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Chapter 19: Problem 20

Consider what happens when a sample of the explosive TNT is detonated under atmospheric pressure. (a) Is the detonation a reversible process? (b) What is the sign of \(q\) for this process? (c) Is w positive, negative, or zero for the process?

Short Answer

Expert verified
The detonation of TNT is an irreversible process. The sign of q (heat) for this process is negative, as it releases heat into the surroundings (exothermic). The sign of w (work) for the process is positive, as the system does work on the surroundings when the explosive expands.

Step by step solution

01

a) Reversible Process

Reversible processes are processes that can be reverted to their original state while also returning the surroundings to their original state. In the case of an explosive such as TNT detonating, it's improbable that the process could revert back to its initial conditions. Thus, the detonation of TNT is an irreversible process.
02

b) Sign of q (Heat)

When the TNT detonates, it releases a large amount of heat into the surroundings. This heat transfer is essentially from the system (TNT) to the surroundings, therefore making it an exothermic process. In an exothermic process, the heat (q) is considered to be negative. Hence, the sign of q for this process is negative.
03

c) Sign of w (Work)

During detonation, the explosive TNT expands rapidly and does work on the surroundings. According to thermodynamics sign conventions, when a system does work on the surroundings, the work (w) is considered to be positive. Therefore, in this case, w is positive for the explosion process.

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Most popular questions from this chapter

(a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of \(\Delta S\) surr?

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and $\Delta S^{\circ}=\( \)+100.0 \mathrm{~J} / \mathrm{K} .$ (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at $298 \mathrm{~K} .(\mathbf{d})\( Is the reaction spontaneous at \)298 \mathrm{~K}$ under standard conditions?

Using \(S^{\circ}\) values from Appendix \(\mathrm{C}\), calculate $\Delta S^{\circ}$ values for the following reactions. In each case, account for the sign of \(\Delta S\). (a) $\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{Cl}^{-}(a q)$ (b) $\mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$ (c) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ (d) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$

Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at $298 \mathrm{~K}$ under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) $\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)$

Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy can only be produced but cannot not be destroyed. (b) In a certain process the entropy of the system changes by $1.2 \mathrm{~J} / \mathrm{K}\( (increase) and the entropy of the surroundings changes by \)-1.2 \mathrm{~J} / \mathrm{K}$ (decrease). Thus, this process must be spontaneous. (c) In a certain process the entropy of the system changes by $1.3 \mathrm{~J} / \mathrm{K}\( (increase) and the entropy of the surroundings changes by \)-1.2 \mathrm{~J} / \mathrm{K}$ (decrease). Thus, this process must be reversible.
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