Question

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanestburning octanes is a compound called 2,3,4-trimethylpentane, which has the following structural formula: CC(C)C(C)C(C)C 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 \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\). (b) 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

The balanced equation for the complete combustion of 1 mol of 2,3,4-trimethylpentane (\(C_8H_{18}\)) is: \[C_8H_{18}(l) + 12.5 O_2(g) \rightarrow 8 CO_2(g) + 9 H_2O(g)\] The standard enthalpy of formation for 2,3,4-trimethylpentane is approximately -209.3 kJ/mol.

Step-by-step solution

Identify the chemical composition

The structural formula for 2,3,4-trimethylpentane is given as: CC(C)C(C)C(C)C. By counting the number of carbon (C) and hydrogen (H) atoms, we can determine its molecular formula: C8H18. ##Step 2: Write the complete combustion reaction##

Balance the equation

To write and balance the combustion reaction, we need to know that complete combustion of a hydrocarbon involves reacting with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). The general form of the combustion reaction for a hydrocarbon is: \[C_xH_y + O_2 \rightarrow CO_2 + H_2O\] In our case, the molecular formula of 2,3,4-trimethylpentane is C8H18. So the unbalanced reaction is: \[C_8H_{18} + O_2 \rightarrow CO_2 + H_2O\] Now we balance the reaction: \[C_8H_{18} + 12.5 O_2 \rightarrow 8 CO_2 + 9 H_2O\] The balanced equation for the complete combustion of 1 mol of 2,3,4-trimethylpentane is: \[C_8H_{18}(l) + 12.5 O_2(g) \rightarrow 8 CO_2(g) + 9 H_2O(g)\] ##Step 3: Calculate the standard enthalpy of formation for 2,3,4-trimethylpentane##

Use Hess's Law and the given information

We are given the standard enthalpy change for the combustion reaction, ΔHc°=-5064.9 kJ/mol. Using Hess's Law, we can calculate the standard enthalpy of formation for 2,3,4-trimethylpentane, ΔHf°, from the following equation: ΔHc° = ΣnΔHf°(products) - ΣmΔHf°(reactants) where n and m are the stoichiometric coefficients of the products and reactants, respectively. We can rewrite the equation specifically for our combustion reaction: ΔHc° = [8ΔHf°(CO2) + 9ΔHf°(H2O)] - [ΔHf°(C8H18) + 12.5ΔHf°(O2)] Since the standard enthalpy of formation for O2 is 0 (ΔHf°(O2) = 0), we can simplify and solve for ΔHf°(C8H18): ΔHf°(C8H18) = [8ΔHf°(CO2) + 9ΔHf°(H2O)] - ΔHc° Now we need to use data from Table 5.3 for the standard enthalpies of formation for CO2 and H2O: ΔHf°(CO2) = -393.5 kJ/mol ΔHf°(H2O) = -241.8 kJ/mol Plugging in the values: ΔHf°(C8H18) = [8(-393.5) + 9(-241.8)] - (-5064.9) ΔHf°(C8H18) = -209.3 kJ/mol The standard enthalpy of formation for 2,3,4-trimethylpentane is approximately -209.3 kJ/mol.

Key concepts

Combustion Reaction

Understanding the process of combustion is essential for anyone studying chemistry. A combustion reaction involves a substance (usually a hydrocarbon) reacting with oxygen to produce carbon dioxide, water, and energy in the form of heat or light. Hydrocarbons are compounds made up of hydrogen and carbon atoms, such as gasoline components like 2,3,4-trimethylpentane. The most important thing in a combustion reaction is that it requires oxygen (O2) and it always yields carbon dioxide (CO2) and water (H2O), along with the release of energy.

For example, when your car burns gasoline, which contains hydrocarbons, it's engaging in a combustion reaction. The energy released from this reaction is what powers the engine. In chemical terms, the reaction is highly exothermic, which means it gives off a lot of heat. This is quantified by the change in enthalpy (ΔH), which, in the case of combustion, is a negative value indicating that energy is being released. The balanced chemical equation for a hydrocarbon undergoing complete combustion is pivotal for understanding the energy transfer in this reaction.

2,3,4-Trimethylpentane

2,3,4-Trimethylpentane is an isomer of octane, which is a hydrocarbon with eight carbon atoms; in this case, it has a structural formula that can be depicted as CC(C)C(C)C(C)C. This molecule is part of what you might fill your car with at the pump - gasoline. Chemically, it is a saturated compound, meaning it consists only of single bonds between its carbon atoms and is packed with hydrogens (C8H18). This structure makes it a stable substance under normal conditions, but very energetic when involved in combustion reactions. The name 'trimethylpentane' indicates that it has three methyl (CH3) groups attached to a pentane (five-carbon) chain, which significantly influences how the molecule behaves during chemical reactions, including combustion.

Knowing the structure of 2,3,4-trimethylpentane is crucial for various applications like the production of fuel, understanding its properties, and analyzing its combustion for energy generation.

Chemical Equation Balancing

Balancing a chemical equation is a fundamental skill in chemistry. It ensures that the Law of Conservation of Mass is adhered to, indicating that matter is neither created nor destroyed in a chemical reaction. To balance a chemical equation, one must make sure that the number of atoms of each element on the reactant side equals the number of atoms on the product side.

For the combustion of 2,3,4-trimethylpentane, this involves ensuring that there are the same numbers of carbon (C), hydrogen (H), and oxygen (O) atoms on both sides of the reaction. In our example, the combustion reaction is balanced by adjusting coefficients to yield 8 CO2 and 9 H2O molecules for every molecule of C8H18 and 12.5 molecules of O2. This step is critical, as it sets the stage for correct calculations involving stoichiometry and thermodynamics.

Hess's Law

Hess's Law is a fundamental principle in thermochemistry that states the total enthalpy change for a chemical reaction is the same no matter how the reaction occurs, step by step or all at once. In simpler terms, it implies that energy changes are state functions and are independent of the path taken.

Using Hess's Law allows us to calculate the standard enthalpy of formation (ΔHf°) for substances like 2,3,4-trimethylpentane, even if the direct measurement is not feasible. By knowing the enthalpies of formation of the products (CO2 and H2O) and the enthalpy change for the combustion reaction (ΔHc°), we can rearrange and solve for the enthalpy of formation of the reactant, which in this case would be the enthalpy of formation of 2,3,4-trimethylpentane. Hess's Law is invaluable for chemists, as it allows them to deduce energy changes for reactions based on known values, without having to perform the reactions themselves.