Reaction Rates, Rate Laws, and Differential Method

1. Overview of Quantitative Rate-Determination Techniques

• Three broad approaches (only 2 covered in this video):

  • Integral (calculus–based) rate analysis: tangent to a concentration-time curve; AP Chem mentions it but does not require performing the calculus.

  • Rate law derived from the balanced chemical equation (stoichiometric relationships of disappearance/appearance).

  • Differential (initial-rate) method based on lab data; main computational focus of the lesson.

  • Next video will add reaction mechanisms & molecularity of the rate-determining step.

2. Rate Laws: Form, Terminology, & Experimental Origin

• Generic form: $\text{rate}=k[\text{reactant}<em>1]^m[\text{reactant}</em>2]^n\cdots$

  • Square brackets denote molar concentration (mol·L⁻¹).

  • $m, n,\dots$ = reaction orders (must be found experimentally, never lifted from coefficients).

  • The expression is interchangeably called rate law, rate equation, or rate expression.

• Five factors influence rate; concentration is the easiest to manipulate in lab, so most rate laws focus on it.

3. The Rate Constant ($k$)

• Encapsulates the other four factors (temperature, catalyst, surface area, nature of reactants).

• For a given reaction:

  • One unique $k$ value exists at each specific temperature.

  • Changing temperature or adding a catalyst changes $k$.

  • Hence an infinite set of $k$’s (one for every possible temperature).

4. Reaction Orders & Their Algebraic Meaning

• First order ($m=1$): doubling a reactant doubles the rate (1 : 1 relationship).

  • $\text{rate}\propto [A]^1 \Rightarrow\text{doubling }[A]\to 2^1=2$ times faster.

• Second order ($m=2$): rate scales with the square of the change.

  • Doubling $[A]$ gives $2^2=4$× rate; tripling → $3^2=9$×; dodecadupling ($\times12$) → $144$×.

• Zero order ($m=0$): reactant concentration has no effect; $[A]^0=1$.

• Overall order = sum of individual orders (used when assigning units to $k$).

• Collision perspective (preview for next video): orders correlate with the molecularity of the slow (rate-determining) elementary step, not with stoichiometric coefficients.

5. Thermodynamics vs Kinetics: Why Stoichiometric Coefficients Don’t Dictate the Rate Law

• Balanced equations represent thermodynamic quantities (how much product forms, not how fast).

• Example showing mismatch:

  • Balanced: $3A+2B\to A<em>3B</em>2$.

  • Possible experimental rate law: $\text{rate}=k[A][B]$ (both first order) or $k[A]^2$ (second order in A, zero order in B). Coefficients 3 & 2 have no automatic bearing on exponents.

6. Relating Measured Rate to Stoichiometry

• Rate can be expressed via any participant; signs/coefficients standardize them:
  $-\dfrac{1}{3}\,\dfrac{d[A]}{dt} = -\dfrac{1}{2}\,\dfrac{d[B]}{dt} = \dfrac{d[A<em>3B</em>2]}{dt}$

• Negative signs = “disappearing” reactants; positive = “appearing” products.

• Multipliers (1/3, 1/2) make all three expressions numerically equal.

• Important AP multiple-choice nuance: negatives merely indicate direction; any equalities are valid as long as directionality is consistent (they might place the minus sign in front of product to trick you).

7. Conceptual Concentration-Change Examples

1. Reaction is 2nd order in $A$, 0th in $B$.

   • Rate law: $k[A]^2$.

   • Doubling both $[A]$ and $[B]$ → rate increases by $2^2=4$× (only $A$ matters).

2. Reaction is 2nd order in $A$, 1st in $B$.

   • Rate law: $k[A]^2[B]$.

   • Tripling $A$ and doubling $B$ → $3^2\times2=18$× faster.

• Strategy for mental checks: set $k=1$, choose simple “baseline” concentrations (usually $1\text{ M}$) so the math is transparent.

8. Differential (Initial-Rate) Method

8.1 Determining Individual Orders

• Pick two trials where one reactant changes and the other(s) stay constant.

