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Bond Energies and Enthalpy


Bond Energies and Enthalpy

Forming chemical bond atoms become more stable and their energies decrease and this energy is released outside. While breaking this bond same amount of energy is required. Energy released during formation of one mol bond and required for breaking one mole bond is called bond energy. They are expressed in kcal/mol.

For example, bond energy of H-H is 104 kcal/mol. This means that, forming one mol H-H bond 104 kcal energy is released or breaking one mol H-H bond 104 kcal energy is required.

Example:

1) 2H(g) → H2(g) ; ∆H=-104 kcal

H2(g) → 2H(g) ; ∆H=104 kcal

2) 2Cl(g) → Cl2(g) ; ∆H=-58 kcal

Cl2(g) → 2Cl(g) ; ∆H=58 kcal

As you can see from the examples above, bond between H atoms is stronger than bonds of Cl atoms. Thus, H2 molecule is more stable than Cl2 molecule.

Chemical reactions occur by breaking bonds between matters and forming new bonds. Thus, there is a relation between bond energies and enthalpy of reactions. Breaking the bond of reactants energy is required and this energy is positive. However, forming new bonds energy is released. This energy is negative.

If we sum these energies, we find enthalpy of reaction.

Reactants → Products ; ∆H=?

∆H=∑(Bond Energies)Reactants-∑(Bond Energies)Products

Where ∑ shows sum of given quantities.

In a reaction If;

  • (Sum of bond energies of reactants) > (Sum of bond energies of products) then, ∆H > 0, in other words reaction is endothermic. Some part of energy required to break bonds of reactants is taken from energy released from formation of bonds of products and some part of it is taken from outside.
  • (Sum of bond energies of reactants) < (Sum of bond energies of products) then, ∆H < 0, in other words reaction is exothermic. Thus, some part of the energy released from forming new bonds in products is used for breaking bond in reactants and some part of energy is released outside.

Example: Find H-Br bond energy by using following reactions;

2H(g) → H2(g)                 ; ∆H=-104 kcal

1/2Br2(g) → Br(g)            ; ∆H= 23 kcal

H2(g) + Br2(g) → 2HBr(g)  ; ∆H=-18 kcal

Solution:

We find bond energy of H-H by reversing first reaction;

H2(g) → 2H(g)     ; ∆H=104 kcal ( since reaction is reversed; ∆H becomes positive)

We find bond energy of Br-Br by multiplying second reaction with 2;

Br2(g) → 2Br(g)            ; ∆H= 46 kcal

Let me say bond energy of H-Br X kcal/mol, we find it by using following formula;

H2(g) + Br2(g) → 2HBr(g)  ; ∆H=-18 kcal

∆H= (Sum of bond energies of reactants) - (Sum of bond energies of products)

-18 = (104 + 46) - 2X

X= 84 kcal/mol

Bond energy of H-Br is 84 kcal

Thermochemistry Exams and Problem Solutions


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Tags: Thermochemistry Cheat Sheet |  Thermochemistry Exams and Problem Solutions |  Thermochemistry Exam1 and Problem Solutions |  Thermochemistry Exam2 and Problem Solutions | 


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