Question

Octane gas \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) at \(25^{\circ} \mathrm{C}\) is burned steadily with 80 percent excess air at \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and 40 percent relative humidity. Assuming combustion is complete and the products leave the combustion chamber at \(1000 \mathrm{K}\), determine the heat transfer for this process per unit mass of octane.

Short answer

The heat transfer per unit mass of octane during the combustion process is -106.52 kJ/g.

Step-by-step solution

Write the balanced chemical equation

For the complete combustion of octane, we need to include the fuel (C8H18), oxygen (O2), and nitrogen (N2) in the reactants side. Since combustion products mostly consist of carbon dioxide (CO2) and water vapor (H2O), we will include them in the products side, along with excess O2 and N2. The balanced chemical equation is: C8H18 + (12.5 + 0.8 * 12.5) * (O2 + 3.76N2) → 8CO2 + 9H2O + excess O2 + excess N2 Where 12.5 is the stoichiometric quotient for the combustion of octane.

Determine the reactants' and products' enthalpies

In this step, we need to determine the enthalpies of the reactants and products. The enthalpy values at different temperatures can be found in standard thermodynamic tables. At 25°C, note the enthalpies for each component: C8H18: \(h_{C8H18} = -249.94\,\text{kJ/mol}\) O2: \(h_{O2} = 0\, \text{kJ/mol}\) N2: \(h_{N2} = 0\, \text{kJ/mol}\) Since the combustion products leave the combustion chamber at 1000 K, we will consider the enthalpies of the products at 1000 K: 8CO2: \(h_{8CO2} = 8 \times 867.9\, \text{kJ/mol}\) 9H2O: \(h_{9H2O} = 9 \times 990.4\, \text{kJ/mol}\) excess O2: \(h_{excess O2} = 0.8 \times 12.5 \times 867.9\, \text{kJ/mol}\) excess N2: \(h_{excess N2} = 0.8 \times 12.5 \times 3.76 \times 866.2\, \text{kJ/mol}\)

Apply the energy conservation equation

We can apply the energy conservation equation for the combustion process as follows: \(\sum m_{reactants} h_{reactants} - \sum m_{products} h_{products} = \text{heat transfer}\) Plugging in the values, we get: $$((-249.94) + 12.5(0) + 0.8 \times 12.5 \times 3.76(0)) - (8 \times 867.9 + 9 \times 990.4 + 0.8 \times 12.5 \times 867.9 + 0.8 \times 12.5 \times 3.76 \times 866.2) = \text{heat transfer}$$ After calculating, heat transfer = -12168.15 kJ/mol Now, we need to find the heat transfer per unit mass of octane. The molar mass of octane (C8H18) is 114.22 g/mol.

Calculate heat transfer per unit mass of octane

To find the heat transfer per unit mass of octane, divide the total heat transfer (in kJ/mol) by the molar mass of octane (in g/mol): $$\frac{\text{heat transfer}}{\text{mass of octane}} = \frac{-12168.15\, \text{kJ/mol}}{114.22\, \text{g/mol}} = -106.52\, \text{kJ/g}$$ So, the heat transfer for this combustion process is -106.52 kJ/g of octane.