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

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The entropy of activation for a bimolecular process is usually negative, as the formation of the activated complex leads to a decrease in the system's entropy due to the reactants becoming more ordered in the transition state.

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01

Understanding entropy

Entropy is a thermodynamic property that measures the degree of disorder or randomness in a system. In a chemical reaction, it is generally observed that a higher entropy system (more disorder) is favored over a lower entropy system (more order). Hence, changes in entropy can indicate the direction of a reaction.
02

Comparing reactants and activated complex

In a bimolecular process, two reactant molecules come together to form an activated complex or transition state. The transition state represents the highest energy state of the system during the reaction and is the point at which the reaction can proceed either forward or backward. During the formation of the activated complex, the reactants come together and get organized into a more ordered structure compared to their initial state. This leads to a decrease in the entropy of the system.
03

Determining the sign of the entropy of activation

Since the entropy of the system decreases during the process of forming the activated complex in a bimolecular reaction, we can conclude that the entropy of activation for a bimolecular process is usually negative. The negative value indicates that the transition state of the system has a lower entropy than that of the separate reactants.

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Most popular questions from this chapter

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of \(76.0 \mathrm{kPa} ?(\mathbf{b})\) If the final pressure on the gas is \(121.6 \mathrm{kPa}\), calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

(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) Which of the thermodynamic quantities \(p, H, q, w,\) and \(G\) are state functions? (b) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(q_{\text {rev }}\) and \(q_{\text {irtev }}\) (c) Consider a system going from state 1 to state 2 in a reversible and an irreversible way. Compare \(w_{\text {rev }}\) and \(w_{\text {trev }}\). (d) For a reversible isothermal process, write an expression for \(\Delta H\) and an expression for \(\Delta G\) in terms of \(q, w\) and \(T, p\) and \(\Delta V\).

Indicate whether each statement is true or false. (a) A reaction that is spontaneous in one direction will be nonspontaneous in the reverse direction under the same reaction conditions. (b) All spontaneous processes are fast. (c) Most spontaneous processes are reversible. (d) An isothermal process is one in which the system loses no heat. (e) The maximum amount of work can be accomplished by an irreversible process rather than a reversible one.

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