AP Chemistry Flashcards: Introduction To Rate Law

What this deck covers

This deck focuses on Introduction To Rate Law, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

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: Identify the rate factor change if $[A]$ is doubled in $\text{rate}=k[A]^3$.

Answer: $8\times$. Rate multiplies by $2^3 = 8$ when concentration doubles.

Flashcard 2: What are the units of $k$ for a zero-order rate law $\text{rate}=k$?

Answer: $\text{mol}\,\text{L}^{-1}\,\text{s}^{-1}$. Zero-order $k$ has same units as rate itself.

Flashcard 3: What are the units of $k$ for a second-order rate law $\text{rate}=k[A]^2$?

Answer: $\text{L}\,\text{mol}^{-1}\,\text{s}^{-1}$. Second-order $k$ needs inverse concentration and time units.

Flashcard 4: What is the general form of a rate law for $aA+bB\rightarrow$ products using orders $m$ and $n$?

Answer: $\text{rate}=k[A]^m[B]^n$. Shows how rate depends on concentrations raised to experimental powers.

Flashcard 5: What does the rate constant $k$ represent in a rate law?

Answer: Proportionality constant relating rate to reactant concentrations. Links reaction rate to concentrations through multiplication.

Flashcard 6: What is the overall reaction order for $\text{rate}=k[A]^m[B]^n$?

Answer: $m+n$. Sum of all individual reaction orders gives overall order.

Flashcard 7: Which statement is correct: reaction orders come from stoichiometric coefficients or from experiment?

Answer: From experiment (not from overall stoichiometric coefficients). Orders must be determined experimentally, not from balanced equation.

Flashcard 8: What does it mean if a reactant is zero order in the rate law?

Answer: Rate is independent of that reactant’s concentration. Zero power means concentration changes don't affect rate.

Flashcard 9: What does it mean if a reactant is first order in the rate law?

Answer: Doubling its concentration doubles the rate. First power means rate changes proportionally with concentration.

Flashcard 10: What does it mean if a reactant is second order in the rate law?

Answer: Doubling its concentration quadruples the rate. Second power means rate changes with concentration squared.

Flashcard 11: What is the rate law for an elementary step $A+B\rightarrow$ products?

Answer: $\text{rate}=k[A][B]$. Elementary steps use stoichiometric coefficients as orders.

Flashcard 12: What is the rate law for an elementary unimolecular step $A\rightarrow$ products?

Answer: $\text{rate}=k[A]$. Single molecule reactions are first order in that reactant.

Flashcard 13: What is the rate law for an elementary termolecular step $A+B+C\rightarrow$ products?

Answer: $\text{rate}=k[A][B][C]$. Three-molecule collisions give third-order rate law.

Flashcard 14: What is the rate law for an elementary step $2A\rightarrow$ products?

Answer: $\text{rate}=k[A]^2$. Two identical molecules reacting gives second-order dependence.

Flashcard 15: What is the definition of rate in terms of concentration change for reactant $A$ in $aA\rightarrow$ products?

Answer: $\text{rate}=-\frac{1}{a}\frac{d[A]}{dt}$. Negative sign shows reactant concentration decreases over time.

Flashcard 16: What is the definition of rate in terms of concentration change for product $P$ in $\rightarrow pP$?

Answer: $\text{rate}=\frac{1}{p}\frac{d[P]}{dt}$. Positive derivative shows product concentration increases over time.

Flashcard 17: What are the units of rate for concentration in $\text{mol}\,\text{L}^{-1}$ and time in seconds?

Answer: $\text{mol}\,\text{L}^{-1}\,\text{s}^{-1}$. Rate measures concentration change per unit time.

Flashcard 18: What are the units of $k$ for a first-order rate law $\text{rate}=k[A]$?

Answer: $\text{s}^{-1}$. First-order $k$ has time inverse units only.

Flashcard 19: Identify the rate factor change if $[A]$ is halved in $\text{rate}=k[A]^2$.

Answer: $\frac{1}{4}\times$. Rate multiplies by $(\frac{1}{2})^2 = \frac{1}{4}$ when concentration halves.

Flashcard 20: Identify the overall order for $\text{rate}=k[A]^{\frac{1}{2}}[B]^{\frac{3}{2}}$.

Answer: $2$. Add exponents: $\frac{1}{2} + \frac{3}{2} = 2$.

Flashcard 21: What are the units of $k$ for a zero-order rate law $\text{rate} = k$?

Answer: $\text{M}\,\text{s}^{-1}$. Units: $k = \text{rate}$ since no concentration dependence.

Flashcard 22: Identify $m$ if doubling $[A]$ makes the rate double (with other reactants constant).

Answer: $m = 1$. Rate doubles means $(2)^m = 2$, so $m = 1$.

Flashcard 23: What is the method called that uses two trials to solve for reaction orders from rate data?

Answer: Method of initial rates. Compares initial rates at different concentrations to find orders.

Flashcard 24: Identify $m$ if doubling $[A]$ makes the rate quadruple (with other reactants constant).

Answer: $m = 2$. Rate quadruples means $(2)^m = 4$, so $m = 2$.

Flashcard 25: Identify $m$ if doubling $[A]$ makes the rate unchanged (with other reactants constant).

Answer: $m = 0$. Rate unchanged means $(2)^m = 1$, so $m = 0$.

Flashcard 26: What is the overall reaction order for $\text{rate} = k[A]^2[B]^1$?

Answer: $3$. Sum all exponents: $2 + 1 = 3$.

Flashcard 27: What does the rate constant $k$ represent in a rate law?

Answer: Proportionality constant relating rate to reactant concentrations. Links reaction rate to concentrations; temperature-dependent.

Flashcard 28: What is the general form of the rate law for $aA + bB \to$ products?

Answer: $\text{rate} = k[A]^m[B]^n$. Exponents $m$ and $n$ are reaction orders, not stoichiometric coefficients.

Flashcard 29: What are the units of reaction rate when concentration is in $\text{M}$ and time is in seconds?

Answer: $\text{M}\,\text{s}^{-1}$. Rate is change in concentration per unit time.

Flashcard 30: What are the units of $k$ for a first-order rate law $\text{rate} = k[A]$?

Answer: $\text{s}^{-1}$. Units: $\frac{\text{M}\,\text{s}^{-1}}{\text{M}} = \text{s}^{-1}$.