AP Chemistry Flashcards: Pre Equilibrium Approximation

What this deck covers

This deck focuses on Pre Equilibrium Approximation, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the main condition for applying pre-equilibrium approximation?

Answer: The early step reaches equilibrium quickly. Fast reversible step must equilibrate before slow step proceeds.

Flashcard 2: What is a critical assumption about intermediate concentration in pre-equilibrium?

Answer: It remains nearly constant. Equilibrium condition maintains stable intermediate concentration.

Flashcard 3: 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 4: 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 5: What is the main condition for applying pre-equilibrium approximation?

Answer: The early step reaches equilibrium quickly. Fast reversible step must equilibrate before slow step proceeds.

Flashcard 6: Identify the technique used to simplify complex reaction mechanisms.

Answer: Pre-equilibrium approximation. Method for deriving rate laws from multi-step mechanisms.

Flashcard 7: What happens to intermediates in pre-equilibrium approximation?

Answer: They quickly reach equilibrium with reactants. Fast equilibrium maintained throughout reaction course.

Flashcard 8: What is one advantage of using pre-equilibrium approximation?

Answer: Simplifies calculation of rate laws. Reduces complex mechanisms to simple rate expressions.

Flashcard 9: 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 10: How can pre-equilibrium approximation help in kinetics?

Answer: Allows simplification of complex mechanisms. Converts complex mechanisms into manageable rate expressions.

Flashcard 11: 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 12: What is the first step in applying pre-equilibrium approximation?

Answer: Identify the fast equilibrium step. Required before deriving rate law from mechanism.

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: Identify the step in a mechanism where pre-equilibrium is assumed.

Answer: The fast initial step. The reversible step that reaches equilibrium quickly.

Flashcard 15: 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 16: In pre-equilibrium approximation, what does the rate of formation equal?

Answer: Rate of formation equals rate of consumption. Equilibrium condition for the intermediate in fast step.

Flashcard 17: State the typical rate law form derived using pre-equilibrium approximation.

Answer: Rate = $k'[\text{reactants}]^{\text{coefficients}}$. Combined rate and equilibrium constants for overall reaction.

Flashcard 18: What is the equilibrium constant expression in pre-equilibrium?

Answer: $K = \frac{[\text{products}]}{[\text{reactants}]}$. Ratio of product to reactant concentrations at equilibrium.

Flashcard 19: What form does the equilibrium constant take in pre-equilibrium approximation?

Answer: $K = \frac{[\text{products}]}{[\text{reactants}]}$. Standard equilibrium expression for the fast step.

Flashcard 20: How does pre-equilibrium approximation affect reaction mechanisms?

Answer: Simplifies complex mechanisms into rate laws. Makes complex multi-step reactions mathematically tractable.

Flashcard 21: In pre-equilibrium, what is the role of the intermediate species?

Answer: Forms and consumes quickly to establish equilibrium. Maintains equilibrium between formation and decomposition.

Flashcard 22: 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 23: What typically characterizes the initial step in pre-equilibrium?

Answer: It is rapid and reversible. Fast forward and reverse rates establish equilibrium.

Flashcard 24: Which assumption in pre-equilibrium approximation relates to reaction speed?

Answer: Initial step is much faster than subsequent steps. Fast equilibrium established before rate-determining step.

Flashcard 25: Why is the pre-equilibrium approximation useful in complex reactions?

Answer: Simplifies the rate law expression. Eliminates need for complex intermediate concentration terms.

Flashcard 26: What equation represents the equilibrium expression in pre-equilibrium?

Answer: $K = \frac{[C]}{[A][B]}$ for $A + B \rightleftharpoons C$. Products over reactants for the equilibrium step.

Flashcard 27: In pre-equilibrium, what balances the rate of intermediate formation and consumption?

Answer: Equilibrium condition. Forward and reverse rates equal in fast step.

Flashcard 28: 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.

Flashcard 29: What is the relationship between $k'$ and $K$ in a derived rate law?

Answer: $k' = k \times K$. Effective rate constant combines kinetic and equilibrium factors.

Flashcard 30: 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 31: What is a key characteristic of the intermediate in pre-equilibrium?

Answer: Rapidly forms and decomposes. Fast equilibration between formation and decomposition reactions.

Flashcard 32: What is an incorrect use of pre-equilibrium approximation?

Answer: Applying to irreversible reactions. Requires reversible steps to establish equilibrium conditions.

Flashcard 33: How is the rate constant $k'$ related to $k$ and $K$ in pre-equilibrium?

Answer: $k' = k \times K$. Product of rate constant and equilibrium constant.

Flashcard 34: 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 35: What is the effect of pre-equilibrium on intermediate concentration?

Answer: Intermediate concentration is assumed constant. Equilibrium maintains constant intermediate levels.

Flashcard 36: 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 37: 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 38: What is the purpose of using pre-equilibrium approximation?

Answer: To simplify rate law derivation. Avoids complex intermediate expressions in rate laws.

Flashcard 39: What is a common misconception about pre-equilibrium approximation?

Answer: That it applies to irreversible reactions. Actually requires reversible steps to establish equilibrium.

Flashcard 40: 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 41: Identify the technique used to simplify complex reaction mechanisms.

Answer: Pre-equilibrium approximation. Method for deriving rate laws from multi-step mechanisms.

Flashcard 42: What is the effect of pre-equilibrium on intermediate concentration?

Answer: Intermediate concentration is assumed constant. Equilibrium maintains constant intermediate levels.

Flashcard 43: What equation represents the equilibrium expression in pre-equilibrium?

Answer: $K = \frac{[C]}{[A][B]}$ for $A + B \rightleftharpoons C$. Products over reactants for the equilibrium step.

Flashcard 44: Which assumption in pre-equilibrium approximation relates to reaction speed?

Answer: Initial step is much faster than subsequent steps. Fast equilibrium established before rate-determining step.

Flashcard 45: Identify the error: 'Pre-equilibrium leads to exact rate laws.'

Answer: Correct: 'leads to approximate rate laws.'. Provides useful approximations for complex reaction systems.

Flashcard 46: Which assumption is critical for pre-equilibrium approximation?

Answer: Intermediate is in equilibrium with reactants. Fast step equilibrium before slow step determines rate.