bond stability: ΔH dissociation vs. ΔH formation. how can you tell which bonds are the most stable?

[LSD-25]

From what I understand:
ΔH dissociation is the positive because energy is required to break bonds. larger ΔH dissociation = more stable stronger shorter bonds.


This is the part i'm not sure about:
ΔH formation: negative because energy is released when bonds are formed. so would a smaller ΔH of formation mean the bonds are shorter and stable and stronger?

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[sazerac]

That depends on your definition of "smaller". Do you mean smaller, as in closer to zero? Or smaller, as in Less Than?

Do you define -5kJ as smaller than -900kJ?

 

[LSD-25]

That depends on your definition of "smaller". Do you mean smaller, as in closer to zero? Or smaller, as in Less Than?

Do you define -5kJ as smaller than -900kJ?

closer to zero (less negative) and if a Δh formation was positive i'm thinking it'd be even more stable is that right?

 

[Cawolf]

The more energy liberated upon bond formation, the stronger the bond. Therefore, a "more negative" enthalpy of formation is indicative of a stronger bond.

Consider what you are stating when you refer to a positive enthalpy of formation. You are saying that to make that bond, you have to add energy. That bond is so unstable, that it will not form without the input of energy.

 

[RH8448]

In other words: a reaction that occurs with a negative enthalpy will produce products that are more stabile than the reactants. A positive enthalpy will produce products that are less stabile than the reactants.

Correct me if I'm wrong, thanks.

 

[Cawolf]

@RH8448

That sounds good to me.

 

[LSD-25]

The more energy liberated upon bond formation, the stronger the bond. Therefore, a "more negative" enthalpy of formation is indicative of a stronger bond.

Consider what you are stating when you refer to a positive enthalpy of formation. You are saying that to make that bond, you have to add energy. That bond is so unstable, that it will not form without the input of energy.

In other words: a reaction that occurs with a negative enthalpy will produce products that are more stabile than the reactants. A positive enthalpy will produce products that are less stabile than the reactants.

Correct me if I'm wrong, thanks.

I'm confused, so is would a more negative enthalpy of formation or more positive one indicate more stable/stronger bonds?

 

[Cawolf]

What is confusing?

Negative ΔHf = exothermic = more stable product
Positive ΔHf = endothermic = less stable product

What you refer to as ΔH Dissociation is more commonly called Bond Dissociation Energy, BDE, or simply D.

The relationships stated above for ΔHf are opposite of BDE. If BDE is large, then it takes a fair amount of energy to break a bond - so that bond is strong. Look at what each term means and then it should be rather intuitive.

ΔH reaction = BDE (reactants) - BDE (products)
ΔH reaction = ΔHf (products) - ΔHf (reactants)