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The alkynes are a homologous family of hydrocarbons - Scottish Highers Chemistry - Question 8 - 2018
Question 8
The alkynes are a homologous family of hydrocarbons. (a) The simplest member of the family is ethyne, C₂H₂, used in welding torches. Ethene can be produced from et... show full transcript
Worked Solution & Example Answer:The alkynes are a homologous family of hydrocarbons - Scottish Highers Chemistry - Question 8 - 2018
Step 1
Calculate the enthalpy change for the reaction.
Answer
To calculate the enthalpy change for the production of ethyne (C₂H₂) from ethane (C₂H₆), we first identify the bond enthalpies involved in the reaction:
- Breaking Bonds: Ethane has 6 C-H bonds and 1 C-C bond.
- Forming Bonds: Ethyne has 3 C-H bonds and 1 C≡C bond.
Using bond enthalpy values, we can sum up the total energy required to break the bonds and the energy released from forming the new bonds.
This leads us to the calculation:
- Total energy for breaking bonds = 6(C-H) + 1(C-C)
- Total energy for forming bonds = 3(C-H) + 1(C≡C)
Consolidating these values gives:
After performing the calculations, the final result for the enthalpy change will be found as 286 kJ mol⁻¹.
Step 2
Calculate the enthalpy change for the formation of propyne from its elements.
Answer
According to Hess's Law, we sum the enthalpy changes of the reactions leading to propyne formation. Using the given reaction equations:
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Start with the formation of carbon from graphite (C(s)) to CO₂(g) and hydrogen from H₂(g) to H₂O(l).
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The reaction for the formation of propyne can then be represented as follows:
With known enthalpy changes from the provided data: -
C(s) + O₂(g) → CO₂(g): ΔH = -394 kJ mol⁻¹
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H₂(g) + ½O₂(g) → H₂O(l): ΔH = -286 kJ mol⁻¹
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C₃H₈(g) + 4O₂(g) → 3CO₂(g) + 2H₂O(l): ΔH = -1939 kJ mol⁻¹
Calculating gives the enthalpy for the reaction as (ΔH = -572 + 394 + 1939):
Thus, we find the enthalpy change is calculated to be +185 kJ mol⁻¹.
Step 3
Calculate the energy released when 1 kg of propyne is burned.
Answer
Given the enthalpy of combustion of propyne is -1939 kJ mol⁻¹, we proceed to calculate the energy released when 1 kg (1000 g) of propyne is combusted:
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First calculate the number of moles of propyne:
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Now, multiply this with the enthalpy of combustion:
Therefore, the energy released is 48475 kJ.
Step 4
Calculate the mass of air required to burn 1 g of propyne.
Answer
Given the combustion reaction:
Using the mass of oxygen from the stoichiometry:
For combustion of 1 g of propyne, with its molar mass 40 g:
- Needed moles of O₂ =
Next, convert moles of O₂ to grams:
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Molar mass of O₂ = 32 g/mol:
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Mass of O₂ = 0.1 × 32 = 3.2 g.
Finally, using the relationship provided:
Therefore, approximately 13.76 g of air is required to burn 1 g of propyne.
Step 5
Discuss why methanol and ethanol require less air to burn.
Answer
Methanol (CH₃OH) and ethanol (C₂H₅OH) both contain hydroxyl (–OH) functional groups in their structures. This means that they already possess oxygen within their molecular structure, reducing the amount of additional oxygen needed for combustion.
In contrast, hydrocarbons like ethane and propane lack this oxygen, meaning they require more external oxygen to ensure complete combustion. Therefore, the presence of the hydroxyl group lowers the overall oxygen demand, thereby requiring less air for combustion.