[xanthiner14]

I know this is pretty straight forward concept, but for some reason I'm still confused. Can someone help me get this stuff straight once and for all?

Using the following rxn:


CH4 + OH ---> CH3 + H2O

It is given that the rxn is exothermic, but the question is why it's exothermic. I narrowed 4 choices to 2.

a) the CH sigma bond is stronger than the OH sigma bond
b) the CH sigma bond is weaker than the OH sigma bond

To decide between the two, I used the fact that in an exothermic rxn the Bond strength of the reactant is less than the Bond strength of its products. Thus I selected (A). But this is wrong. It's B. What is wrong with my logic here?

[Jorriss]

That reaction is exothermic?

Did you write that correctly?

[Schrödinger]

a) the CH sigma bond is stronger than the OH sigma bond
b) the CH sigma bond is weaker than the OH sigma bond

To decide between the two, I used the fact that in an exothermic rxn the Bond strength of the reactant is less than the Bond strength of its products. Thus I selected (A). But this is wrong. It's B. What is wrong with my logic here?

Does your choice support your reasoning?

[tamim83]

The reaction is indeed exothermic, I checked.  

Your reasoning is correct, you just need to take it a step further.  How do you calculate the reaction enthalpy from bond dissociation energies?  If the bond dissociation energy of a reactant bond is less than that of a product bond, how does this effect the sign of the enthalpy?  

[xanthiner14]

Well, given;
A) if an exothermic reaction is indicated by (-)Delta H
B) Delta Hf= (sum of BDE's of P's) -(BDE's of R's)

From simple math, it would seem that the reactants would require higher BDE than the BDE of the products. I checked a chart and did the math and indeed found this to be true and arrived at a (-)Delta H. Great! BUT...does this mean that bond strength is the inverse of BDE?

Since the because the rxn is exothermic, and I know that in this case the Bond strength of the reactants is less than the products, and that the Energy released to form the products is greater than the Energy required to break the reactant bonds. This makes sense (assuming its right). I think it was the way the answers were worded that threw me off a bit.

Could someone please confirm that this thought process is correct? Perhaps add some insight that would help me to solidify this concept once and for all!

[tamim83]

B) Delta Hf= (sum of BDE's of P's) -(BDE's of R's)

If you are using BDEs, the enthalpy of formation is actually the reverse (reactants - products).  All BDEs are positive since breaking a bond is endothermic (it takes energy).  Forming a bond releases energy, which is why you must subbtract the sum of the BDEs for bonds formed from the sum of the BDEs for bonds broken. 

Great! BUT...does this mean that bond strength is the inverse of BDE?

No, since all of the BDEs are positive, a higher BDE means a stronger bond. 

I think your reasoning is mostly correct, you just need to iron out things.  Perhaps you are confusing calculating the reaction enthalpy using BDE and enthalpy of formation, which is common.