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Coke, which is essentially pure carbon, is widely used as a fuel - VCE - SSCE Chemistry - Question 3 - 2005 - Paper 1

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Coke, which is essentially pure carbon, is widely used as a fuel. Its complete combustion can be represented by the following equation. $$C(s) + O_2(g) \rightarrow ... show full transcript

Worked Solution & Example Answer:Coke, which is essentially pure carbon, is widely used as a fuel - VCE - SSCE Chemistry - Question 3 - 2005 - Paper 1

Step 1

Calculate the energy for 80% oxidation to CO2

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Answer

First, we convert the mass of coke to moles:

2.00 tonne=2.00×106 g2.00 \text{ tonne} = 2.00 \times 10^6 \text{ g}

Using the molar mass of carbon, which is approximately 12 g/mol:

n=massmolar mass=2.00×106 g12 g/mol=1.67×105 moln = \frac{mass}{molar \, mass} = \frac{2.00 \times 10^6 \, g}{12 \, g/mol} = 1.67 \times 10^5 \, mol

Now, we find the moles for the 80% oxidised to CO2:

n(CO2)=0.80×1.67×105 mol=1.33×105 moln(CO_2) = 0.80 \times 1.67 \times 10^5 \, mol = 1.33 \times 10^5 \, mol

Using the enthalpy change for the complete combustion:

ΔHCO2=−393 kJ/mol\Delta H_{CO_2} = -393 \, kJ/mol

The energy released is:

EnergyCO2=1.33×105 mol×(−393 kJ/mol)=−5.24×107 kJEnergy_{CO_2} = 1.33 \times 10^5 \, mol \times (-393 \, kJ/mol) = -5.24 \times 10^7 \, kJ

Step 2

Calculate the energy for 20% oxidation to CO

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Answer

For the 20% that is oxidised to CO:

n(CO)=0.20×1.67×105 mol=0.33×105 moln(CO) = 0.20 \times 1.67 \times 10^5 \, mol = 0.33 \times 10^5 \, mol

Using the enthalpy change for the incomplete combustion:

ΔHCO=−232 kJ/mol\Delta H_{CO} = -232 \, kJ/mol

The energy released is:

EnergyCO=0.33×105 mol×(−232 kJ/mol)=−0.76×107 kJEnergy_{CO} = 0.33 \times 10^5 \, mol \times (-232 \, kJ/mol) = -0.76 \times 10^7 \, kJ

Step 3

Total energy calculation

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Answer

Total energy released when coal burns:

Total Energy=EnergyCO2+EnergyCOTotal \: Energy = Energy_{CO_2} + Energy_{CO}

Substituting previously calculated values:

Total Energy=−5.24×107 kJ+(−0.76)×107 kJ=−6.00×107 kJTotal \: Energy = -5.24 \times 10^7 \, kJ + (-0.76) \times 10^7 \, kJ = -6.00 \times 10^7 \, kJ

Thus, the total energy released is approximately −5.63×107-5.63 \times 10^7 kJ.

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