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👩🏽‍🔬Honors Chemistry

Kinetics Equations

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Why This Matters

Kinetics is where chemistry gets predictive—it's not enough to know what products form, you need to understand how fast they form and why. On your Honors Chemistry exam, you're being tested on your ability to connect mathematical relationships to molecular behavior: why does doubling concentration sometimes double the rate and sometimes quadruple it? Why does a 10°C temperature increase often double reaction speed? These equations aren't just formulas to memorize—they're tools for explaining reaction mechanisms, energy barriers, and molecular collisions.

The equations in this guide fall into distinct categories: rate laws that describe concentration dependence, integrated rate laws that track concentration over time, and temperature relationships that explain why heat speeds things up. Don't just memorize the math—know what each equation tells you about molecular behavior. When you see a kinetics problem, ask yourself: Am I finding a rate? Tracking concentration change? Comparing temperatures? That question points you to the right equation every time.


Rate Laws: How Concentration Drives Reaction Speed

Rate laws connect what you can measure (concentration) to what you want to know (how fast the reaction proceeds). The mathematical relationship between concentration and rate reveals the reaction mechanism.

General Rate Law

  • rate=k[A]m[B]n\text{rate} = k[A]^m[B]^n—the foundation of all kinetics calculations, showing rate as a function of reactant concentrations
  • Reaction orders (m and n) must be determined experimentally; they are not the same as stoichiometric coefficients
  • Rate constant k is temperature-dependent and unique to each reaction—it's your bridge to the Arrhenius equation

Zero-Order Integrated Rate Law

  • [A]=kt+[A]0[A] = -kt + [A]_0—concentration decreases linearly with time, giving a straight line when plotting [A][A] vs. tt
  • Rate is constant regardless of how much reactant remains; often seen when a catalyst surface is saturated
  • Slope = k-k when graphed, making experimental determination of the rate constant straightforward

First-Order Integrated Rate Law

  • ln[A]=kt+ln[A]0\ln[A] = -kt + \ln[A]_0—plot ln[A]\ln[A] vs. tt for a straight line; this is your go-to for radioactive decay problems
  • Rate depends on one reactant's concentration; doubling [A][A] doubles the rate
  • Most common reaction order on exams—if you're unsure, check first-order graphing first

Second-Order Integrated Rate Law

  • 1[A]=kt+1[A]0\frac{1}{[A]} = kt + \frac{1}{[A]_0}—plot 1[A]\frac{1}{[A]} vs. tt for a straight line with positive slope
  • Rate depends on concentration squared; doubling [A][A] quadruples the rate
  • Slope = kk (positive, unlike zero and first order), which helps distinguish it graphically

Compare: First-order vs. second-order reactions—both depend on concentration, but first-order plots ln[A]\ln[A] while second-order plots 1[A]\frac{1}{[A]}. On graphing questions, try both transformations and see which gives a straight line.


Half-Life: Timing Reaction Progress

Half-life equations let you predict how long reactions take without tracking exact concentrations. The mathematical form of half-life reveals the reaction order.

First-Order Half-Life

  • t1/2=ln(2)kt_{1/2} = \frac{\ln(2)}{k}—half-life is constant regardless of starting concentration, a unique feature of first-order kinetics
  • Independent of [A]0[A]_0 means each successive half-life takes the same amount of time
  • Essential for radioactive decay and drug metabolism calculations—if half-life stays constant, the reaction is first-order

Compare: Zero-order half-life (t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}) depends on initial concentration and gets shorter as the reaction proceeds. First-order half-life stays constant. This difference is a common exam question—if they give you changing half-lives, it's not first-order.


Temperature and Energy: Why Heat Speeds Reactions

These equations explain the molecular reason behind the rule "reactions go faster when heated." Temperature increases the fraction of molecules with enough energy to overcome the activation energy barrier.

