💯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 Energies

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