Chapter U5, Problem 14STP

Interpretation Introduction

Interpretation:

The energy change during combustion of propane with the help of bond energies needs to be determined.

${\text{C}}_{\text{3}}{\text{H}}_{\text{8(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 3 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Concept introduction:

The bond energy of a chemical bond can be defined as the energy required to break that chemical bond. The bond energy that is needed to break the bonds in reactant molecule and the energy released to form chemical bonds in product can be used to calculate the $\Delta \text{H}$ of the chemical reaction.

$\Delta \text{H=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}$

Answer to Problem 14STP

Option (A) is correct answer.

$\text{ΔH= -2060 kJ/mol}$

Explanation of Solution

Reason for correct option:

${\text{C}}_{\text{3}}{\text{H}}_{\text{8(g) }}{\text{ + 5 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 3 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

For the given reaction:

$\begin{array}{l}\text{ΔH=}\sum {\text{BE}}_{\text{reactant }}\text{- }\sum {\text{BE}}_{\text{product }}\\ \text{ΔH=}3\times {\text{BE}}_{\text{C-C }}\text{+}8\times {\text{BE}}_{\text{C-H }}\text{+ }5\times {\text{BE}}_{\text{O=O }}\text{- }\left({\text{6×BE}}_{\text{C}=O}+{\text{8×BE}}_{\text{O-H}}\right)\\ \text{ΔH=}3\times \text{347 +}8\times \text{413+ }5\times \text{495-}\left(\text{ 6×}799+\text{8×467}\right)\\ \text{ΔH= -2060 kJ/mol}\end{array}$

Conclusion

Reasons for incorrect options: The difference of the bond energies of reactant and product is equal to the combustion enthalpy of the reaction thus option (b), (c) and (d) are incorrect.