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When discussing the combustion of ethanol, writing a balanced chemical equation is crucial. Ethanol reacts with oxygen to produce carbon dioxide and water. The chemical equation for this reaction is: \[ C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g) \] This equation shows that one molecule of ethanol reacts with three molecules of oxygen. In return, it forms two molecules of carbon dioxide and three molecules of water. Understanding how to balance chemical equations ensures that the number of atoms for each element is equal on both sides.

• Identify each reactant: ethanol and oxygen.
• Determine the products: carbon dioxide and water.
• Ensure the atoms on both sides are equal: two carbons, six hydrogens, and seven oxygens.

The standard enthalpy change of a reaction tells us about the energy exchange occurring during a chemical reaction. It is defined as the energy change when reactants transform into products under standard conditions. For combustion of ethanol, the equation used is: \[ \Delta H^\circ = \sum n \Delta H^\circ_f(\text{products}) - \sum m \Delta H^\circ_f(\text{reactants}) \] Where \( \Delta H^\circ_f \) represents standard enthalpies of formation, and \( n \) and \( m \) are the stoichiometric coefficients. The provided values were:

• \( \Delta H^\circ_f(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -277.7 \text{kJ/mol} \)
• \( \Delta H^\circ_f(\mathrm{O}_2) = 0 \text{kJ/mol} \) (as it is a natural state element).
• \( \Delta H^\circ_f(\mathrm{CO}_2) = -393.5 \text{kJ/mol} \)
• \( \Delta H^\circ_f(\mathrm{H}_2\mathrm{O}) = -241.8 \text{kJ/mol} \)

Substituting these, the calculated standard enthalpy change was: \[ \Delta H^\circ = -1366.9 \text{kJ/mol} \] This negative value indicates that the reaction releases energy; it's exothermic.

Heat produced in a chemical reaction is a fundamental part of understanding energy flows. To calculate the heat produced by burning ethanol, we start by considering its density, which is 0.789 g/mL. For one liter of ethanol, the mass is: \[ \text{Mass} = 0.789 \text{g/mL} \times 1000 \text{mL} = 789 \text{g} \] Then calculate the moles of ethanol using its molar mass (46.08 g/mol): \[ \text{Moles of ethanol} = \frac{789 \text{g}}{46.08 \text{g/mol}} \approx 17.13 \text{mol} \] Now, using the standard enthalpy change determined earlier: \[ \text{Heat produced} = 17.13 \text{mol} \times (-1366.9 \text{kJ/mol}) \approx -23414 \text{kJ/liter} \] This calculation shows the energy released when one liter of ethanol is combusted. It's vital for applications like fuel efficiency understanding.

Calculating carbon dioxide emissions from the combustion of ethanol helps us understand environmental impacts. Since the reaction produces CO2, it's important to quantify this. From the balanced equation, we know 2 moles of CO2 are produced per mole of ethanol. Hence, for our 17.13 moles of ethanol: \[ \text{Moles of CO}_2 = 17.13 \text{mol} \times 2 = 34.26 \text{mol} \] Then, convert the moles to the mass of CO2 using its molar mass (44.01 g/mol): \[ \text{Mass of CO}_2 = 34.26 \text{mol} \times 44.01 \text{g/mol} \approx 1507.5 \text{g} \] Finally, we want to find the mass per kJ of heat emitted: \[ \frac{\text{Mass of CO}_2}{\text{kJ of heat}} = \frac{1507.5 \text{g}}{23414 \text{kJ}} \approx 0.064 \text{g CO}_2/\text{kJ} \] Understanding these emissions per energy unit is essential for assessing the environmental efficiency of fuels.