Arrhenius Equation

  • k=AeEa/RTk = Ae^{-E_a/RT}—the master equation connecting rate constant to temperature and activation energy
  • Pre-exponential factor A represents collision frequency and proper molecular orientation; it's the "best-case scenario" rate
  • Exponential term shows that small changes in EaE_a or TT cause large changes in kk—this is why catalysts are so powerful

Two-Temperature Arrhenius Form

  • k2k1=eEa/R(1/T21/T1)\frac{k_2}{k_1} = e^{-E_a/R(1/T_2 - 1/T_1)}—use this when comparing rate constants at two different temperatures
  • Rearranges to lnk2k1=EaR(1T11T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) for easier calculation
  • Common exam application: given kk at two temperatures, solve for EaE_a, or predict kk at a new temperature

Maxwell-Boltzmann Distribution

  • f(E)=4π(m2πkT)3/2E1/2eE/kTf(E) = 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} E^{1/2} e^{-E/kT}—describes how molecular energies are distributed at a given temperature
  • Only molecules above EaE_a can react; the area under the curve past EaE_a represents reactive collisions
  • Higher temperature shifts the curve right and flattens it, increasing the fraction of molecules that can overcome the activation barrier

Compare: Arrhenius equation vs. Maxwell-Boltzmann—Arrhenius tells you how much the rate constant changes with temperature, while Maxwell-Boltzmann explains why (more molecules have sufficient energy). FRQs often ask you to connect these concepts qualitatively.


Catalysis and Collision Theory: The Molecular View

These relationships explain reaction rates in terms of what molecules actually do—collide, orient, and overcome energy barriers.

Collision Theory Equation

  • k=pZeEa/RTk = pZe^{-E_a/RT}—rate constant depends on collision frequency (ZZ), orientation factor (pp), and the Boltzmann factor
  • Steric factor p accounts for the fact that molecules must collide in the correct orientation; typically much less than 1
  • Collision frequency Z increases with temperature and concentration, explaining part of why reactions speed up when heated

Catalyst Effect on Activation Energy

  • Ea(catalyst)<Ea(uncatalyzed)E_a(\text{catalyst}) < E_a(\text{uncatalyzed})—catalysts provide an alternative reaction pathway with a lower energy barrier
  • Catalysts are not consumed and don't appear in the overall rate law; they work by stabilizing the transition state
  • Exponential impact: even a small decrease in EaE_a dramatically increases kk due to the exponential relationship in Arrhenius

Compare: Increasing temperature vs. adding a catalyst—both speed up reactions, but temperature increases molecular energy while catalysts lower the energy barrier. On the Maxwell-Boltzmann curve, heating shifts the distribution right; catalysis moves the EaE_a line left.


Quick Reference Table

ConceptKey Equations
Rate dependence on concentrationrate=k[A]m[B]n\text{rate} = k[A]^m[B]^n
Zero-order kinetics[A]=kt+[A]0[A] = -kt + [A]_0; plot [A][A] vs. tt
First-order kineticsln[A]=kt+ln[A]0\ln[A] = -kt + \ln[A]_0; t1/2=ln2kt_{1/2} = \frac{\ln 2}{k}
Second-order kinetics1[A]=kt+1[A]0\frac{1}{[A]} = kt + \frac{1}{[A]_0}
Temperature-rate relationshipk=AeEa/RTk = Ae^{-E_a/RT}; two-point form for comparing temperatures
Energy distributionMaxwell-Boltzmann distribution
Molecular collision requirementsk=pZeEa/RTk = pZe^{-E_a/RT}
Catalyst mechanismLowers EaE_a, provides alternative pathway

Self-Check Questions

  1. You plot concentration data three ways: [A][A] vs. tt, ln[A]\ln[A] vs. tt, and 1[A]\frac{1}{[A]} vs. tt. Only the ln[A]\ln[A] plot is linear. What is the reaction order, and how would you find kk from this graph?

  2. A first-order reaction has a half-life of 20 minutes. If you start with 0.80 M reactant, what concentration remains after 60 minutes? How would your approach change if this were zero-order?

  3. Compare and contrast how increasing temperature and adding a catalyst both increase reaction rate. Use the Arrhenius equation and Maxwell-Boltzmann distribution in your explanation.

  4. Two reactions have the same pre-exponential factor A, but Reaction 1 has Ea=50 kJ/molE_a = 50 \text{ kJ/mol} and Reaction 2 has Ea=75 kJ/molE_a = 75 \text{ kJ/mol}. Which reaction is more sensitive to temperature changes? Explain using the Arrhenius equation.

  5. Given rate constant values at 300 K and 350 K, describe the steps you would use to calculate the activation energy. Which form of the Arrhenius equation would you use, and why?