Problem 79 - Chemistry: The Central Science

Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is currently blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

Step-by-Step Solution

Answer

The balanced equation for the combustion of ethanol in air is: \[C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\] The standard enthalpy change for this reaction is \(-1367.3 \ \mathrm{kJ/mol}\). The heat produced per liter of ethanol by combustion under constant pressure is \(-23419 \ \mathrm{kJ/L}\). The mass of CO₂ produced per kJ of heat emitted is \(-0.0327 \ \mathrm{g/kJ}\).

Step 1: (a) Balanced equation for combustion of ethanol

For the combustion of ethanol, we will need oxygen from the air. The products of ethanol's combustion are carbon dioxide (CO2) and water (H2O). The balanced equation for the combustion of ethanol can be written as: \[C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\]

Step 2: (b) Calculate the standard enthalpy change

To find the standard enthalpy change for the reaction, we can use the heat of formation values of the reactants and the products. The standard heat of formation (ΔHf°) values are: Ethanol: \(-277.7 \ \mathrm{kJ/mol}\) Oxygen: \(0 \ \mathrm{kJ/mol}\) Carbon dioxide: \(-393.5 \ \mathrm{kJ/mol}\) Water (gas): \(-241.8 \ \mathrm{kJ/mol}\) Using the standard enthalpy change formula, we have: ΔH° = [Sum of (ΔHf° of products)] - [Sum of (ΔHf° of reactants)] ΔH° = [2(-393.5) + 3(-241.8)] - [(-277.7) + 3(0)] ΔH° = \(-1367.3 \ \mathrm{kJ/mol}\)

Step 3: (c) Calculate the heat produced per liter of ethanol

We are given the density of ethanol as \(0.789 \ \mathrm{g/mL}\). To find the heat produced per liter of ethanol, we first need to convert the density to grams per liter, then use the molar mass of ethanol and the standard enthalpy change to find the heat produced. 1: Convert the density of ethanol to grams per liter: Density = \(\frac{0.789 \ \mathrm{g} }{\mathrm{mL}} × 1000 \ \mathrm{mL/L} = 789 \ \mathrm{g/L}\) 2: Use the molar mass of ethanol (C: 12.01g/mol, H: 1.01g/mol, and O: 16.00g/mol) to find moles per liter of ethanol: Moles per liter = \(\frac{789 \ \mathrm{g/L}}{2(12.01) + 6(1.01) + 1(16.00) \ \mathrm{g/mol}} = \frac{789}{46.07} = 17.13 \ \mathrm{mol/L}\) 3: Use the standard enthalpy change and moles per liter to find the heat produced per liter of ethanol: Heat produced = \(\frac{-1367.3 \ \mathrm{kJ/mol}}{1 \ \mathrm{mol}} × 17.13 \ \mathrm{mol/L} = -23419 \ \mathrm{kJ/L}\)

Step 4: (d) Calculate the mass of CO2 produced per kJ of heat emitted

To find the mass of CO2 produced per kJ of heat emitted, we can use the stoichiometry of the balanced equation (2 moles of CO2 produced per mole of ethanol) and the molar mass of CO2. 1: Calculate moles of CO2 produced per kJ of heat emitted: Moles of CO2 = \(\frac{2 \ \mathrm{mol} \ CO_2}{1 \ \mathrm{mol} \ C_{2}H_{5}OH} × -\frac{1 \ \mathrm{mol} \ C_{2}H_{5}OH}{1367.3 \ \mathrm{kJ}} = -\frac{2}{1367.3} \ \mathrm{mol/kJ}\) 2: Calculate the mass of CO2 produced per kJ of heat emitted using the molar mass of CO2 (C: 12.01g/mol, O: 16.00g/mol): Mass of CO2 per kJ = \(-\frac{2}{1367.3} \ \mathrm{mol/kJ} × (12.01 + 2(16.00)) \ \mathrm{g/mol} = -0.0327 \ \mathrm{g/kJ}\)