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

Worked Solution & Example Answer:Methanol, CH3OH(l), 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

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

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Answer

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

E=VimesIimestE = V imes I imes t

Where:

  • VV = potential (volts) = 5.25 V
  • II = current (amperes) = 1.50 A
  • tt = time (seconds) = 3.00 minutes = 3.00 × 60 s = 180 s

Now, substituting the values into the formula:

E=5.25imes1.50imes180=1.42imes103extJE = 5.25 imes 1.50 imes 180 = 1.42 imes 10^3 ext{ J}

Next, we can find the calibration constant using the formula:

extCalorimeterconstant=EriangleT ext{Calorimeter constant} = \frac{E}{ riangle T}

Where:

  • EE = energy supplied = 1.42imes103extJ1.42 imes 10^3 ext{ J}
  • riangleT riangle T = temperature change = 0.593 °C

Now substituting these values:

extCalorimeterconstant=1.42imes1030.593extJ°C−1=2.39extkJ°C−1 ext{Calorimeter constant} = \frac{1.42 imes 10^3}{0.593} ext{ J °C}^{-1} \\ = 2.39 ext{ kJ °C}^{-1}

Step 2

ii. 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

To find ΔH for the combustion of methanol, we need to use the calorimeter constant we just calculated, along with the temperature change observed when combusting the methanol.

The energy released during the combustion is calculated using:

E=CimesriangleTE = C imes riangle T

Where:

  • CC = calorimeter constant = 2.39 kJ °C⁻¹
  • riangleT riangle T = temperature increase = 8.63 °C

Calculating energy:

E=2.39extkJ°C−1imes8.63ext°C=20.62extkJE = 2.39 ext{ kJ °C}^{-1} imes 8.63 ext{ °C} = 20.62 ext{ kJ}

Next, to calculate the molar heat of combustion, we need to divide the energy by the number of moles of methanol burned.

First, calculate the moles of methanol:

  • Molar mass of methanol (CH3OH) ≈ 32.04 g/mol

extMolesofmethanol=0.934extg32.04extg/mol=0.0292extmol ext{Moles of methanol} = \frac{0.934 ext{ g}}{32.04 ext{ g/mol}} = 0.0292 ext{ mol}

Now we calculate ΔH:

ΔH=Eextmoles=20.62extkJ0.0292extmol≈705.47extkJ/molΔH = \frac{E}{ ext{moles}} = \frac{20.62 ext{ kJ}}{0.0292 ext{ mol}} \\ ≈ 705.47 ext{ kJ/mol}

Thus, the value of ΔH for the combustion of methanol is approximately 705.47 kJ/mol.

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