Question

A large SUV has a mass of 2400 kg .

Calculate the mass of CO2 emitted into the atmosphere upon accelerating the SUV from 0.0 to 66.0 mph . Assume that the required energy comes from the combustion of octane (ΔH∘f=−250.1 kJ) with 30% efficiency. (Hint: Use KE=12mv2 to calculate the kinetic energy required for the acceleration.)

Answer 1

The first thing to do is to calculate the kinetic energy with the available information.

KE = 1/2 mv²

the mass given is 2400 kg, the speed is 66 mph, change this to m/s

66 mph * 1609.34 meters / miles * 1 hour / 3600 s

66 mph = 29.5 m/s

Apply the equation to get:

KE = 1/2 * (29.5)² * 2400 = 1044300 Joules = 1 ,044.3 KJ

Now consider the heat of combustion of octane, which i get a value of -5470 KJ / mol, the value -250 KJ / mol is the standard enthalpy of formation so.

The first thing to do is to balance the combustion of octane, but we need the balance chemical equation to do that so:

C₈H₁₈ + O₂ ==== CO₂ + H₂O , i will assume a complete combustion, balance the number of carbons, add 8 to the right

C₈H₁₈ + O₂ ==== 8CO₂ + H₂O , now balance the Hydrogens, you have 18 on the left and 2 on the right, add a 9 on the water molecule

C₈H₁₈ + O₂ ==== 8CO₂ + 9H₂O , now balance the oxygen, you have 16+9=27 on the right, add a 27/2 on the left

C₈H₁₈ + 27/2 O₂ ==== 8CO₂ + 9H₂O , now multiply all the equation by 2

2 C₈H₁₈ + 25 O₂ ==== 16 CO₂ + 18 H₂O, so we know that 1 mole of octane provides 5740 KJ of energy , then 2 moles of octane will provide 11, 480 KJ

since we only have 30% efficiency then

11 480 * 0.3 = 3 444 KJ for to moles

for one mole we will have 1 722 KJ

So let's calculate the ammount required

the KE is 1 044.3 KJ,

moles of octane required:

1044.3 / 1722 = 0.6064 moles of octane required

1 mole of octane produces 8 moles of CO2 so

0.6064 moles of octane * 8 moles CO2 / 1 mole octane = 4.851 moles of CO₂

Calculate the mass

moles = mass / molar mass

mass = moles * molar mass

mass = 4.851 * 44 g/mol = 213.44 grams of CO₂ produced