Problem 16 - General Chemistry: Principles and Modern Applications

Question

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

Step-by-Step Solution

Answer

\[ \mathrm{C}_{3} \mathrm{H}_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l), \Delta H = -29.4 \mathrm{kJ}\] \[ \mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(s) + 15O_{2}(g) \rightarrow 10CO_{2}(g) + 8H_{2}O(l) , \Delta H = -5.27 \mathrm{kJ}\] \[ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(l) + 4O_{2}(g) \rightarrow 3CO_{2}(g) + 3H_{2}O(l) , \Delta H = -58.3 \mathrm{kJ}\]

Step 1: Combustion of propane

Balance the following equation: \(\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l)\) and append the enthalpy change: \(\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l) \quad \Delta H = -29.4 \mathrm{kJ}\)

Step 2: Combustion of camphor

Balance the equation: \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(s) + 15O_{2}(g) \rightarrow 10CO_{2}(g) + 8H_{2}O(l)\) and append the enthalpy change: \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(s) + 15O_{2}(g) \rightarrow 10CO_{2}(g) + 8H_{2}O(l) \quad \Delta H = -5.27 \mathrm{kJ}\)

Step 3: Combustion of acetone

Balance the equation: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(l) + 4O_{2}(g) \rightarrow 3CO_{2}(g) + 3H_{2}O(l)\) and append the enthalpy change: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(l) + 4O_{2}(g) \rightarrow 3CO_{2}(g) + 3H_{2}O(l) \quad \Delta H = -58.3 \mathrm{kJ}\)

Key Concepts

Enthalpy Change

In chemical reactions, the enthalpy change (\( \Delta H \)) plays a crucial role in understanding the energy dynamics involved. Enthalpy is a measure of total energy within a system and is observed during reactions when a change occurs. If the system releases heat, the reaction is exothermic, and \( \Delta H \) is negative. Conversely, if the system absorbs heat, the reaction is endothermic, and \( \Delta H \) is positive.
This concept is essential in combustion reactions, where substances react with oxygen, producing energy—usually in the form of heat—and other byproducts like carbon dioxide and water. When analyzing reactions like those of propane, camphor, or acetone, the measure of \( \Delta H \) provides insight into the energy transfer within the system.
The listed enthalpy changes for each are negative, indicating that these combustion reactions release heat. Understanding this not only tells us about the energy produced but also helps in comparing the efficiency of different fuels.

Balanced Chemical Equations

A balanced chemical equation is a fundamental aspect of chemistry that ensures mass conservation in reactions. This means the number of atoms for each element on the reactant side should equal the number on the product side.
The combustion equations for substances like propane (\( \mathrm{C}_3\mathrm{H}_8 \)), camphor (\( \mathrm{C}_{10}\mathrm{H}_{16}\mathrm{O} \)), or acetone (\( \left(\mathrm{CH}_3\right)_2 \mathrm{CO} \)) all demonstrate this principle.

• Propane combustion: \( \mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \)
• Camphor combustion: \( \mathrm{C}_{10}\mathrm{H}_{16}\mathrm{O} + 15\mathrm{O}_2 \rightarrow 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \)
• Acetone combustion: \( \left(\mathrm{CH}_3\right)_2 \mathrm{CO} + 4\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \)

Balancing involves proper counting and adjustment of the coefficients (numbers placed before formulas) to maintain equal quantities of elements. This not only respects the laws of chemistry but also provides a foundation for calculating other important measures, such as the enthalpy change.

Thermochemistry

Thermochemistry is a branch of chemistry that focuses on the energy and heat associated with chemical reactions. Understanding thermochemical principles is pivotal when examining the energy changes in reactions like combustion.

• In combustion reactions, a fuel reacts with oxygen to produce energy, often in the form of heat and light. This is where thermochemistry overlaps with real-world applications, such as energy production and fuel efficiency.
• Using concepts like enthalpy (\( \Delta H \)), thermochemistry enables us to quantify the energy released or absorbed during reactions.

Thermochemistry is invaluable in everyday life, influencing how we choose fuels, predict the outcomes of mixing substances, and design energy solutions for industries. It helps predict product stability and understand reactions at both theoretical and practical levels. By examining exercises that involve combustion and enthalpy changes, learners can gain insights into these practical applications and the broader impact of chemical reactions on energy use and efficiency.