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For the unbalanced combustion reaction shown, \(1 \mathrm{~mol}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) releases \(1370 \mathrm{~kJ}:\) $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ (a) Write a balanced equation for the combustion reaction. (b) What is the sign of \(\Delta H\) for this reaction? (c) How much heat (in kilocalories) is released from the combustion of \(5.00 \mathrm{~g}\) of ethanol? (d) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) must be burned to raise the temperature of \(500.0 \mathrm{~mL}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) (The specific heat of water is \(4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) See Section 1.11.) (e) If the density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL},\) calculate the combustion energy of ethanol in kilojoules/milliliter.

Step by step solution

01

Balance the Combustion Reaction

To balance the combustion reaction of ethanol, start by writing the unbalanced equation:\( \text{C}_2 \text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \).Now, balance carbon atoms first:\( \text{C}_2 \text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \).Then, balance hydrogen:\( \text{C}_2 \text{H}_5\text{OH} + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).Finally, balance oxygen:\( \text{C}_2 \text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).So the balanced equation is:\( \text{C}_2 \text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \).

02

Identify the Sign of ΔH

Since the reaction releases 1370 kJ of energy per mole of ethanol, the reaction is exothermic. Thus, the sign of \( \Delta H \) is negative: \( \Delta H = -1370 \, \text{kJ/mol} \).

03

Calculate Heat Released for 5.00 g Ethanol

First, calculate the number of moles in 5.00 g of ethanol.Ethanol \( \text{C}_2 \text{H}_5\text{OH} \) has a molar mass of 46.08 g/mol.\[ \text{Moles of Ethanol} = \frac{5.00 \text{ g}}{46.08 \text{ g/mol}} = 0.1085 \text{ mol} \]Heat released:\[ \text{Heat} = 0.1085 \text{ mol} \times (-1370 \text{ kJ/mol}) = -148.64 \text{ kJ} \]Convert kJ to kcal (1 kcal = 4.184 kJ):\[ \text{Heat} = \frac{-148.64 \text{ kJ}}{4.184 \text{ kJ/kcal}} = -35.52 \text{ kcal} \]

04

Calculate Mass to Heat Water

Calculate the energy required to heat 500.0 mL \(\text{H}_2\text{O}\) from 20°C to 100°C:1mL water = 1g\( \text{q} = m \times c \times \Delta T \)where \( m = 500 \text{ g} \), \( c = 4.184 \text{ J/g°C} \), \( \Delta T = 80 \text{°C} \).\[ \text{q} = 500 \text{ g} \times 4.184 \text{ J/g°C} \times 80 °C = 167360 \text{ J} = 167.36 \text{ kJ} \]Calculate mass of ethanol:Number of moles to provide \( 167.36 \text{ kJ} \):\[ \text{Moles of Ethanol} = \frac{167.36 \text{ kJ}}{1370 \text{ kJ/mol}} = 0.1221 \text{ mol} \]Mass:\[ \text{Mass} = 0.1221 \text{ mol} \times 46.08 \text{ g/mol} = 5.62 \text{ g} \]

05

Calculate Combustion Energy in kJ/mL

Given density \( 0.789 \text{ g/mL} \), find the kJ released per mL ethanol.Molar volume of ethanol:\[ \text{Volume per mole} = \frac{46.08 \text{ g/mol}}{0.789 \text{ g/mL}} = 58.41 \text{ mL/mol} \]Energy per mL:\[ \text{Energy} = \frac{1370 \text{ kJ/mol}}{58.41 \text{ mL/mol}} = 23.46 \text{ kJ/mL} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reaction

A combustion reaction involves a substance combining with oxygen to produce heat and light, typically resulting in the formation of water and carbon dioxide. In the specific case of ethanol combustion, the unbalanced equation is: \( \text{C}_2 \text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \). To balance this equation, we first ensure that the number of each type of atom is equal on both sides of the equation. This involves:

• Balancing the carbon atoms: From ethanol to carbon dioxide.
• Balancing the hydrogen atoms: From ethanol to water.
• Balancing the oxygen atoms: Ensuring adequate oxygen molecules are present.

The balanced combustion equation for ethanol, therefore, is: \( \text{C}_2 \text{H}_5\text{OH} + 3 \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \), indicating that each molecule of ethanol requires three molecules of oxygen to fully combust.

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), is a measure of heat absorbed or released during a chemical reaction at constant pressure. In an exothermic reaction, such as the combustion of ethanol, heat is released to the surroundings, resulting in a negative \( \Delta H \). For ethanol, the enthalpy change \( \Delta H = -1370 \, \text{kJ/mol} \) indicates that 1370 kilojoules are released per mole of ethanol combusted.Exothermic reactions like this are crucial in applications where heat generation is beneficial, such as heating systems and energy generation. The negative sign of \( \Delta H \) signifies that energy flows from the system to the surroundings, validating the energy-releasing characteristic of combustion reactions.

Heat Calculation

Calculating the heat involved in a reaction involves understanding the relationship between mass, concentration, and energy release. For ethanol, given information such as its molar mass (46.08 g/mol) and the known \( \Delta H \), we can determine heat for different amounts and forms of ethanol.To find the heat released by 5.00 grams of ethanol:

• Calculate the number of moles using the molar mass: \( \frac{5.00 \, \text{g}}{46.08 \, \text{g/mol}} = 0.1085 \, \text{mol} \).
• Use the \( \Delta H \) to find heat: \( 0.1085 \, \text{mol} \times (-1370 \, \text{kJ/mol}) = -148.64 \, \text{kJ} \).
• Convert to kilocalories: \( \frac{-148.64 \, \text{kJ}}{4.184 \, \text{kJ/kcal}} = -35.52 \, \text{kcal} \)

This method showcases how manual calculations allow us to understand the energy implications in chemical reactions.

Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius. For water, it is \( 4.184 \, \text{J/g°C} \). This property is vital when determining how much energy is necessary to change the temperature of a substance.For example, to calculate the energy needed to raise the temperature of 500.0 mL (or 500 g) of water from 20°C to 100°C:

• The temperature change (\( \Delta T \)) is 80°C.
• Use the formula \( q = m \times c \times \Delta T \), where \( m = 500 \, \text{g} \), \( c = 4.184 \, \text{J/g°C} \), and \( \Delta T = 80 \text{°C} \).
• Thus, \( q = 500 \, \text{g} \times 4.184 \, \text{J/g°C} \times 80 \, \text{°C} = 167360 \, \text{J} = 167.36 \, \text{kJ} \).

Understanding specific heat capacity helps connect the thermal properties of water with the energy released during ethanol's combustion, showing how substances interact thermally with their environments.