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Isooctane \((2,2,4\) -trimethylpentane), one of the many hydrocarbons that makes up gasoline, burns in air to give water and carbon dioxide. $$ \begin{aligned} 2 \mathrm{C}_{8} \mathrm{H}_{18}(\ell)+25 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\ell) \\\ \Delta H^{\circ}=&-10,922 \mathrm{~kJ} \end{aligned} $$ What is the enthalpy change if you burn \(1.00 \mathrm{~L}\) of isooc- $$ \text { tane }(\text { density }=0.69 \mathrm{~g} / \mathrm{mL}) ? $$

Short Answer

Expert verified
The enthalpy change is approximately \(-32,977\text{ kJ}\).

Step by step solution

01

Determine Mass of Isooctane

To find the mass of isooctane, use the density formula: \( ext{density} = \frac{\text{mass}}{\text{volume}} \). Rearrange this to find mass: \( \text{mass} = \text{density} \times \text{volume} \). Substitute the values: \( 0.69 \text{ g/mL} \times 1000 \text{ mL} = 690 \text{ g} \). Thus, the mass of isooctane is 690 g.
02

Calculate Moles of Isooctane

First, calculate the molar mass of isooctane \( (\text{C}_8\text{H}_{18}) \): \( 8 \times 12.01 + 18 \times 1.008 = 114.224 \text{ g/mol} \). Next, find moles: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{690 \text{ g}}{114.224 \text{ g/mol}} \approx 6.04 \text{ moles} \).
03

Find Enthalpy Change for Combustion

Using the balanced chemical equation, note that 2 moles of isooctane release \(-10,922\text{ kJ}\). Therefore, the enthalpy change per mole is \( \frac{-10,922 \text{ kJ}}{2} = -5,461 \text{ kJ/mol} \).
04

Calculate Total Enthalpy Change

Using the number of moles of isooctane (6.04 moles), calculate the total enthalpy change: \( 6.04 \text{ moles} \times (-5,461 \text{ kJ/mol}) \approx -32,977 \text{ kJ} \). Thus, the enthalpy change is approximately \(-32,977\text{ kJ}\).

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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 is a chemical process where a substance reacts with oxygen to produce heat and light, often resulting in the formation of carbon dioxide and water. It's a type of exothermic reaction, meaning it releases energy in the form of heat. In the case of isooctane burning, the reaction with oxygen gives off a large amount of energy, as demonstrated by the enthalpy change (\(\Delta H^{\circ}=-10,922 \, \text{kJ}\)).
Key aspects of combustion reactions include:
  • Reactants: Typically a hydrocarbon and oxygen.
  • Products: Carbon dioxide, water, and energy.
  • The reaction is often vigorous and produces a flame.
The chemical equation for the combustion of isooctane helps us understand the stoichiometry of the reaction, indicating that two moles of isooctane react with twenty-five moles of oxygen, to form sixteen moles of carbon dioxide and eighteen moles of water. Understanding this allows us to appreciate how much energy can be released, which is crucial for applications like fuel consumption and energy production.
Moles Calculation
Calculating moles is an essential skill in chemistry for understanding the quantitative aspects of chemical reactions. The concept of "moles" allows chemists to count particles by weighing them and is based on the molar mass of substances.
To calculate moles, follow these steps:
  • Find the molar mass of the compound. For isooctane \((\text{C}_8\text{H}_{18})\), the molar mass is calculated by adding the atomic masses: \(8 \times 12.01 + 18 \times 1.008 = 114.224 \, \text{g/mol}.\)
  • Use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}.\)
  • Substitute the mass of isooctane (690 g in this case) to find the moles: \(\frac{690 \, \text{g}}{114.224 \, \text{g/mol}} \approx 6.04 \, \text{moles}.\)
Understanding how to calculate moles aids in predicting how substances will react and allows calculation of energy changes in reactions. It's a fundamental concept that connects the macroscopic world with the molecular level.
Density Formula
The density formula is a vital tool used to convert between mass, volume, and density of a substance. The formula is typically written as:\[\text{density} = \frac{\text{mass}}{\text{volume}}.\]
This formula can be rearranged depending on which quantity you need to find:
  • When you have density and volume, you can find mass: \(\text{mass} = \text{density} \times \text{volume}.\)
  • When you have mass and density, you can find volume: \(\text{volume} = \frac{\text{mass}}{\text{density}}.\)
In the exercise about isooctane, we used the density formula to calculate how much 1 liter of isooctane weighs. Knowing that the density is \(0.69 \text{ g/mL},\) and the volume is \(1000 \text{ mL},\) the mass is calculated as:\[ 0.69 \text{ g/mL} \times 1000 \text{ mL} = 690 \text{ g}. \] This is an important calculation because it helps translate volume (a frequently more easily measured property) into mass, which is essential for chemical calculations such as finding moles or energy changes.

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Most popular questions from this chapter

Why is chemical energy classified as a form of potential energy?

What characteristic does every exothermic reaction have?

The law of Dulong and Petit states that the heat capacity of metallic elements is approximately \(25 \mathrm{~J} / \mathrm{mol} \cdot{ }^{\circ} \mathrm{C}\) at \(25^{\circ} \mathrm{C}\). In the 19 th century, scientists used this relationship to obtain approximate atomic masses of metals, from which they determined the formulas of compounds. Once the formula of a compound of the metal with an element of known atomic mass is known, the mass percentage composition of the compound is used to find the atomic mass of the metal. The following example shows the calculations involved. (a) Experimentally, the specific heat of a metal is found to be \(0.24 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Use the law of Dulong and Petit to calculate the approximate atomic mass of the metal. (b) An oxide of this element is \(6.90 \%\) oxygen by mass. Use the molar mass of \(16.00 \mathrm{~g} / \mathrm{mol}\) for oxygen and the approximate atomic mass found in part (a) to determine the subscripts \(x\) and \(y\) in the formula of the oxide, \(\mathrm{M}_{x} \mathrm{O}_{y} .\) (The mole ratio of the elements you find will not be exactly whole numbers, so considerable rounding is needed to obtain whole numbers in the formula.) (c) From the formula established in part (b), \(x\) mol \(M\) are combined with \(y\) mol \(O .\) Calculate the mass of the metal that is combined with \(y \mathrm{~mol} \mathrm{O}\), using the percent composition of the oxide, and find the atomic mass of the metal. What is the element \(\mathrm{M}\) ?

Under what circumstances is the heat of a process equal to the enthalpy change for the process?

When a \(2.30-\mathrm{g}\) sample of magnesium dissolves in dilute hydrochloric acid, \(16.25 \mathrm{~kJ}\) of heat is released. Determine the enthalpy change for the thermochemical equation $$ \begin{aligned} \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \\ \Delta H=? \end{aligned} $$

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