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Chapter 17: Problem 12

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.

Short Answer

Expert verified
The dispersion of toxic substances in the environment aligns with the second law of thermodynamics, as it increases the disorder (entropy) in the system. In terms of the second law, preventing contamination before it occurs requires less work compared to cleaning the environment after it has been contaminated, since prevention does not demand reversing the natural tendency of increasing entropy.

Step by step solution

01

Understanding the Second Law of Thermodynamics

The second law of thermodynamics states that in any natural process, the total entropy of a closed system will always increase over time, or remain constant in ideal cases where the system is in a steady state (equilibrium). Entropy can be viewed as a measure of disorder or randomness in a system. In simpler terms, this law suggests that it is a fundamental principle of nature that systems tend towards chaos and disorder.
02

Connecting the Second Law and Dispersion of Toxic Substances

Now, let's connect this idea with the dispersion of toxic substances in the environment. When a toxic substance is released into the environment, it begins to spread and disperse, eventually reaching a uniform distribution. This action increases the disorder and randomness in the system (the environment), which is consistent with the second law of thermodynamics. Thus, the dispersion of toxic substances aligns with the natural tendency of increasing entropy in the environment.
03

Analyzing Which Requires Less Work - Prevention or Cleanup

According to the second law, processes that increase entropy are more likely to occur spontaneously. Hence, once a toxic substance has been released and dispersed into the environment, trying to reverse that process (cleaning up) would require a significant amount of work, as it would involve decreasing the system's entropy. On the other hand, preventing the contamination from occurring in the first place doesn't require a decrease in entropy in any system. Instead, it aims to maintain the prevailing level of order in the environment. Consequently, prevention involves far less work when compared to the energy-intensive process of cleaning up contamination after it has occurred.
04

Conclusion

In conclusion, the dispersion of toxic substances in the environment is consistent with the second law of thermodynamics, as it increases the disorder (entropy) of the system. In terms of the second law, preventing contamination before it occurs requires less work compared to cleaning the environment after it has been contaminated, since it does not demand reversing the natural tendency of increasing entropy.

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Most popular questions from this chapter

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})\) at \(1.00\) atm and \(1.00 \mathrm{~mol} \mathrm{~B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).
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