If a gasoline-powered vehicle requires primary energy of 3.7 MJ/km from
the combustion of gasoline (mostly octane) and emits 1.7 g/km of CO, what fraction of the carbon in the octane is not fully oxidized?

Question

Quizplus · Accepted Answer

The Answer is The complete combustion of octane is

2 \mathrm { C } _ { 8 } \mathrm { H } _ { 18 } + 25 \mathrm { O } _ { 2 } ightarrow 16 \mathrm { CO } _ { 2 } + 18 \mathrm { H } _ { 2 } \mathrm { O } + 46.8 \mathrm { MJ } / \mathrm { kg }

For

1 \mathrm {~km}

the amount of octane required is

( 3.7 \mathrm { MJ } / \mathrm { km } ) / ( 46.8 \mathrm { MJ } / \mathrm { kg } ) = 0.079 \mathrm {~kg} / \mathrm { km }

The molecular mass of octane is

8 imes 12 + 18 imes 1 = 114 \mathrm {~g} / \mathrm { mol }

of which

8 imes 12 = 96 \mathrm {~g} / \mathrm { mol }

is carbon. So

0.079 \mathrm {~kg}

octane contains

( 0.079 \mathrm {~kg} ) imes ( 96 \mathrm {~g} / \mathrm { mol } ) / ( 114 \mathrm {~g} / \mathrm { mol } ) = 0.0665 \mathrm {~kg}

carbon. The CO emission is

0.0017 \mathrm {~kg} / \mathrm { km }

. The molecular mass of

\mathrm { CO }

is

12 + 16 = 28

\mathrm { g } / \mathrm { mol }

of which

12 \mathrm {~g} / \mathrm { mol }

is carbon, or

0.0017 \mathrm {~kg} \mathrm { CO }

contains

( 0.0017 \mathrm {~kg} ) imes ( 12 \mathrm {~g} / \mathrm { mol } ) / ( 28

\mathrm { g } / \mathrm { mol } ) = 7.29 imes 10 ^ { - 4 } \mathrm {~kg}

carbon. So the percent carbon in the emission that is in the form of

\mathrm { CO }

is

100 imes \left( 7.29 imes 10 ^ { - 4 } \mathrm {~kg} ight) / ( 0.0665 \mathrm {~kg} ) = 1.1 \%

If a Gasoline-Powered Vehicle Requires Primary Energy of 3