The following hydrocarbons can all be used as fuels - Leaving Cert Chemistry - Question 8 - 2002
Question 8
The following hydrocarbons can all be used as fuels.
methane (CH₄)
butane (C₄H₁₀)
2,2,4-trimethylpentane (C₈H₁₈)
a) Butane is a major component of LPG. What do t... show full transcript
Worked Solution & Example Answer:The following hydrocarbons can all be used as fuels - Leaving Cert Chemistry - Question 8 - 2002
Step 1
a) What do the letters LPG stand for?
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Answer
LPG stands for Liquefied Petroleum Gas.
Step 2
a) Draw two structural isomers of butane.
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The two structural isomers of butane (C₄H₁₀) are:
n-butane:
CH₃−CH₂−CH₂−CH₃
Isobutane (2-methylpropane):
CH₃−CH(CH₃)−CH₃
Step 3
b) Why are mercaptans often added to natural gas?
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Mercaptans are added to natural gas primarily to give it an odor, allowing for detection of gas leaks. This is crucial for safety as natural gas is odorless and can pose explosion risks.
Step 4
b) Apart from leaking gas pipes, name a major source from which methane is released to the atmosphere.
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A major source of methane released to the atmosphere is agricultural activities, particularly from livestock digestion (enteric fermentation) and manure management.
Step 5
c) What structural feature of 2,2,4-trimethylpentane results in it having a high octane rating?
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The presence of branching in the structure of 2,2,4-trimethylpentane contributes to its high octane rating. Branched hydrocarbons tend to burn more smoothly and are less likely to knock.
Step 6
c) Give one other structural feature which increases the octane number of a hydrocarbon.
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The presence of methyl branches or a short carbon chain structure also increases the octane number of hydrocarbons.
Step 7
d) Define heat of combustion of a compound.
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The heat of combustion of a compound is the amount of energy released when one mole of the substance undergoes complete combustion with oxygen, forming carbon dioxide and water.
Step 8
e) Calculate the heat of combustion of butane.
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To calculate the heat of combustion, use the formula:
extΔHcomb=extΣΔHf(products)−extΣΔHf(reactants)
From the balanced equation:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
The heats of formation are:
ΔH_f(C₄H₁₀) = -125 kJ/mol,
ΔH_f(CO₂) = -394 kJ/mol,
ΔH_f(H₂O) = -286 kJ/mol.