[Munch]

HI
 
I have a problem and i really have tried to tackle it and i thought i have the answer until i was told by my teacher that i am incorrect so some light on this matter would be excellent.

Iodoethane reacts with water to form ethanol and hydrogen iodide

C2H5I + H2O -> C2H5OH + HI gives + 38kJ mol-1

C-H = 413
C- C = 347
H - O = 464
H-I = 298
C-O = 358
C- I = ?

So we have to find out the mean C- I bond enthalpy

I was guessing that you add up the first of the equation and make it 3340 (without the C-I) of course. and then you add up the other side which is -3532. Now i thought that you take the sum of the bonds broken from tghe the sum of bonds formed so

3340 - 3532 = -192

But then i was told this was incorrect. Do i have to do something with the standard enthalpy of +38kJmol-1 like subtracting it from -192 giving an answer of -230??

Thanks for your help

[plu]

You are on the right track but you're missing a step!  You know that the energy needs to break the bonds on the left hand side of the equation is 3340 kJ + C-I and you know that the energy released from forming the bonds on the right hand side of the equation is 3532 kJ.  However, you also know that the net energy change for the reaction has to be +38 kJ from the enthalpy of reaction.  Therefore:

38 = 3340 + C-I - 3532