• Generic ratio equation: $\left(\dfrac{[A]<em>\text{high}}{[A]</em>\text{low}}\right)^{x}=\dfrac{\text{rate}<em>\text{high}}{\text{rate}</em>\text{low}}$

  • Solve exponent $x$ → order in $A$ (repeat for each reactant).

  • Works even when numbers are not neat integers (on AP/Honors data often round to whole orders).

8.2 Example 1: Both Reactants 3rd Order (Overall 6th)

• Data (mol·L⁻¹):

  • Trial 1 $[A]=2\,,\,[B]=2\,,\,\text{rate}=2.02$

  • Trial 2 $[A]=4\,,\,[B]=2\,,\,\text{rate}=16.15$

  • Trial 3 $[A]=2\,,\,[B]=6\,,\,\text{rate}=54.57$

• Using trials 1&2 (vary $A$ only): $2^x=8\Rightarrow x=3$.

• Using trials 1&3 (vary $B$ only): $3^x=27\Rightarrow x=3$.

• Rate law: $\text{rate}=k[A]^3[B]^3$.

• Overall order $=3+3=6$.

8.3 Calculating the Rate Constant $k$

• Rearranged: $k=\dfrac{\text{rate}}{[A]^3[B]^3}$.

• Substitute any complete trial (e.g., Trial 1):
  $k=\dfrac{2.02}{2^3\times2^3}=0.032\;\text{L}^5\,\text{mol}^{-5}\,\text{s}^{-1}$.

• Units rule: for overall order $n$,
  $k:\; \text{L}^{\,(n-1)}\,\text{mol}^{-(n-1)}\,\text{s}^{-1}$ (here $n=6\Rightarrow\text{L}^5\text{mol}^{-5}\text{s}^{-1}$).

8.4 Predicting a New Rate

• Plug desired $[A]=3.2\,\text{M},\,[B]=5.3\,\text{M}$:
  $\text{rate}=0.032\times3.2^3\times5.3^3\approx1.56\times10^{2}\;\text{mol·L}^{-1}\text{s}^{-1}$.

8.5 Example 2: 2nd Order in $A$, Zero Order in $B$

• Orders obtained: $m=2,\,n=0,$ overall $n_{\text{tot}}=2$.

• Rate law: $\text{rate}=k[A]^2$.

• Units for $k$: $\text{L·mol}^{-1}\text{s}^{-1}$.

• Calculated $k=4.02\,\text{L·mol}^{-1}\text{s}^{-1}$.

• New conditions $[A]=2.4\,\text{M}$ (ignore $B$): $\text{rate}=4.02\times2.4^2\approx23\;\text{mol·L}^{-1}\text{s}^{-1}$.

8.6 Example 3: 1st Order in $A$, 2nd in $B$

• Rate law: $k[A][B]^2$ (overall 3rd order).

• $k=0.016\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}$ (units $\text{L}^{n-1}$ rule).

• Predictive calculation yielded $\approx8.8\times10^{-2}\,\text{mol·L}^{-1}\text{s}^{-1}$ for given concentrations.

9. Shortcut for $k$ Units

• Memorize: $k$’s units flip the $\text{mol/L}$ fraction and reduce the exponent by one.

  • Overall order $n$ ⇒ $k$ units $=\text{L}^{(n-1)}\text{mol}^{-(n-1)}\text{s}^{-1}$.

• 50 % of AP free-response credit often assigned just for correct units.

10. Frequent AP/Honors Traps & Pro Tips

• Don’t infer exponents from coefficients; always use data or mechanism.

• Zero-order species still appear in the balanced equation yet have no kinetic influence; raising any number to the 0 power gives 1.

• When comparing disappearance/appearance rates, coefficients must normalize the numerical values.

• Negative sign placement in rate equality statements only indicates direction, not magnitude; AP may test this.

• Always square/cube the numeric value when an exponent is present—students often write $[B]^2$ in the algebra but forget to square the measured number.

• Use ultraconvenient values (e.g., $k=1$, concentrations $=1\,\text{M}$) for conceptual “factor change” problems.

11. Looking Ahead

• Next lesson: deriving rate laws from multi-step mechanisms, understanding molecularity, and linking orders to the slow (rate-determining) elementary step.

All equations employ LaTeX formatting per instructions; units consistently shown to solidify dimensional analysis practice.