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  • Determine the enthalpy (in kJ) of the following reaction: <img alt="CH_3CH = CH + \frac{9}{2}O_2 \rightarrow 3CO_2 + 3H_2O" src="http://entrancecorner.oncodecogs.com/gif.latex?CH_3CH%20%3

Determine the enthalpy (in kJ) of the following reaction:

CH_3CH = CH + \frac{9}{2}O_2 \rightarrow 3CO_2 + 3H_2O

The bond energy of different bonds is given in table 

Bond

C-C

 C=C C-H C=O O=O

O-H

 

Bond Energy (kJ/mol)

 347 611 414 736 498

464

 

Option: 1

-1931


Option: 2

1931


Option: 3

12469


Option: 4

-12469


Answers (1)

best_answer

Let’s count the number of bonds in reactants and products

Reactant

No. of C=C bonds  = 1

No. of C-C bonds  = 1

No. of C-H bonds  = 5

No. of O=O bonds  = 4.5

Product

No. of C=O bonds  = 6

No. of O-H bonds  = 6

 

\\\Delta H = \Sigma E_{\textup{reactant bonds broken}} - \Sigma E_{\textup{product bonds broken}} \\\Delta H = [(1) (611) + (1) (347) + (5) (414) + (4.5) (498)] - [(6) (736) + (6) (464)] \\\Delta H = 5269 - 7200 = -1931 kJ

Posted by

Pankaj Sanodiya

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