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Methanol, CH₃OH, undergoes combustion according to the equation 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g) In an experiment to determine its suitability as a fuel, a sample of methanol underwent complete oxidation in a bomb calorimeter - VCE - SSCE Chemistry - Question 6 - 2012 - Paper 1

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Methanol,-CH₃OH,-undergoes-combustion-according-to-the-equation--2CH₃OH(l)-+-3O₂(g)-→-2CO₂(g)-+-4H₂O(g)--In-an-experiment-to-determine-its-suitability-as-a-fuel,-a-sample-of-methanol-underwent-complete-oxidation-in-a-bomb-calorimeter-VCE-SSCE Chemistry-Question 6-2012-Paper 1.png

Methanol, CH₃OH, undergoes combustion according to the equation 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g) In an experiment to determine its suitability as a fuel, a s... show full transcript

Worked Solution & Example Answer:Methanol, CH₃OH, undergoes combustion according to the equation 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g) In an experiment to determine its suitability as a fuel, a sample of methanol underwent complete oxidation in a bomb calorimeter - VCE - SSCE Chemistry - Question 6 - 2012 - Paper 1

Step 1

Determine the calibration constant, in kJ °C⁻¹, for the calorimeter and its contents.

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Answer

To find the calibration constant of the calorimeter, we will first calculate the energy supplied to the system using the formula:

E=VimesIimestE = V imes I imes t

Where:

  • EE is the energy in joules (J)
  • VV is the voltage applied (5.25 V)
  • II is the current (1.50 A)
  • tt is the time in seconds (3.00 minutes = 180 seconds)

Substituting in the values: E=5.25imes1.50imes180=1425extJE = 5.25 imes 1.50 imes 180 = 1425 ext{ J}

Next, we will calculate the calibration constant using the temperature change (ΔT = 0.593 °C):

ext{Calibration Constant} = rac{E}{ ext{ΔT}}

ewline ext{Calibration Constant} = 2390.39 ext{ J °C}^{-1} = 2.39 ext{ kJ °C}^{-1}$$ Thus, the calibration constant is 2.39 kJ °C⁻¹.

Step 2

Use this experimental data to determine the value of ΔH for the combustion of methanol given by the following equation.

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Answer

We will use the calibration constant and the temperature change for the combustion of methanol to find ΔH.

From the previous calculation, we found the calibration constant to be 2.39 kJ °C⁻¹. Given that the temperature of the water increased by 8.63 °C, we can calculate the energy released during the combustion:

Q=extCalibrationConstantimesextΔTQ = ext{Calibration Constant} imes ext{ΔT}

Substituting the values gives: Q=2.39extkJ°C−1imes8.63ext°C=20.6377extkJQ = 2.39 ext{ kJ °C}^{-1} imes 8.63 ext{ °C} = 20.6377 ext{ kJ}

To find ΔH, note that since 0.934 g of methanol has been used, we should calculate the molar heat of combustion. The molar mass of methanol (CH₃OH) is approximately 32.04 g/mol, so:

  1. Calculate the number of moles of methanol burned:
ightarrow n ext{ (moles)} ≈ 0.0292 ext{ mol}$$ 2. Then, the molar enthalpy change (ΔH) for combustion is given by: $$ ext{ΔH} = rac{Q}{n} = rac{20.6377 ext{ kJ}}{0.0292 ext{ mol}} ≈ 706.1 ext{ kJ/mol}$$ Thus, the value of ΔH for the combustion of methanol is approximately 706.1 kJ/mol.

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