Problem 79 - Chemistry The Central Science In Si Units

Question

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 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

1. Combustion: \( \mathrm{C}_2\mathrm{H}_5\mathrm{OH(l)} + 3\mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g) + 3\mathrm{H}_2O(g) \) 2. \( \Delta H^0 = -1234.8 \, \text{kJ/mol} \) 3. Heat per liter: -21135 kJ 4. Mass of \( CO_2 \) per kJ: 0.0713 g/kJ.

Step 1: Write the Balanced Chemical Equation

To write the balanced equation for the combustion of ethanol, identify the reactants and products. Ethanol (C_2H_5OH) combusts in air (oxygen, O_2) to form carbon dioxide (CO_2) and water vapor (H_2O(g)). The balanced equation is:\[ \mathrm{C}_2\mathrm{H}_5\mathrm{OH(l)} + 3\mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g) + 3\mathrm{H}_2O(g) \]

Step 2: Calculate Standard Enthalpy Change

Find the standard enthalpy of formation (\Delta H_f^) for each substance. Using these, calculate the standard enthalpy change (\Delta H^0) using the equation:\[ \Delta H^0 = \sum \Delta H_f^0(\text{products}) - \sum \Delta H_f^0(\text{reactants}) \]Using standard tables:- \Delta H_f^0 of C_2H_5OH(l) = -277.38 kJ/mol- \Delta H_f^0 of CO_2(g) = -393.5 kJ/mol- \Delta H_f^0 of H_2O(g) = -241.8 kJ/molCalculate:\[ \Delta H^0 = [2(-393.5) + 3(-241.8)] - [-277.38 + 0] = -1234.8 \, \text{kJ/mol} \]

Step 3: Calculate Heat Produced Per Liter of Ethanol

First, convert the density of ethanol to match 1 liter:1 L = 1000 mL, so the mass of 1 L ethanol = 1000 \, \text{mL} \times 0.789 \, \text{g/mL} = 789 \, \text{g}Calculate moles of ethanol using its molar mass (46.07 g/mol):\[ \text{Moles of ethanol} = \frac{789 \, \text{g}}{46.07 \, \text{g/mol}} = 17.12 \, \text{mol} \]Using the enthalpy change per mole from Step 2:\[ \text{Heat produced} = 17.12 \, \text{mol} \times (-1234.8 \, \text{kJ/mol}) = -21135 \, \text{kJ} \]

Step 4: Calculate Mass of CO₂ Produced Per kJ of Heat Emitted

Determine the total CO₂ produced per reaction (2 moles for every mole of ethanol). Therefore, mass of CO_2 per mole of ethanol:\[ \text{Mass of } CO_2 = 2 \, \text{mol} \times 44.01 \, \text{g/mol} = 88.02 \, \text{g} \]Find total mass of CO_2 from Step 3's molar amount:\[ 17.12 \, \text{mol ethanol} \times 88.02 \, \text{g/mol} = 1507.7 \, \text{g} \]Calculate mass of CO_2 per kJ:\[ \text{Mass of } CO_2 \, \text{per kJ} = \frac{1507.7 \, \text{g}}{21135 \, \text{kJ}} = 0.0713 \, \text{g/kJ} \]

Key Concepts

Balanced Chemical Equations

Understanding a balanced chemical equation is like fitting puzzle pieces together. In chemistry, every element needs to be accounted for on both sides of the equation. This means the number of atoms of each element must remain equal before and after the reaction. When ethanol (\(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\)) combusts, it reacts with oxygen (\(\mathrm{O}_2\)) in the air to produce carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).When writing the balanced equation for the combustion of ethanol, the equation initially might look simple with ethanol combining with oxygen. However, to balance it:

• Start with ethanol, the fuel, which has \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\), indicating 2 carbon and 6 hydrogen atoms.
• On the product side, 2 \(\mathrm{CO}_2\) molecules will balance the carbon atoms.
• Similarly, 3 \(\mathrm{H}_2\mathrm{O}\) molecules will balance the hydrogen atoms.
• Finally, ensure oxygen atoms are balanced by adding 3 \(\mathrm{O}_2\) on the reactant side.

The balanced equation reads as:\[\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\mathrm{(l)} + 3\mathrm{O}_2\mathrm{(g)} \rightarrow 2\mathrm{CO}_2\mathrm{(g)} + 3\mathrm{H}_2O\mathrm{(g)}\] Understanding balanced equations is crucial for accurately modeling combustion processes, including estimating emissions and energy production.

Standard Enthalpy Change

Standard enthalpy change is a pivotal concept when understanding chemical reactions because it helps us know the energy changes involved. Essentially, it measures the heat released or absorbed during a chemical reaction at constant pressure.To find the standard enthalpy change (\(\Delta H^0\)) for the combustion of ethanol, you must use the standard heats of formation (\(\Delta H_f^0\)) of all reactants and products:

• The sum of the heat of formation for products (\(\mathrm{2CO}_2(g)\) and \(\mathrm{3H}_2O(g)\)) is calculated first.
• The sum of the heat of formation for reactants (\(\mathrm{C}_2\mathrm{H}_5\mathrm{OH(l)}\) and \(\mathrm{O}_2(g)\)) follows.

We apply the formula:\[\Delta H^0 = \sum \Delta H_f^0(\text{products}) - \sum \Delta H_f^0(\text{reactants})\]Plugging in the values:\[\Delta H^0 = [2(-393.5) + 3(-241.8)] - [-277.38 + 0] = -1234.8\, \text{kJ/mol}\]This negative enthalpy value indicates that the reaction releases heat, making it an exothermic process. Understanding these energy changes is key to evaluating the efficiency and environmental impact of fuel combustion.

Density of Ethanol

The density of a substance is its mass per unit volume. For ethanol, the density is given as \(0.789\,\text{g/mL}\), which is crucial when measuring energy output related to volume.Let's convert the density into practical terms to understand quantities:For 1 liter of ethanol:

• Convert the volume to milliliters: \(1\,\text{L} = 1000\,\text{mL}\)
• Calculate the mass using density: \(0.789\,\text{g/mL} \times 1000\,\text{mL} = 789\,\text{g}\)

Knowing how much each liter weighs allows us to calculate how much energy can be released per liter. This is critical for applications in fuel usage and comparison with other energy sources. Understanding the density helps to bridge the gap from theoretical calculations to practical, usable figures.

Molar Mass Calculations

Calculating molar mass is an essential step in chemical equations and energy calculations. The molar mass of a substance is the mass of one mole of its particles, measured in grams per mole (g/mol).Ethanol (\(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\)) has a molar mass calculated as:

• Carbon: 2 atoms \(\times 12.01\,\mathrm{g/mol} = 24.02\,\mathrm{g/mol}\)
• Hydrogen: 6 atoms \(\times 1.008\,\mathrm{g/mol} = 6.048\,\mathrm{g/mol}\)
• Oxygen: 1 atom \(\times 16.00\,\mathrm{g/mol} = 16.00\,\mathrm{g/mol}\)

Adding these gives the molar mass of ethanol:\[\text{Total: } 24.02 + 6.048 + 16.00 = 46.07\,\mathrm{g/mol}\]Molar mass is essential for converting mass into moles, which are used in calculating energy changes and reactant-product relationships. It's the first step to transitioning from theoretical equations to practical chemical applications, like determining how much ethanol can produce certain amounts of heat.