Question

1. a. Calculate the number of pounds of CO₂ produced from a weekend trip to San Fransisco. Assume 275 miles one-way and an average fuel efficiency of 22 miles per gallon. Assume gasoline to be pure octane (C₈H₁₈) with a density of .70 g/mL. (Hint: start your work by writing out the balanced equation for the combustion of octane.)

b. A past electric bill showed that my housegold used 380 Kwh (kilowatt-hours) of power over a period of a month. If a Kw is equal to a kJ/s (kilojoule per second) and the enthalpy (heat) of combustion of octane (deltaH_comb.)= - 5980 kJ/mol (C₈H₁₈), calculate the number of moles of octane that it would take to make this electricity. Convert the moles of octane into pounds of CO₂. (Although power plants do not use octane to make electricity, other reduced carbon fuels are common. Also, we must remember that the conversion of chemical energy within the power plant to electrical energy and its delivery to your household outlet is a process that is, at best, 40% efficient. Factor this efficiency into your final answer.)

c. Compare your answers from Part A and Part B. What can you conclude about the amount of energy needed to take a "day trip down the coast" and the energy necessary to live in a house for a month? Explain.

Answer 1

1 gallon = 3.785 L

1 kg = 2.2 Pound

a) Total distance covered in a round trip = 275*2 = 550 miles

Total fuel used = 550/22 = 25 gallon (22 miles per gallon is given)

= 25* 3.785L (1 gallon = 3.785 L)

= 94.625L = 94625ml

mass of Octane = 0.7*94625 = 66237.5g (density of octane = 0.7 g/ml)

molar mass of octane(C8H18) = 12*8 + 1*18 = 96+18 = 114g

No. of moles of octane burnt = 66237.5/114 = 581 moles.

Now, writing balanced chemical equation for combustion of octane(C8H18)

2C₈H₁₈ + 25O₂   16CO₂ + 18H₂O

we can say that 2 moles of octane will give 16 moles of CO₂

So, 1 mole of octane will give 8 moles of CO₂

So, 581 moles of octane will give 8*581 = 4648 moles of CO₂

= 4648*12 = 55776g of CO₂ =

= 55.776kg of CO₂

= 55.776*2.2 Pound of CO₂ = 122.7 pound CO₂ in a round trip.

b) Total Monthly electric consumption = 380 kWH

= 380*3600 kJ (1H = 3600s)

= 1368000 kJ

when 1 mole of Octane reacts, energy released = 5980 kJ (Given H_c = -5980 kJ/mol)

So, moles of octane reacted = 1368000 / 5980

= 228.76 moles of Octane required.

Now, since 581 moles of Octane was giving 122.7 pound CO₂

So, 228.76 mole of Octane will give = 122.7*228.76 / 581

= 48.3 pound of CO₂

Now, as the problem says that that the power plant is working at 40% efficiency only. so to produce the same amount of electricity we have to burn more fuel and more CO₂ will form.

lets say the actual amount of CO₂ formed is y pound.

so, y*0.4 = 48.3

So, y = 48.3/0.4 = 120.78 pounds of CO₂ formed to get electricity for one month.

c) The CO₂ produced in a round trip = 122.7 pound CO₂

The CO₂ formed to get electricity for one month = 120.78 pounds.

We can see both the amount of CO₂ produced are nearly equal. So, one round trip down the coast will cost same environmental hazard as using electricity for a whole month.