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Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

Short Answer

Expert verified
The detonation of a TNT sample is a spontaneous exothermic process, characterized by a negative Gibbs free energy (ΔG), a negative heat exchange (q), and a negative work done (w). The change in internal energy (ΔE) for the process is also negative, indicating a decrease in the system's internal energy during the detonation.

Step by step solution

01

(a) Spontaneity of Detonation

To determine if the detonation of the TNT sample is a spontaneous process, we must consider the overall change in Gibbs free energy (ΔG) for the process. A spontaneous process has a negative ΔG. The detonation of TNT is known to be spontaneous, as it occurs without any external influence once initiated. Thus, the detonation of the TNT is a spontaneous process.
02

(b) Sign of q (Heat Exchanged)

The detonation of TNT is characterized by an exothermic reaction, wherein the chemical energy stored in the TNT molecules is released as heat. In an exothermic reaction, the system releases heat to the surroundings, meaning q < 0. Therefore, the sign of q for the detonation of the TNT process is negative.
03

(c) Sign of w (Work Done)

Work done (w) can be described as the expansion or compression of gases during a process. In the detonation of TNT, the rapid release of energy causes gas molecules to be produced and displaced. As a result, there is an expansion of gases which results in work being done by the system on its surroundings. When the system does work on the surroundings, w < 0. In conclusion, the sign of w for the detonation process is negative.
04

(d) Sign of ΔE (Change in Internal Energy)

The change in internal energy (ΔE) of a system can be determined from the equation: \[ ΔE = q + w \] From previous parts, we know that q < 0 (exothermic reaction) and w < 0 (system does work on the surroundings). The combination of these two negative values results in a negative ΔE. Therefore, the sign of ΔE for the detonation process is negative. This means that the internal energy of the system decreases during the detonation process.

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Key Concepts

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

Spontaneity of Reactions
In chemistry, the spontaneity of a reaction refers to whether a process will occur on its own without the need for continuous external input. This is often determined by looking at the change in Gibbs free energy (\(\Delta G\)) of the system. If \(\Delta G < 0\), the reaction is said to be spontaneous. Spontaneous reactions are energetically favorable, implying that they proceed towards more stability.

In the original exercise, the detonation of TNT is described as spontaneous. This means that once initiated, TNT will explode without needing additional help. The negative Gibbs free energy indicates that the reaction naturally progresses towards completion. Spontaneous reactions can be hectic and often very dangerous due to their rapid release of energy. By understanding the concept of spontaneity, chemists can predict and control reactions more effectively.
Exothermic Processes
An exothermic process is one where energy, in the form of heat, is released into the surrounding environment. This release of energy is characteristic of many chemical reactions, especially those that involve combustion or explosions, such as the detonation of TNT.

In an exothermic reaction, the reactants have higher stored energy than the products. During the reaction, this excess energy is expelled, leading to an increase in the temperature of the surroundings. For the detonation of TNT, it is clear from the original solution that the reaction is exothermic because the system releases heat, indicated by \(q < 0\).

Understanding whether a process is exothermic is crucial for several purposes:
  • It helps in predicting the temperature changes in the surroundings.
  • It allows engineers to design appropriate containers and environments for safely conducting these reactions.
  • It plays a vital role in industrial applications where heat output is a critical factor.
Gibbs Free Energy
Gibbs free energy (\(G\)) is a thermodynamic potential that indicates the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. It is derived from the enthalpy (\(H\)), temperature (\(T\)), and entropy (\(S\)) of a system through the equation \(G = H - TS\).

The change in Gibbs free energy (\(\Delta G\)) for a process indicates whether a reaction is spontaneous, as mentioned earlier. A negative \(\Delta G\) means the process will happen spontaneously under the given conditions.

Connecting this to the original exercise, for the TNT explosion, \(\Delta G\) being negative illustrates why the reaction proceeds without further energy input after initiation. It provides insights into understanding both the energy dynamics and the feasibility of chemical processes. Scientists and engineers often calculate \(\Delta G\) to analyze and design chemical reactions, aiming for efficient and safe applications.

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Most popular questions from this chapter

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K .\) (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(\underline{K}\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

How does the entropy of the system change when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

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