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The alkyes are a homologous family of hydrocarbons - Scottish Highers Chemistry - Question 8 - 2018

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The alkyes are a homologous family of hydrocarbons. (a) The simplest member of the family is ethyne, C₂H₂, used in welding torches. Ethylene can be produced from e... show full transcript

Worked Solution & Example Answer:The alkyes are a homologous family of hydrocarbons - Scottish Highers Chemistry - Question 8 - 2018

Step 1

Using bond enthalpies to calculate enthalpy change

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Answer

To calculate the enthalpy change for the reaction, we consider the bond enthalpies involved in the breaking and forming of bonds:

  1. Calculate the total bond energy for the reactants (ethane).
  2. Calculate the total bond energy for the products (ethyne).
  3. The enthalpy change (ΔH) can be calculated as:
ΔH=Total energy of reactants−Total energy of productsΔH = \text{Total energy of reactants} - \text{Total energy of products}

Using values from the data book:

  • C-C bond in ethane is approximately 348 kJ/mol.
  • C-H bonds are approximately 412 kJ/mol.
  • Total for ethane is (3 × 412) + 348 = 1584 kJ/mol.
  • Ethyne has a triple bond (C≡C) which is approximately 839 kJ/mol and 6 C-H at 412 kJ/mol (6 × 412 = 2472 kJ/mol).
  • Now, plugging these values:
ΔH=1584−2472=−888extkJmol−1ΔH = 1584 - 2472 = -888 ext{ kJ mol}^{-1}

Subtracting gives:

ΔH=286extkJmol−1ΔH = 286 ext{ kJ mol}^{-1}

Step 2

Calculate the enthalpy change for the formation of propyne

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Answer

Using Hess's Law:

  1. Start with the formation reaction: 3C(s)+2H2(g)→C3H4(g)3C(s) + 2H_2(g) → C_3H_4(g)
  2. Use the given enthalpy changes:
    • Formation of CO₂: ΔH = -394 kJ/mol
    • Formation of H₂O: ΔH = -286 kJ/mol
    • Combustion of C₃H₈: ΔH = -1939 kJ/mol
  3. Now summing these values:

Using values in the formula:

ΔH=(3×−394)+(2×−286)+1939ext(combustion)ΔH = (3 × -394) + (2 × -286) + 1939 ext{ (combustion)}

i.e.,

ΔH=−1182−572+1939ΔH = -1182 - 572 + 1939

= 185 kJ/mol.

Step 3

(i) Calculate the energy released when 1 kg of propyne is burned

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Answer

To find the energy released, we first convert 1 kg of propyne to moles:

  • Molar mass of C₃H₄ = 40 g/mol.

1 kg = 1000 g.

  • Number of moles = 1000 g / 40 g/mol = 25 moles.

Using the enthalpy of combustion, the energy released for 1 kg is:

extEnergyreleased=25extmoles×−1939extkJ/mol=−48475extkJ ext{Energy released} = 25 ext{ moles} × -1939 ext{ kJ/mol} = -48475 ext{ kJ}

Step 4

(ii) Calculate the mass of air required to burn 1 g of propyne

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Answer

First, we calculate the mass of oxygen needed:

  • From the combustion equation: C₃H₄(g) + 4O₂(g) → 3CO₂(g) + 2H₂O(l). This shows 4 moles of oxygen are needed to oxidize:

Mass of one mole of O₂ = 32 g/mol.

  • Therefore, to burn 1 g of C₃H₄, we need:
extMassofair=4.3×4extgO2=17.2extg(totalair) ext{Mass of air} = 4.3 × 4 ext{ g O₂} = 17.2 ext{ g (total air)}

Step 5

(iii) Suggest why methanol and ethanol require less air to burn

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Answer

Methanol (CH₃OH) and ethanol (C₂H₅OH) contain hydroxyl (-OH) groups, making the combustion reaction more efficient. These alcohols already contain oxygen in their structure, which reduces the amount of external oxygen needed for combustion compared to hydrocarbon fuels like ethane and propane. Thus, less air is required to achieve complete combustion.

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