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 \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .(\mathbf{b})\) 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

The short answer to the given question is: (a) The balanced equation for the combustion of 1 mol of C₈H₁₈ is: \[ C_{8}H_{18}(l) + \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g) \] (b) The balanced equation for the formation of C₈H₁₈ from its elements is: \[ 8C(graphite) + 9H_{2}(g) \rightarrow C_{8}H_{18}(l) \] (c) The standard enthalpy of formation, ΔHf°, for 2,3,4 trimethylpentane is -259.3 kJ/mol.

Step-by-step solution

Combustion Reaction Equation

The combustion of a hydrocarbon involves a reaction with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). When balanced, the equation for the combustion of 2,3,4 trimethylpentane can be written as: \[ C_{8}H_{18}(l) + \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g) \] Step 2: Write the balanced equation for the formation of C₈H₁₈ from its elements

Formation Reaction Equation

To write the balanced equation for the formation of C₈H₁₈ from its elements, we need to combine carbon (C) and hydrogen (H₂) to produce C₈H₁₈. Since carbon is in its standard state as graphite and hydrogen is diatomic, the balanced equation is: \[ 8C(graphite) + 9H_{2}(g) \rightarrow C_{8}H_{18}(l) \] Step 3: Calculate the standard enthalpy of formation, ΔHf°, of 2,3,4 trimethylpentane

Calculating ΔHf° for C₈H₁₈

We are given the enthalpy change for the combustion reaction, which is -5064.9 kJ/mol. We also need the enthalpies of formation for CO₂ and H₂O from Table 5.3. Here they are: ΔHf°(CO₂) = -393.5 kJ/mol ΔHf°(H₂O) = -241.8 kJ/mol The reaction enthalpy for the formation of C₈H₁₈ can be expressed as follows: ΔHf°(C₈H₁₈) = Σ[ΔHf°(products)] - Σ[ΔHf°(reactants)] Using the given data (-5064.9 kJ/mol), data from Table 5.3, and our balanced equations, we can calculate the ΔHf° for C₈H₁₈: ΔHf°(C₈H₁₈) = [8ΔHf°(CO₂) + 9ΔHf°(H₂O)] - [ΔHrxn°] ΔHf°(C₈H₁₈) = [8(-393.5 kJ/mol) + 9(-241.8 kJ/mol)] - [-5064.9 kJ/mol] ΔHf°(C₈H₁₈) = (-3148 kJ +(-2176.2 kJ)) + 5064.9 kJ/mol ΔHf°(C₈H₁₈) = -259.3 kJ/mol So, the standard enthalpy of formation for 2,3,4 trimethylpentane is -259.3 kJ/mol.

Key concepts

Understanding Combustion Reaction Equations

A combustion reaction is an exothermic chemical reaction where a substance, typically a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy in the form of heat or light.

In the given exercise, we're looking at the combustion of a hydrocarbon named 2,3,4 trimethylpentane. The general formula for the combustion of a hydrocarbon is \[ \text{Hydrocarbon} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \.\]
The combustion of 2,3,4 trimethylpentane is specifically represented by the equation:\
\[ C_{8}H_{18}(l) + \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(g) \]
This balanced equation shows that one mole of 2,3,4 trimethylpentane reacts with oxygen to produce eight moles of carbon dioxide and nine moles of water vapor.

An essential aspect of understanding these reactions is ensuring that they are properly balanced, meaning the number of atoms for each element is equal on both sides of the equation. For students struggling with this concept, remember that coefficients can be used to balance the atoms, as demonstrated in the step-by-step solution.

Formation Reaction Equations Made Simple

A formation reaction involves the production of a compound from its elements in their standard states. The standard state is the form in which the element is most stable under standard conditions (1 atm pressure and usually at 25°C).

For the formation of an organic compound like 2,3,4 trimethylpentane, the equation involves carbon and hydrogen. Here, carbon is typically represented as graphite, the most stable form of carbon under standard conditions, and hydrogen is diatomic (H₂).
Thus, the balanced formation reaction for 2,3,4 trimethylpentane from its elements is given by:\
\[ 8C(graphite) + 9H_{2}(g) \rightarrow C_{8}H_{18}(l) \]
This equation shows that eight atoms of carbon from graphite react with 18 atoms of hydrogen from diatomic hydrogen molecules to form one mole of 2,3,4 trimethylpentane liquid.

To help students grasp the balancing aspect, focus on the fact that each carbon atom will form one-eighth of the hydrocarbon compound, and that two hydrogen atoms will form one molecule of diatomic hydrogen, H₂. It is important to match both the types and numbers of atoms on each side of the equation for it to be balanced.

Standard Enthalpy Change Explained

The standard enthalpy change, \( \Delta H^\circ \), represents the heat exchanged in a reaction at constant pressure when all reactants and products are in their standard states.

For reactions, the standard enthalpy change can be calculated using Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps that lead to the overall reaction.
In the context of our exercise, the standard enthalpy of formation, \( \Delta H_f^\circ \), for 2,3,4 trimethylpentane can be calculated by considering the standard enthalpy changes of the combustion products and then rearranging the equation to solve for the unknown formation enthalpy. The equation for this process is: \[ \Delta H_f^\circ(\text{compound}) = \Sigma[\Delta H_f^\circ(\text{products})] - \Sigma[\Delta H_f^\circ(\text{reactants})] \]
With given values for the combustion of 2,3,4 trimethylpentane and the formation enthalpies of CO₂ and H₂O, we deduce the heat needed to form 2,3,4 trimethylpentane.

If students find this calculation challenging, it may be helpful to visualize the reaction in terms of bond breaking and forming — breaking bonds requires energy (endothermic) and forming bonds releases energy (exothermic). This visualization can make understanding enthalpy changes more intuitive.