Question

Some automobiles and buses have been equipped to burn propane $\left(C_3{H}_8\right)$ as a fuel. Compare the amount of energy that can be obtained per gram of $C_3{H}_8\left(g\right)$ with that per gram of gasoline, assuming that gasoline is octane $\left[C_8{H}_{18}\left(l\right)\right].$ (See Example 9.8.) Look up the physical properties of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Short answer

The enthalpy of combustion of propane is $-50.465kJ/g$ and for octane is $-47.90kJ/g.$

Step-by-step solution

Combustion of propane

The combustion reaction of propane is,

$C_3{H}_8\left(g\right)+5O_2\left(g\right)\to 3C{O}_2\left(g\right)+4H_{2}O\left(l\right)$

The standard enthalpy for combustion of propane is,

$$\begin{gathered} \Delta H^o=\Delta H_f^o\left(product\right)-\Delta H_f^o\left(reactant\right) \\ \Delta H^o=\left[3\times \Delta H_f^o\left(C{O}_2\right)+4\times \Delta H_f^o\left(H_{2}O\right)\right]-\left[\Delta H_f^o\left(C_3{H}_8\right)+5\times \Delta H_f^o\left(O_2\right)\right] \\ \Delta H^o=\left[2\times \left(-393.5kJ\right)+4\times \left(-286.0kJ\right)\right]-\left[\left(-104.0kJ\right)+\left(5\times 0\right)\right] \\ \Delta H^o=-2220.5kJ/mol \end{gathered}$$

1 mol of propane = 44 g

Heat released by propane per gram is,

$Heat=\frac{-2220.5kJ}{44g}=-50.465kJ/g$

Hence, the enthalpy of combustion of propane is role="math" localid="1658169974814" $-50.465kJ/g.$

Combustion of octane

The combustion reaction of octane is,

$C_8{H}_{18}\left(g\right)+\frac{25}{2}{O}_2\left(g\right)\to 8C{O}_2\left(g\right)+9H_{2}O\left(l\right)$

The standard enthalpy for combustion of propane is,

$$\begin{gathered} \Delta H^o=\Delta H_f^o\left(product\right)-\Delta H_f^o\left(reactant\right) \\ \Delta H^o=\left[8\times \Delta H_f^o\left(C{O}_2\right)+9\times \Delta H_f^o\left(H_{2}O\right)\right]-\left[\Delta H_f^o\left(C_8{H}_{18}\right)+\frac{25}{2}\times \Delta H_f^o\left(O_2\right)\right] \\ \Delta H^o=\left[8\times \left(-393.5kJ\right)+9\times \left(-286.0kJ\right)\right]-\left[\left(-250.0kJ\right)+\left(\frac{25}{2}\times 0\right)\right] \\ \Delta H^o=-5472.0kJ/mol \end{gathered}$$

1 mol of octane = 114.22 g

Heat released by octane per gram is,

$Heat=\frac{-5472.0kJ}{114.22g}=-47.90kJ/g$

Hence, the enthalpy of combustion of octane is -47.90 kJ/g.

Disadvantages of using propane as fuel

The boiling point of propane is lower than octane. Because of the low boiling point, there are extra safety hazards associated with using high-pressure gas tanks.