Chapter 3: Energy from fuel Flashcards
Mass to mass calculation
- Determine the molar mass of the known and unknown species
- Convert the known substance mass to moles using the formula n = m/M.
- Use the stoichiometric ratio from the balanced equation to determine the number of moles of the unknown substance.
- If needed, convert the number of moles of the unknown substance to a mass using the formula m = n × M.
- Ensure the final answer has the appropriate units and number of significant figures.
EXAMPLE
3.00 kg of ethane gas undergoes combustion in an excess of oxygen according to the following equation. C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g) Calculate the mass, in grams, of CO2 produced if the reaction goes to completion.
m(C2H6) = 3.00 kg = 3000 g
M(C2H6) = 2 * 12 + 6 * 1 = 30.0 g mol-1
n(C2H6) = 3000 / 30.0
= 100 mol
n(C2H6) : n(CO2)
1 : 2
⇒ n(C2H6) = 200 mol
M(CO2) = 44gmol-1
m(CO2) = 44 * 200
= 8800 g
Express the answer to the least amount of sf in the question
8.80 × 10^3 g of CO2 is produced
Mass to volume calculations
- similar to mass to mass
- use the molar volume of gases (24.8 L mol-1) and the formula n = V/Vm
EXAMPLE
At SLC, methane gas undergoes complete combustion according to the following reaction. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) In a reaction vessel at SLC, 2.50 g of H₂O is produced. Calculate the volume of methane that must have been present in the reaction vessel.
n(H₂O) = 2.50 / 18.0
= 0.1388 mol
n(H₂O) : n(CH₄)
2 : 1
⇒ n(H₂O) = 0.1388 / 2 * 1
= 0.06944 mol
V(CH₄) = n × Vₘ
= 0.06944 * 24.8 L mol-1
= 1.72 L
Calculating quantity of greenhouse gases produced by combustion reactions
- Calculate the total mass of greenhouse gas products
- Calculate the total mass of greenhouse gas reactant
- Calculate the net change in mass of greenhouse gases during the reaction
net amount = amount produced - amount consumed
EXAMPLE
Under non-standard conditions, methane gas undergoes complete combustion in an excess of oxygen according to the following reaction. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) In a reaction vessel under these conditions, 10.0 g of CH4 is combusted. Calculate the net change in mass of greenhouse gases as a result of this reaction.
The products CO2 and H2O are greenhouse gases.
n(CH4) = 10.0 / 16.0
= 0.625 mol
n(CO2) : n(CH4)
1 : 1
⇒ n(CO2) = 0.625 mol
m(CO2 = n x M
= 0.625 mol × 44.0 g mol-1
= 27.5 g
n(H2O) : n(CH4)
2 : 1
⇒ n(CO2) = n(CH4) × 2
= 1.25 mol
m(H2O) = n x M
= 1.25 mol × 18.0 g mol-1
= 22.5 g
Total = m(CO2) + m(H2O)
= 27.5g + 22.5 g
= 50.0 g
The reactant CH4 is a greenhouse gas
m(CH4) = 10.0 g (given by the question)
Net change = mass produced − mass reacted
= 50.0 g − 10.0 g
= 40.0 g
What are the major greenhouse gases involved in combustion reactions
- water vapour (H2O)
- carbon dioxide (CO2)
- methane (CH4)
Volume to volume calculations
- all gases occupy the same volume per mole at a given temperature and pressure
Given a balanced combustion equation, conversion can be found using:
V(unknown) = coefficient of unknown volume/ coefficient of known volume × V(known)
Calculating amount of heat produced
energy released = heat of combustion × n(fuel)
Limiting reactant vs limiting reagent
A reactant is a substance that gets consumed in a chemical reaction
A reagent is a substance or compound added to check if a reaction occurs
* The limiting reagent is a reactant that is completely consumed in the chemical reaction
* When there is a limiting reagent in a reaction, there will always be an excess reagent present
* Reagents are not necessarily consumed during a reaction but reactants are
Enthalpy
the total energy stored in a substance
* also known as the heat content of a substance
Eenergy content of food/fuel that isn’t pure
energy content = energy transferred to water/change in mass of fuel during combustion
* q/ΔM
* ΔM = initial mass - final mass
Heat of combustion of pure fuels
q/n
* q is energy absorbed by water
* n is number of moles of fuel
Minimising heat loss
- put a lid on container holding water
- insulate beaker of water w/ flameproof material
- placing insulation around the burning fuel
Effect of heat loss
* water temperature doesn’t increase as much due to heat loss to surroundings
* lower temp change ΔT → lower q value
Specific heat capacity
describes the amount of energy required (measured in joules, J) to raise the temperature of 1 gram of a substance by 1°C
* given the symbol c
* measured in Jg−1°C− or J/gºC
Specific heat capacity of water
Water has a relatively high specific heat capacity of 4.18 Jg−1°C
* specific heat capacity depends on the strength of intermolecular forces.
* strong hydrogen bonds between water molecules allow it to absorb a large amount of thermal energy before increasing in temperature.
Calculating using specific heat capacity
q = mcΔT
- q = the amount of heat energy (J)
- m= the mass (g) (of the water heated)
- c = the specific heat capacity ($J g^-1 ºC^-1$) (of the water heated)
- ΔT = the temperature change (°C)
Limitations of experimental determination of heat of combustion
- A large quantity of heat energy released from the combustion of the fuel or food
(often ~90%) is lost to the surroundings, producing inaccurate results - The temperature of the laboratory affects the amount by which the water temperature increases, limiting reproducibility of results
- Not all fuels or foods are available in a form that can be safely and feasibly combusted
- The mass of water being heated will actually decrease due to evaporation
