Question

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest-burning octanes is a compound called \(2,3,4\) -trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) leads to \(\Delta H^{\circ}=-5064.9 \mathrm{~kJ} / \mathrm{mol}\). (a) Write a balanced equation for the combustion of 1 mol of \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .(\mathrm{b})\) Write a balanced equation for the formation of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3\), calculate \(\Delta H_{f}^{\circ}\) for \(2,3,4\) -trimethylpentane.

Short answer

a) The balanced combustion equation for 1 mol of 2,3,4-trimethylpentane (C8H18) is: \(\text{C}_8\text{H}_{18}(\text{l}) + \frac{25}{2} \text{O}_2(\text{g}) \rightarrow 8 \text{CO}_2(\text{g}) + 9\text{H}_2\text{O}(\text{g})\) b) The balanced formation equation for 2,3,4-trimethylpentane from its elements is: \(8 \text{C}(\text{s, graphite}) + 9\text{H}_2(\text{g}) \rightarrow \text{C}_8\text{H}_{18}(\text{l})\) c) Using Hess's Law and the given data, the standard enthalpy of formation for 2,3,4-trimethylpentane is: ΔHf° [C8H18(l)] = -3150.4 kJ/mol

Step-by-step solution

Balanced combustion equation

Let's first write the balanced combustion equation for 1 mol of 2,3,4-trimethylpentane (C8H18). Combustion of hydrocarbons involves reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). \(\text{C}_8\text{H}_{18}(\text{l}) + \frac{25}{2} \text{O}_2(\text{g}) \rightarrow 8 \text{CO}_2(\text{g}) + 9\text{H}_2\text{O}(\text{g})\)

Balanced formation equation

Now, write a balanced equation for the formation of 2,3,4-trimethylpentane from its elements (carbon and hydrogen) in their standard states. \(8 \text{C}(\text{s, graphite}) + 9\text{H}_2(\text{g}) \rightarrow \text{C}_8\text{H}_{18}(\text{l})\)

Calculate ΔHf° using Hess's Law

To calculate ΔHf° for 2,3,4-trimethylpentane, we will use Hess's Law. We are given the heat of combustion for C8H18 (-5064.9 kJ/mol). We also need the standard enthalpy of formation for CO2(g) and H2O(g). These can be found in Table 5.3. From Table 5.3, we find: ΔHf° [CO2(g)] = -393.5 kJ/mol ΔHf° [H2O(g)] = -241.8 kJ/mol Now, we can use Hess's law to find ΔHf° [C8H18(l)]: ΔHf° [C8H18(l)] + (-5064.9 kJ/mol) = 8 * ΔHf° [CO2(g)] + 9 * ΔHf° [H2O(g)] ΔHf° [C8H18(l)] + (-5064.9 kJ/mol) = 8 * (-393.5 kJ/mol) + 9 * (-241.8 kJ/mol) Solve for ΔHf° [C8H18(l)]: ΔHf° [C8H18(l)] = -3150.4 kJ/mol So, the standard enthalpy of formation for 2,3,4-trimethylpentane is -3150.4 kJ/mol.

Key concepts

Combustion of Hydrocarbons

Understanding the combustion of hydrocarbons is crucial for students who are studying chemical reactions, energy production, and environmental science. Hydrocarbons are compounds consisting solely of carbon (C) and hydrogen (H) atoms. When they combust, or burn in the presence of oxygen (O₂), they form carbon dioxide (CO₂) and water (H₂O), releasing energy in the form of heat.

For example, the combustion of octane, a typical hydrocarbon found in gasoline, can be depicted through a chemical equation that must be balanced to reflect the law of conservation of matter. This law states that atoms are neither created nor destroyed in a chemical reaction. Therefore, the number of atoms of each element in the reactants must equal the number of atoms in the products.

A balanced equation ensures that this law is upheld. For instance, the balanced combustion equation for octane is written as:
\[ \text{{C}}_8\text{{H}}_{18}(\text{{l}}) + \frac{{25}}{{2}} \text{{O}}_2(\text{{g}}) \rightarrow 8 \text{{CO}}_2(\text{{g}}) + 9\text{{H}}_2\text{{O}}(\text{{g}}) \]
Each side of the equation contains 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms. Balancing combustion equations provides a clear picture of the stoichiometry and is essential for calculating the energy released (enthalpy change) in such reactions.

Hess's Law

Hess's Law is a powerful tool in thermochemistry that allows us to calculate the enthalpy change (\(\Delta H\)) of a reaction even when it cannot be measured directly. The fundamental assertion of Hess's Law is that the total enthalpy change for a chemical reaction is the same, regardless of the pathway by which the reaction occurs, as long as the initial and final conditions are the same.

This principle is particularly useful when determining standard enthalpy of formation (\(\Delta H_f^\circ\)), which is the heat change that results when one mole of a compound is formed from its elements in their standard states. To employ Hess's Law effectively, one must write and manipulate balanced chemical equations and use given enthalpy values to solve for the unknown enthalpy.

In the provided textbook exercise, the student uses Hess's Law to deduce the standard enthalpy of formation for 2,3,4-trimethylpentane from the known enthalpies of combustion for the hydrocarbon and the formation enthalpies of the combustion products, CO₂ and H₂O. By rearranging the equations and using algebra, the enthalpy of formation for the hydrocarbon can be calculated, demonstrating the practical application of Hess's Law in chemistry.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry. It involves adjusting the coefficients (numbers in front of chemical formulas) to ensure that the number of atoms of each element is equal on both sides of the equation, fulfilling the law of conservation of mass.

In the context of the sample problem mentioned, the balanced chemical equation for the formation of 2,3,4-trimethylpentane from carbon and hydrogen is:
\[8 \text{{C}}(\text{{s, graphite}}) + 9\text{{H}}_2(\text{{g}}) \rightarrow \text{{C}}_8\text{{H}}_{18}(\text{{l}})\]
This equation indicates that 8 carbon atoms combine with 18 hydrogen atoms (9 H₂ molecules) to form one molecule of 2,3,4-trimethylpentane. The skill of balancing equations not only upholds the laws governing chemical reactions but also facilitates the quantitative analysis of reactants and products—vital for applications such as calculating reaction yields, gauging reactant quantities, and complying with safety regulations. It is an indisputable aspect of chemical literacy and an indispensable tool for any student delving into the field of chemistry.