All flashcards
Flashcard 1: What do we assume about intermediate concentration in pre-equilibrium approximation?
Answer: Concentration remains constant over time. Equilibrium established faster than consumption in slow step.
Flashcard 2: In pre-equilibrium, what balances the rate of intermediate formation and consumption?
Answer: Equilibrium condition. Forward and reverse rates equal in fast step.
Flashcard 3: What is the first step in applying pre-equilibrium approximation?
Answer: Identify the fast equilibrium step. Required before deriving rate law from mechanism.
Flashcard 4: What role does the equilibrium constant K play in pre-equilibrium approximation?
Answer: Relates intermediate and reactant concentrations. Links intermediate concentration to initial reactants.
Flashcard 5: How is the rate constant k′ related to k and K in pre-equilibrium?
Answer: k′=k×K. Product of rate constant and equilibrium constant.
Flashcard 6: What is a critical assumption about intermediate concentration in pre-equilibrium?
Answer: It remains nearly constant. Equilibrium condition maintains stable intermediate concentration.
Flashcard 7: What is a key characteristic of the intermediate in pre-equilibrium?
Answer: Rapidly forms and decomposes. Fast equilibration between formation and decomposition reactions.
Flashcard 8: Which assumption is critical for pre-equilibrium approximation?
Answer: Intermediate is in equilibrium with reactants. Fast step equilibrium before slow step determines rate.
Flashcard 9: What do we assume about intermediate concentration in pre-equilibrium approximation?
Answer: Concentration remains constant over time. Equilibrium established faster than consumption in slow step.
Flashcard 10: Find the error: 'Pre-equilibrium approximation assumes steady state.'
Answer: Correct: 'assumes equilibrium, not steady state.'. Pre-equilibrium involves equilibrium, not steady-state conditions.
Flashcard 11: State the typical rate law form derived using pre-equilibrium approximation.
Answer: Rate = k′[reactants]coefficients. Combined rate and equilibrium constants for overall reaction.
Flashcard 12: What is the relationship between k′ and K in a derived rate law?
Answer: k′=k×K. Effective rate constant combines kinetic and equilibrium factors.
Flashcard 13: Find the error: 'Pre-equilibrium applies to irreversible reactions.'
Answer: Correct: 'applies to reversible reactions.'. Pre-equilibrium requires reversible steps to establish equilibrium.
Flashcard 14: What is one advantage of using pre-equilibrium approximation?
Answer: Simplifies calculation of rate laws. Reduces complex mechanisms to simple rate expressions.
Flashcard 15: Identify the error: 'Pre-equilibrium applies to steady-state conditions.'
Answer: Correct: 'applies to equilibrium conditions.'. Pre-equilibrium requires equilibrium conditions, not steady-state.
Flashcard 16: Why is the pre-equilibrium approximation useful in complex reactions?
Answer: Simplifies the rate law expression. Eliminates need for complex intermediate concentration terms.
Flashcard 17: Identify the step in a mechanism where pre-equilibrium is assumed.
Answer: The fast initial step. The reversible step that reaches equilibrium quickly.
Flashcard 18: What happens to intermediates in pre-equilibrium approximation?
Answer: They quickly reach equilibrium with reactants. Fast equilibrium maintained throughout reaction course.
Flashcard 19: Which condition must be met for using pre-equilibrium approximation?
Answer: Fast step reaches equilibrium before slow step occurs. Fast equilibrium must precede slow rate-determining step.
Flashcard 20: What typically characterizes the initial step in pre-equilibrium?
Answer: It is rapid and reversible. Fast forward and reverse rates establish equilibrium.
Flashcard 21: What is the principle of pre-equilibrium approximation?
Answer: Assumes early equilibrium in a reaction mechanism. Fast step reaches equilibrium before proceeding to slow step.
Flashcard 22: Identify the key condition for using pre-equilibrium in a mechanism.
Answer: Initial step is fast and reversible. Fast reversible step required for equilibrium assumption.
Flashcard 23: How does pre-equilibrium approximation affect reaction mechanisms?
Answer: Simplifies complex mechanisms into rate laws. Makes complex multi-step reactions mathematically tractable.
Flashcard 24: What is an incorrect use of pre-equilibrium approximation?
Answer: Applying to irreversible reactions. Requires reversible steps to establish equilibrium conditions.
Flashcard 25: What is the role of K in pre-equilibrium rate laws?
Answer: Relates initial concentrations and intermediate. Connects initial species concentrations to intermediate levels.
Flashcard 26: State the main difference in assumptions between pre-equilibrium and steady-state.
Answer: Equilibrium vs. constant concentration assumptions. Different underlying principles for each approximation method.
Flashcard 27: What is the equilibrium constant expression in pre-equilibrium?
Answer: K=[reactants][products]. Ratio of product to reactant concentrations at equilibrium.
Flashcard 28: In pre-equilibrium, how is the overall rate law determined?
Answer: Derived from the slow step using equilibrium expressions. Rate law comes from slow step with equilibrium substitution.
Flashcard 29: How can pre-equilibrium approximation help in kinetics?
Answer: Allows simplification of complex mechanisms. Converts complex mechanisms into manageable rate expressions.
Flashcard 30: What is the primary difference between steady-state and pre-equilibrium approximations?
Answer: Steady-state assumes constant intermediate concentration. Pre-equilibrium assumes equilibrium, not steady state.