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Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short Answer

Expert verified
For the given reactions, the entropy change (ΔS) would be: (a) Negative, as the number of gas molecules decrease from 4 to 2. (b) Positive, since there is an increase in the number of gas molecules from 0 to 1. (c) Negative, as the number of gas molecules decrease from 3 to 1. (d) Positive, as there is no change in the number of gas molecules, but the complexity of molecules increases in products.

Step by step solution

01

Count gas molecules

On the reactant side, there is 1 molecule of N2 and 3 molecules of H2, which results in a total of \(1 + 3 = 4\) gas molecules. On the product side, there are 2 molecules of NH3.
02

Predict sign of entropy change

Since the number of gas molecules decreases from 4 to 2, the entropy change for this reaction is negative (ΔS < 0). #b) CaCO3(s) -> CaO(s) + CO2(g)#
03

Count gas molecules

On the reactant side, there is 1 molecule of solid compound (CaCO3), and there are no gas molecules. On the product side, there is 1 molecule of CO2 gas.
04

Predict sign of entropy change

Since there is an increase in the number of gas molecules (from 0 to 1), the entropy change for this reaction is positive (ΔS > 0). #c) 3C2H2(g) -> C6H6(g)#
05

Count gas molecules

On the reactant side, there are 3 molecules of C2H2 gas. On the product side, there is 1 molecule of C6H6 gas.
06

Predict sign of entropy change

Since the number of gas molecules decreases from 3 to 1, the entropy change for this reaction is negative (ΔS < 0). #d) Al2O3(s) + 3H2(g) -> 2Al(s) + 3H2O(g)#
07

Count gas molecules

On the reactant side, there are 3 molecules of H2 gas. On the product side, there are 3 molecules of H2O gas.
08

Predict sign of entropy change

Since there is no change in the number of gas molecules (from 3 to 3), we need to analyze other factors. Both reactants and products have one solid compound, so the only difference lies in the complexity of the molecules. Al2O3 is a more complex molecule than Al, and H2 is a simpler molecule than H2O. So, the reaction results in more complex molecules, leading to an increase in entropy. Therefore, the entropy change for this reaction is positive (ΔS > 0).

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physical science that deals with heat and its relation to other forms of energy and work. It describes how thermal energy is converted to and from other forms of energy and how it affects matter. The four laws of thermodynamics lay the foundation for understanding various interactions involving heat transfer, energy transformations, and the directionality of physical processes.

One key aspect of thermodynamics is the study of entropy, which is a measure of the disorder or randomness in a system. Entropy is often associated with the amount of energy in a system that is no longer capable of doing work. As a system becomes more disordered, its entropy increases. This concept is crucial for predicting the spontaneous direction of chemical reactions and understanding the efficiency of energy conversions.
Chemical Reactions
Chemical reactions involve the transformation of one or more substances into different substances. During these transformations, energy is either absorbed or released, and changes occur in the arrangement of atoms and molecules. The role of entropy in chemical reactions is to provide a direction to these transformations. In thermodynamics, the second law states that the total entropy of an isolated system can never decrease over time. This means that, in a spontaneous chemical reaction, the entropy of the universe (system plus surroundings) should increase or, at least, remain unchanged.

Understanding the changes in entropy during chemical reactions helps scientists predict whether a reaction will occur spontaneously. A positive change in entropy suggests that the disorder of the universe increases, often indicating a spontaneous reaction. Conversely, a negative change signals that a reaction is not spontaneous, unless it is coupled with enough energy to drive the process.
Gaseous Molecules
Gaseous molecules have higher levels of disorder compared to liquids and solids because they occupy a greater volume and their particles have more freedom to move around. Consequently, the entropic state of gaseous molecules is higher than that of the same molecules in a different state. When predicting entropy changes in chemical reactions, particularly those involving gases, a useful heuristic is to consider the total count of gas molecules before and after the reaction.

Typically, if the number of gaseous molecules increases during a reaction, the entropy increases; conversely, if the number decreases, the entropy decreases. This is because more gas molecules mean greater dispersion of energy and increased disorder. However, other factors, such as molecular complexity and energy content, can also influence the entropy changes in reactions involving gases.
Predicting Entropy
Predicting entropy changes during chemical reactions can be complex because entropy is influenced by several factors such as temperature, pressure, and molecular characteristics. However, some general trends can assist in determining whether entropy is likely to increase or decrease in a given reaction.

As a starting point, the number of particles should be considered, with more particles usually correlating to higher entropy. Physical state changes, like solid to gas, generally result in increased entropy. Additionally, the complexity of molecules, with larger or more complex structures typically having higher entropy than smaller or simpler ones, should be taken into account. These considerations are particularly useful in exercises where students are asked to predict the entropy change's sign, helping to contextualize the abstract concept of entropy with tangible examples from chemical reactions.

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Most popular questions from this chapter

(a) What sign for \(\Delta S\) do you expect when the volume of \(0.200\) mol of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L}\) ? (b) If the final volume is \(18.5 \mathrm{~L}\) calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{5}\). Did your classmate make a mistake in the calculation? Explain.

(a) What sign for \(\Delta S\) do you expect when the pressure on \(0.600 \mathrm{~mol}\) of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of \(0.750 \mathrm{~atm}\) ? (b) If the final pressure on the gas is \(1.20 \mathrm{~atm}\), calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H\), we can know absolute values of \(S\). (b) If you heat a gas such as \(\mathrm{CO}_{2}\) you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(\mathrm{g})\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

When most elastomeric polymers (e.g-, a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

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