💯An Ingelious Way to Solve Chemical Energetics Bond Energy Questions

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Chemical Energetics Bond Energies

Here’s an “ingelious” way to solve bond energies questions related to chemical energetics.
Let’s look at a typical JC A Level H2 Chemistry question:

Find the enthalpy change of reaction for
NH_3(g) + 3F_2(g) \rightarrow NF_3(g) + 3HF(g)\qquad \Delta\, H_f^\ominus = \, ? kJ mol^{-1}

Bond energy N-H 390 kJ mol -1
Bond energy F-F 158 kJ mol -1
Bond energy H-F 562 kJ mol -1
Bond energy N-F 272 kJ mol -1

Avoid using
\Delta\, H_f^\ominus = \Sigma \Delta\, H_f^\ominus(products) - \Sigma \Delta\, H_f^\ominus(reactants)
\Delta\, H_c^\ominus =  \Sigma \Delta\,H_c^\ominus(reactants) - \Sigma \Delta\, H_c^\ominus(products)
\Delta\, H_r^\ominus =  \Sigma Bond\, energy\, (bonds\, broken) - \Sigma Bond\, energy\, (bonds\, formed)

I know these are the standard formulae provided in your lecture notes.
However, these equations are:

  • confusing and easy to mix them up
  • can only be used if all the equations are of the same type (all enthalpy changes of formation, or enthalpy change of combustion)
  • need to memorise 3 sets of equations.

Instead I will use the following method:

1) Consider the total energy taken in to break bonds and total energy given out to form bonds.
2) Take the difference in magnitude.
3) Then add “+” if the overall energy is endothermic or “-” if the overall enthalpy is exothermic. The idea is to decouple magnitude and sign convention.

Total energy absorbed = 3(390) + 3(158) = 1644 kJ mol -1
Total energy released = 3(272) + 3(562) = 2502 kJ mol -1
Overall difference = 2502 – 1644 = 858 kJ mol -1
Since total energy released > total energy absorbed, overall reaction is exothermic.
\therefore \Delta\, H_r^\ominus = -858 kJ mol^{-1}

Chemical energetics bond energy 1

Chemical energetics bond energy 2

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Chemical energetics bond energy 6

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