Reaction Rates, Rate Laws, and Differential Method

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.

Generic form: \text{rate}=k[\text{reactant}1]^m[\text{reactant}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.

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).

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.

Balanced equations represent thermodynamic quantities (how much product forms, not how fast).
Example showing mismatch:
- Balanced: 3A+2B\to A3B2.
- 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.

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[A3B2]}{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).

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).
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.

Pick two trials where one reactant changes and the other(s) stay constant.
Generic ratio equation: \left(\dfrac{[A]\text{high}}{[A]\text{low}}\right)^{x}=\dfrac{\text{rate}\text{high}}{\text{rate}\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).

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.

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}).

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}.

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}.

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.

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.

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.

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.