Use the Table of Bond Energies to Find the Ho for the Reaction
Chemistry Honors
Chapter 8
Charts
Information obtained from
Bond length (pm) and bond energy (kJ/mol)Bond / Length / Energy / Bond / Length / Energy
H--H / 74 / 436 / H--C / 109 / 413
C--C / 154 / 348 / H--N / 101 / 391
N--N / 145 / 170 / H--O / 96 / 366
O--O / 148 / 145 / H--F / 92 / 568
F--F / 142 / 158 / H--Cl / 127 / 432
Cl-Cl / 199 / 243 / H--Br / 141 / 366
Br-Br / 228 / 193 / H--I / 161 / 298
I--I / 267 / 151
C--C / 154 / 348
C--C / 154 / 348 / C=C / 134 / 614
C--N / 147 / 308 / CC / 120 / 839
C--O / 143 / 360
C--S / 182 / 272 / O--O / 148 / 145
C--F / 135 / 488 / O=O / 121 / 498
C--Cl / 177 / 330
C--Br / 194 / 288 / N--N / 145 / 170
C--I / 214 / 216 / NN / 110 / 945
Bond dissociation energy
The amount of energy required to break a bond is called bond dissociation energy or simply bond energy. Since bond lengths are consistent, bond energies of similar bonds are also consistent.
Use the table of bond energies to find the Ho for the reaction:
H2(g) + Br2(g) 2 HBr(g)
Solution
From the Table of bond length and bond energy given above, a table below is obvious:
Changes / HoH-H H + H / 436
Br-Br Br + Br / 193
H + H + Br + Br 2 H-Br / 2*(-366) = -732
Overall (add up)
H-H + Br-Br 2 H-Br / -103
Another approach is shown below. Write the bond energy below the formula, and then apply the principle of conservation of energy.
Bonds broken / Bonds formedH-H + Br-Br / 2 H-Br
Ho / 436 + 193 / -2*366 energy released
Ho = 436 + 193 - 2*366 = -103
If the change in enthalpy is less than zero the reaction is exothermic and energy is given off by the reaction. If the change in enthalpy is greater than zero the reaction is endothermic and energy is taken in to the reaction.
Electronegativity and bond polarity
We can use the difference in electronegativity between two atoms to gauge the polarity of the bonding between them
Compound / F2 / HF / LiFElectronegativity Difference / 4.0 - 4.0 = 0 / 4.0 - 2.1 = 1.9 / 4.0 - 1.0 = 3.0
Type of Bond / Nonpolar covalent / Polar covalent / Ionic (non-covalent)
- In F2 the electrons are shared equally between the atoms, the bond is nonpolar covalent
- In HF the fluorine atom has greater electronegativity than the hydrogen atom.
The sharing of electrons in HF is unequal: the fluorine atom attracts electron density away from the hydrogen(the bond is thus a polar covalent bond)
The H-F bond can thus be represented as:
- The 'd+' and 'd-' symbols indicate partial positive and negative charges.
- The arrow indicates the "pull" of electrons off the hydrogen and towards the more electronegative atom
- In lithium fluoride the much greater relative electronegativity of the fluorine atom completely strips the electron from the lithium and the result is an ionic bond (no sharing of the electron)
A general rule of thumb for predicting the type of bond based upon electronegativity differences:
- If the electronegativities are equal (i.e. if the electronegativity difference is 0-0.5), the bond is non-polar covalent
- If the difference in electronegativities between the two atoms is greater than 0.5, but less than 1.5, the bond is polar covalent
- If the difference in electronegativities between the two atoms is 1.5, or greater, the bond is ionic
