Question

Propane has the structure H₃C-CH₂-CH₃. Use average bond enthalpies from the given data to estimate the change in enthalpy $\Delta H^{\circ }$for the reaction:$C_3{H}_8\left(g\right)+5O_2\left(g\right)\to 3C{O}_2\left(g\right)+4H_{2}O\left(g\right)$

Short answer

The answer is $\Delta H=-1.58\times {10}^{3}kJ.$

Step-by-step solution

Given data

Bond	Enthalpy
O=O	498
C=O	728
C-H	413
C-C	348
O-H	463

Concept of Enthalpy change

Enthalpy change is the amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in a chemical reaction.

It is represented by $\Delta H$.

By the use of bond enthalpies, the change in enthalpy has been calculated.

The structures of the participants of this reaction are given below.

The reactants:

$C_3{H}_8:8\hspace{0.33em}C-HBonds,2C-Cbondsand5O_2:5\hspace{0.33em}O=Obonds.$

The products:

$3C{O}_2:6C=OBondsand{H}_{2}O:4O-Hbonds.$

Calculate the enthalpy change of the products

By the use of bond enthalpies, the product enthalpy change has been calculated.

$$\begin{gathered} 3C{O}_2:6C=OBondsand{H}_{2}O:4O-Hbonds. \\ \Delta H_{products}=6\times \Delta H(C=O)+8\times \Delta H(O-H) \\ \Delta H_{products}=6\times 728kJ+8\times 463kJ \\ \Delta H_{products}=8072kJ \end{gathered}$$

 Step 4: Calculate the enthalpy change of the reactants

By the use of bond enthalpies, the product enthalpy change has been calculated.

$$\begin{gathered} C_3{H}_8:8\hspace{0.33em}C-HBonds,2C-Cbondsand5O_2:5\hspace{0.33em}O=Obonds. \\ \Delta H_{reac\tan ts}=8\times \Delta H(C-H)+2\times \Delta H(C-C)+5\times \Delta H(O=O) \\ \Delta H_{reac\tan ts}=8\times 413+2\times 348kJ+5\times 498kJ \\ \Delta H_{reac\tan ts}=6490kJ \end{gathered}$$

Calculate the overall energy change

Overall energy:

$$\begin{gathered} \Delta H_{reac\tan ts}=6490kJ \\ \Delta H_{products}=8072kJ \end{gathered}$$

Apply these values to calculate the overall energy change.

$$\begin{gathered} \Delta H=\Delta B{E}_{reac\tan ts}-\Delta B{E}_{products} \\ \Delta H=6490\hspace{0.33em}kJ-8072\hspace{0.33em}kJ \\ \Delta H=-1.58\times {10}^{3}kJ \end{gathered}$$