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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 leads to an increase in entropy, which is consistent with the second law of thermodynamics, as it favors higher disorder or randomness. Preventing contamination before it occurs requires less work compared to cleaning the environment after it has been contaminated. This is because preventing contamination aligns with the natural tendency of the system to increase in entropy, whereas cleaning up contamination goes against this tendency and requires additional work to decrease 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 the universe increases. Entropy is a measure of disorder or randomness of a system. The second law implies that natural processes tend to move towards a state of higher entropy or disorder.
02

Relating toxic dispersion to the second law of thermodynamics

When a toxic substance contaminates the environment, it tends to disperse and spread out. This dispersion leads to an increase in the entropy of the system (in this case, the environment). The toxic substance spreads into the environment, increasing the disorder of the system, which is consistent with the second law of thermodynamics.
03

Cleaning the environment after it has been contaminated

Cleaning the environment after contamination involves decreasing the entropy of the system. This means that work is required to restore order or to decrease the randomness in the environment. In terms of thermodynamics, this process would be against the natural tendency of the system, as the second law favors an increase in entropy.
04

Preventing contamination before it occurs

Preventing contamination is a proactive measure that does not require decreasing the entropy of the system. Since no increase in disorder has occurred, there is no need to reverse this disorder, which would require additional work. In other words, keeping the environment clean and avoiding contamination doesn't go against the second law of thermodynamics, as it would if cleaning up the contamination afterward.
05

Comparing energy requirements for cleaning and prevention

Based on the second law of thermodynamics, preventing contamination before it occurs requires less work compared to cleaning the environment after it has been contaminated. Preventing contamination aligns with the natural tendency of the system to increase in entropy, whereas cleaning up contamination goes against this tendency and requires additional work to be done to decrease entropy.

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Most popular questions from this chapter

Using the following data, calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), one of the least soluble of the common nitrate salts. $$ \begin{array}{lc} \text { Species } & \Delta G_{\mathrm{f}}^{\circ} \\ \mathrm{Ba}^{2+}(a q) & -561 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{NO}_{3}^{-}(a q) & -109 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(s) & -797 \mathrm{~kJ} / \mathrm{mol} \\ \hline \end{array} $$

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ} $$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) $$ \begin{aligned} \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned} $$

Consider the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C}\), predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{~atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{~atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{~atm}\) c. \(P_{\mathrm{NO},}=0.29 \mathrm{~atm}, P_{\mathrm{N}, \mathrm{O}_{4}}=1.6 \mathrm{~atm}\)

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$ \mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) $$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 79.)
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