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One of the steps in the cracking of petroleum into gasoline in. volves the thermal breakdown of large hydrocarbon molecules into smaller ones. For example, the following reaction might occur: $$\mathrm{C}_{11} \mathrm{H}_{24} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{C}_{4} \mathrm{H}_{8}+\mathrm{C}_{3} \mathrm{H}_{6}$$ Is \(\Delta S\) for this reaction likely to be positive or negative? Explain.

Short Answer

Expert verified
The entropy change ( \( \Delta S \) ) is likely positive due to increased number of molecules.

Step by step solution

01

Understand Entropy Concept

Entropy (\( \Delta S \)) is a measure of disorder or randomness in a system. Generally, when a reaction produces more molecules as products than there were as reactants, the entropy increases (positive \( \Delta S \)).
02

Analyze Molecules Involved

In the given reaction, one large hydrocarbon molecule \( \mathrm{C}_{11} \mathrm{H}_{24} \) breaks down into three smaller molecules: \( \mathrm{C}_{4} \mathrm{H}_{10} \), \( \mathrm{C}_{4} \mathrm{H}_{8} \), and \( \mathrm{C}_{3} \mathrm{H}_{6} \).
03

Count Number of Molecules

Initially, there is 1 molecule of \( \mathrm{C}_{11} \mathrm{H}_{24} \), and after the reaction, there are 3 new molecules. This means that the total number of gaseous molecules has increased from 1 to 3.
04

Evaluate Change in Entropy

Given that the number of molecules has increased, the system has become more disordered, which implies an increase in entropy. Thus, \( \Delta S \) is likely positive.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy Change
Entropy, denoted as \( \Delta S \), is a crucial concept in understanding chemical reactions. At its core, entropy is a measure of disorder or randomness within a system. When we talk about entropy change in a chemical reaction, we are interested in how the disorder changes as reactants transform into products.
  • A positive entropy change (\( \Delta S > 0 \)) indicates an increase in disorder.
  • A negative entropy change (\( \Delta S < 0 \)) suggests a decrease in disorder.
In the context of the exercise, the breakdown of a single large hydrocarbon molecule into several smaller ones leads to an increase in the total number of molecules, increasing disorder in the system, and therefore, \( \Delta S \) is positive. This is because multiple smaller molecules are typically more disordered and occupy more spatial configurations.
Thermal Breakdown
Thermal breakdown is a process where molecules decompose into smaller fragments under the influence of heat. It's a key reaction in the field of chemical engineering, especially when converting complex hydrocarbons in petroleum to usable products like gasoline. During thermal breakdown, heat energy provides the necessary activation energy to break chemical bonds.
  • This energy causes large hydrocarbon molecules to split into multiple smaller molecules.
  • The increased entropy (or disorder) often contributes to the favorability of the reaction.
This process is not only aimed at breaking down molecules but also at optimizing the distribution of desired products through temperature control and reaction time.
Hydrocarbon Molecules
Hydrocarbons are organic molecules consisting entirely of carbon and hydrogen. They are the fundamental components of petroleum and natural gas.In the cracking process, hydrocarbons like \( \mathrm{C}_{11} \mathrm{H}_{24} \) are broken down to form smaller hydrocarbons such as \( \mathrm{C}_{4} \mathrm{H}_{10} \), which are often more useful for various industrial applications.
  • Smaller hydrocarbons typically have higher demand as they form the base for fuels and chemicals.
  • The process increases the overall surface area and reactivity of hydrocarbons.
Understanding the behavior and breakdown of hydrocarbon molecules is essential for producing efficient fuels and minimizing wastage in chemical processes.

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Most popular questions from this chapter

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\), whose aqueous solutions are known as vinegar, is prepared by reaction of ethyl alcohol ( \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ) with oxygen: $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)+\mathrm{H}_{2} \mathrm{O}(l)$$ Use the following data to calculate \(\Delta H^{\circ}\) in kilojoules for the reaction: $$\begin{aligned}&\Delta H_{\mathrm{f}}\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\right]=-277.7 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H^{\circ} \mathrm{f}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)\right]=-484.5 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)\right]=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

What does entropy measure?

Chloroform has \(\Delta H_{\text {vaporimation }}=29.2 \mathrm{~kJ} / \mathrm{mol}\) and boils at \(61.2{ }^{\circ} \mathrm{C}\). What is the value of \(\Delta S_{\text {vaporization }}\) for chloroform?

The familiar "ether" used as an anesthetic agent is diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\). Its heat of vaporization is \(+26.5 \mathrm{~kJ} / \mathrm{mol}\) at its boiling point. How much energy in kilojoules is required to convert 100 mL of diethyl ether at its boiling point from liquid to vapor if its density is \(0.7138 \mathrm{~g} / \mathrm{mL}\) ?

When \(2.00 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in a bomb calorimeter, the temperature of the water and calorimeter rises by \(18.6^{\circ} \mathrm{C}\). Assuming that the bath contains \(250.0 \mathrm{~g}\) of water and that the heat capacity for the calorimeter is \(623 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), calculate combustion energies \((\Delta E)\) for glucose in both \(\mathrm{kJ} / \mathrm{g}\) and \(\mathrm{Cal} / \mathrm{g}\).

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