kJ mol^{-1}


Option: 2

-120.0 kJ mol^{-1}


Option: 3

-553.0 kJ mol^{-1}


Option: 4

-1523.6 kJ mol^{-1}


Answers (1)

best_answer

\Delta H=B.E._{\left ( reactant \right )}-B.E._{\left ( product \right )}

 

=\left [ 4\times B.E_{\left ( C-H \right )}+1\times B.E_{\left ( c=c \right )}+1\times B.E\left ( H-H \right )-\left ( 6\times B.E_{\left ( C-H \right )}+1\times B.E_{\left ( C-C \right )} \right ) \right ]

\left [ 4\times 410.5+1\times 606.10+1\times 431.37 \right ]

-\left [ 6\times 410.5+1\times 336.49 \right ]

-120kj/mol

Option (2) is correct answer.

Posted by

Suraj Bhandari

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