3.1.4.4 Bond enthalpies
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Mean bond enthalpy
The enthalpy needed to break the covalent bond into gaseous atoms, averaged over different molecules.
Why are all mean bond enthalpies positive values??
It is the energy required to break a bond.
When does the mean bond enthalpy definition apply?
When the substances and products are all in a gaseous state.
Why may the mean bond enthalpy differ from the calculated bond energy of a bond in a molecule?
Every single bond in a compound has a slightly different bond enthalpy- breaking each one requires a different amount of enthalpy.
However, the average value is calculated for mean bond enthalpy from across all molecules with that bond.
The relationship between sum of bond enthalpies of products and reactants in an exothermic reaction.
In an exothermic reaction, the sum of bond enthalpies of the reactants is less than that of the products.
The relationship between sum of bond enthalpies of products and reactants in an exothermic reaction.
In an endothermic reaction, the sum of bond enthalpies of the products is less than that of the reactants.
Equation representing relationship between enthalpy change and the bond enthalpies of the products and reactants
enthalpy change = sum of bond enthalpies broken - sum of bond enthalpies made
sum of bond enthalpies broken = sum of bond enthalpies in the reactants
sum of bond enthalpies made = sum of bond enthalpies in the products
Why is the enthalpy change calculated from mean bond enthalpies less accurate than using formation or combustion data?
Mean bond enthalpies are used which aren’t exact and differ between compounds.
How and why do enthalpies of combustion change with successive members of a homologous series?
Each member in the series increases by CH2 / has one more C-C and two more C-H
Combustion of each successive member in a homologous series means one more C-C and two more C-H bonds are broken AND one more mole of CO2 and one more mole of H20 is formed.
Why are calculated and experimental results from using a calorimeter different?
Experimental results will be much lower than the calculated ones due to heat loss and incomplete combustion.
Note 
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