1 / 6

# Enthalpy and Bond Energies

Enthalpy and Bond Energies. For reactions: a. With simple molecules or atoms b. That occur only in the gas phase, with little attraction between particles. The change in enthalpy for a reaction can be estimated using Bond Energies. Télécharger la présentation ## Enthalpy and Bond Energies

E N D

### Presentation Transcript

1. Enthalpy and Bond Energies For reactions: a. With simple molecules or atoms b. That occur only in the gas phase, with little attraction between particles. The change in enthalpy for a reaction can be estimated using Bond Energies. a. Bond Energies are classified based on the two atoms involved in the bond and the order of the bond. b. Bond Energies are the average dissociation energy of all the covalent bonds of the same class in kJ mol -1.

2. Bond Energy Table – Average Bond Dissociation Energies for Selected Covalent Bonds The higher the Bond Energy, the stronger the bond. A stronger bonds indicates a more stable, lower energy state for the atoms in the bond.

3. Estimated Change in Enthalpy for a Reaction ( H) • H = Sum of all the Bond Energies of Bonds Broken • (positive) plus the sum of all the Bond Energies • Formed (negative).

4. Estimate the value of H for the reaction: 2 HCOOH + O2  2 H2O + 2 CO2 • Reactants • 2 C – H bonds ( 2 x 413 ) = 826 • 2 C = O bonds ( 2 x 805 ) = 1610 • C – O bonds ( 2 x 358 ) = 716 • 2 O – H bonds ( 2 x 464 ) = 928 • 3 O = O bonds ( 1 x 498 ) = 498 • Total = + 4578 kJ Products 4 H – O bonds ( 4 x 464 ) = 1856 4 C = O bonds ( 4 x 805 ) = 3320 Total = - 5076 kJ Hrxn = Products + Reactants = (4578) + (-5176) = - 498 kJ

5. Estimate the value of H for the reaction: 2 H2S + 3 O2  2 H2O + 2 SO2 Reactants 4 H – S bonds ( 4 x 347 ) = 1388 3 O = O bonds ( 3 x 498 ) = 1194 Total = + 2882 kJ Products 4 H – O bonds ( 4 x 464 ) = 1856 2 S – O bonds ( 2 x 265) = 530 2 S = O bonds ( 2 x 523) = 1046 Total = - 3432 kJ Hrxn = Products + Reactants = (2882) + (-3432) = - 550 kJ

6. Estimate the value of H for the reaction: 4 NH3 + 7 O2  6 H2O + 4 NO2 Reactants 12 N – H bonds ( 12 x 393 ) = 4716 7 O = O bonds ( 7 x 498 ) = 3486 Total = + 8202 kJ Products 12 H – O bonds ( 12 x 464 ) = 5568 4 N = O bonds ( 4 x 805 ) = 2428 4 N – O bonds ( 4 x 201 ) = 804 Total = - 8800 kJ Hrxn = Products + Reactants = (8202) + (-8800) = - 598 kJ

More Related