# enthalpy reaction

2C2H6 + 7O2 --->  4CO2 + 6H2O.

According to the table in the book:

C-H bond = 414 x2
O2 = 0
C-O double bond = 745 x 4
H2O bond = 460 x 6

The answer should be - 2759 and I am way off.
Thank you.

### Re: enthalpy reaction

kingchemist Tue, 01/12/2010 - 16:31

According to the equation you have
1. Each C atom in C2H6 has 3 H atoms attached. So there are 6 x C-H bonds to break in each C2H6 and you have two of these. So 12 mole of C-H bonds to be broken
2. 7 x O=O bonds to be broken and the value is certainly not 0 - I found a value of 498kJ per mole of O=O bonds.
3. In each O=C=O molecule you have to make 2 x C=O bonds, so in 4CO2 you will have 8 moles of C=O bonds to make
4. In each H2O there are 2 x H-O bonds to make, so in 6H2O there are 12 x O-H bonds to make.

### Re: enthalpy reaction

gmg12 Tue, 01/12/2010 - 16:35

kingchemist wrote:

According to the equation you have
1. Each C atom in C2H6 has 3 H atoms attached. So there are 6 x C-H bonds to break in each C2H6 and you have two of these. So 12 mole of C-H bonds to be broken
2. 7 x O=O bonds to be broken and the value is certainly not 0 - I found a value of 498kJ per mole of O=O bonds.
3. In each O=C=O molecule you have to make 2 x C=O bonds, so in 4CO2 you will have 8 moles of C=O bonds to make
4. In each H2O there are 2 x H-O bonds to make, so in 6H2O there are 12 x O-H bonds to make.

Ok, thanks. I thought when an element was by itself it would be 0 regardless of the coeffecient?

### Re: enthalpy reaction

kingchemist Tue, 01/12/2010 - 17:14

Any bonds that have to be broken or made require/release energy. Bond breaking is an endothermic process and bond making is an exothermic process

### Re: enthalpy reaction

kingchemist Tue, 01/12/2010 - 18:17

You need also to break one C-C bond in each ethane molecule, so in 2C2H6 there are 2 x C-C bonds to be broken

### Re: enthalpy reaction

gmg12 Wed, 01/13/2010 - 14:22

I am still not getting -2759....

C-C bond  347x2 = 694
C-H bond 414 x 12 = 4968
O=O bond 498.7 x 7 = 3490.9
C=O bond 745x8 = 5960
H-O bond 460x12= 5520

694+4968+3490.9  -  5960 + 5520 = -2327.1  ???

### Re: enthalpy reaction

kingchemist Wed, 01/13/2010 - 14:38

Energy needed
12 x 414 for C-H bonds = 4968
2 x 347 for C-C bonds = 694
7 x 498.7 for O=O bonds = 3490.9

Total energy needed = 9152.9 kJ

Energy released
8 x 745 for C=O bonds = 5960
12 x 460 for O-H bonds = 5520

Total energy released = 11480 kJ

Overall energy released = 11480 - 9152.9 = 2327.1kJ

That's what Iget