Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1
Question 6
Consider the following information.
Ethanol burns in excess air according to the following equation.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
ΔH = −1364 kJ mol−1
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Worked Solution & Example Answer:Consider the following information - VCE - SSCE Chemistry - Question 6 - 2007 - Paper 1
Step 1
Calculate the minimum amount of energy, in kJ, required to heat 550 g of water and the pot from 18.5°C to 100.0°C.
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Answer
To calculate the energy required, we can use the formula:
Calculate the mass, in g, of ethanol that needs to be completely burnt to provide this energy.
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Answer
Using the enthalpy change for ethanol combustion:
ΔH=−1364kJ mol−1
Energy needed is approximately 1423.74 kJ.
Calculate moles of ethanol required:
Moles of ethanol, n=∣ΔH∣Q=13641423.74=1.04mol
Calculate mass of ethanol:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Mass = n×Molar Mass=1.04×46.07=47.94g.
Step 3
Only 35% of the energy released by the combustion of ethanol is transferred to the cooking pot and contents. Calculate the mass, in g, of ethanol that needs to be burnt in practice to heat the water and the pot from 18.5°C to 100.0°C.
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Answer
Calculate the effective energy used:
Effective energy = 0.35×1423.74=497.31kJ.
Calculate the moles of ethanol needed for this energy:
n=1364497.31=0.365mol.
Calculate the mass of ethanol:
Mass = 0.365×46.07=16.80g.
Step 4
Calculate the magnitude of ΔH, in kJ mol−1, for the reaction.
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Answer
Using the fact that 10.0 g of ethanol releases the same energy as 6.00 g of butane:
Calculate energy released by 10.0 g of ethanol:
ΔH=1364kJ mol−1 is the energy for 1 mol.
For 10.0 g: Moles of ethanol = 46.0710=0.217mol, then energy = 0.217×1364=296.9kJ.
Since 6.00 g of butane releases the same energy, we can set up:
ΔH for butane corresponds to this energy, confirmed through stoichiometry in the reaction.