How Do You Calculate Enthalpy Change Using Bond Energies?

[Priyadarshini]

Homework Statement

Homework Equations

The Attempt at a Solution

6C-H bonds= 410 x 6 = 2460 kJ/mol
3C-C bonds
H of formation= +53.3 kJ/mol
H of atomisation = +717 kJ/mol
H-H = 436 kJ/mol

as H of formation is 53.3, can't I just do:
2460+3x=53.3?
But then I am not using a lot of the values provided and the answer in not in the options.[/B]

 

[Bystander]

You have to make a "good faith" effort.

 

[Priyadarshini]

You have to make a "good faith" effort.

Actually, I don't know what to do at all. I know that I have make a Hess's cycle so that I can form equations. But I can't figure out the Hess's cycle here.

 

[Bystander]

What is Hess's cycle?

 

[Priyadarshini]

What is Hess's cycle?

The total enthalpy change is the same for all the routes the reaction can take to produce the product.

 

[Bystander]

Correct. What route are you taking?

 

[Priyadarshini]

Correct. What route are you taking?

Let the C-C bond energy be x
(3*-717) + (-436*3) + (410*6) + 3x = 53.3
-2151-1308+2460+3x=53.3
-999+3x=53.3
3x=1052.3
x=350.767
approx. 351 kJ/mol

But the answer should be 315kJ/mol.

 

[Bystander]

Were you to use conventional notation, you might be more successful.

 

[Priyadarshini]

Were you to use conventional notation, you might be more successful.

What do you mean by conventional notation?