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Chapter 19: Problem 104

In chemical kinetics, the entropy of activation is the entropy change for the process in which the reactants reach the activated complex. Predict whether the entropy of activation for a bimolecular process is usually positive or negative.

Short Answer

Expert verified
The entropy of activation (\( \Delta S^\ddag \)) for a bimolecular process is usually negative, as the formation of the activated complex involves a decrease in available configurations and freedom of motion when two individual molecules collide and form a single structure.

Step by step solution

01

Understanding a Bimolecular Process

A bimolecular process is a chemical reaction that proceeds through the direct collisions between two reacting molecules or species. These reactions have a rate-determining step in which two molecules collide and form a single activated complex or a transition state. Examples of bimolecular reactions include SN2 substitution reactions, some elimination reactions, and particular reactions involving radicals.
02

Understanding Entropy of Activation

Entropy of activation (\( \Delta S^\ddag \)) is an important factor in determining the rate of a chemical reaction. It represents the change in entropy as the reacting species progress from the reactants to the activated complex or transition state. It is a measure of the degree of disorder during the transition from reactants to the activated complex. According to the transition state theory, an activated complex is a higher-energy, less stable, and short-lived species that is formed between the process of reactants converting into products. The activated complex can either revert to the reactants or continue to form products.
03

Entropy Change in Bimolecular Processes

In a bimolecular process, two individual molecules collide and form a single activated complex. Prior to the collision, these molecules can be present in various orientations and with different energies. However, when they collide and form the activated complex, they are now confined within a single structure, leading to a decrease in the number of available configurations and freedom of motion. As a result, the formation of the activated complex in bimolecular reactions generally results in a decrease in the entropy. This suggests that the entropy of activation for bimolecular processes is typically negative because the螖S's will usually be negative (\( \Delta S^\ddag < 0 \)).
04

Conclusion

The entropy of activation for a bimolecular process is generally negative. This conclusion is based on the fact that the formation of the activated complex, which is a higher-energy and less stable species, requires the collision of two individual molecules and results in a decrease in the number of available configurations and freedom of motion.

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Most popular questions from this chapter

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is a constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ} .\)

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?

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Consider the reaction $$ \mathrm{PbCO}_{3}(s) \rightleftharpoons \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) $$ Using data in Appendix C, calculate the equilibrium pres- sure of \(\mathrm{CO}_{2}\) in the system at (a) \(400^{\circ} \mathrm{C}\) and \((\mathbf{b}) 180^{\circ} \mathrm{C} .\)
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