Mastering the Arrhenius Equation & Reaction Kinetics
A comprehensive guide to understanding the Arrhenius equation and solving zero and first-order reaction kinetics problems. Enhance your problem-solving skills with step-by-step solutions and real-world applications.
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Importance of Reaction Kinetics in Chemistry
Predict Reaction Behaviors
Kinetics provides quantitative tools to predict how quickly reactants transform into products, essential for industrial process optimization and quality control.
Design Chemical Processes
Understanding reaction rates allows chemists to design efficient reaction conditions, saving energy and resources in manufacturing.
Drug Development
Pharmaceutical companies rely on kinetics to determine drug stability, shelf-life, and metabolic pathways in the body.
Overview: Zero and First Order Reactions
Zero Order Reactions
Rate is independent of reactant concentration
Linear decrease in concentration over time
Rate = k (constant)
Half-life increases as reaction proceeds
First Order Reactions
Rate is directly proportional to reactant concentration
Exponential decrease in concentration over time
Rate = k[A]
Constant half-life regardless of concentration
The Arrhenius Equation: Definition
k = Ae^(-Ea/RT)
The Arrhenius equation mathematically describes how reaction rates depend on temperature and activation energy. Developed by Swedish chemist Svante Arrhenius in 1889, it establishes that reaction rates increase exponentially with temperature and decrease exponentially with higher activation energy barriers.
This fundamental relationship helps explain why heating accelerates chemical reactions and provides a quantitative framework for predicting rate changes under varying conditions.
Physical Meaning of the Arrhenius Equation
Thermal Energy Distribution
Molecules possess a distribution of kinetic energies; higher temperatures mean more molecules have sufficient energy to react.
Energy Barrier Concept
The equation visualizes reactions requiring molecules to overcome an energy "hill" (activation energy) to transform into products.
Molecular Collision Theory
Reflects the probability of molecular collisions with sufficient energy and proper orientation to result in product formation.
Variables in the Arrhenius Equation
Activation Energy: Concept and Significance
Energy Threshold
Minimum energy required for reactants to form products
Reaction Control
Higher Ea means more temperature-sensitive reactions
Catalyst Effect
Catalysts lower Ea, increasing reaction rate without being consumed
Experimental Determination
Calculated from rate measurements at different temperatures
The Role of Temperature in Reaction Rates
Low Temperature
Fewer molecules possess sufficient energy to overcome Ea
Reaction proceeds slowly
Moderate Temperature
More molecules exceed activation threshold
Moderate reaction rate
High Temperature
Significant fraction of molecules exceed Ea
Reaction proceeds rapidly
Pre-exponential Factor (A) Explained
10^10
Collision Frequency
Typical order of magnitude (s⁻¹) for molecular collisions
10⁻²
Orientation Factor
Fraction of collisions with proper molecular alignment
10¹³
Overall Factor Range
Common range for pre-exponential factor in solution reactions
The pre-exponential factor A represents the frequency of molecular collisions and the probability of favorable orientation between colliding molecules. It's sometimes called the "frequency factor" and remains relatively constant with temperature changes, unlike the exponential term.
Arrhenius Equation: Mathematical Form

Standard Form
k = Ae^(-Ea/RT)
Logarithmic Form
ln(k) = ln(A) - Ea/RT
Linear Form
ln(k) = -Ea/R × (1/T) + ln(A)
The logarithmic transformation converts the exponential relationship to a linear form, making it easier to analyze experimental data. This allows chemists to plot ln(k) against 1/T, creating a straight line with slope -Ea/R.
Arrhenius Plot: log(k) vs 1/T
The Arrhenius plot transforms the exponential relationship between temperature and rate constant into a linear form by plotting ln(k) versus 1/T. This straight-line graph has a slope equal to -Ea/R, providing a straightforward method for determining activation energy from experimental data.
Calculating Activation Energy using Arrhenius Graphs
Plot Experimental Data
Create a graph of ln(k) versus 1/T using rate constants measured at different temperatures.
Determine Slope
Calculate the slope of the best-fit line through data points (Slope = -Ea/R).
Calculate Activation Energy
Multiply the negative of the slope by the gas constant: Ea = -slope × R
For example: With slope = -12,000 K, Ea = 12,000 K × 8.314 J/(mol·K) = 99.8 kJ/mol
Zero Order Reactions: Definition
Rate = k
In zero-order reactions, the rate of reaction is independent of the concentration of reactants. This means the reaction proceeds at a constant rate until the reactant is depleted, regardless of how much reactant remains in the system.
The rate is determined solely by the rate constant k, which represents the constant amount of reactant being consumed per unit time. This behavior typically occurs when a critical factor other than concentration limits the reaction rate.
Characteristics of Zero Order Reactions
Linear Concentration Decrease
Reactant concentration decreases linearly with time, not exponentially as in higher-order reactions.
Variable Half-Life
Half-life increases as the reaction proceeds, unlike first-order reactions where it remains constant.
Saturation Phenomena
Often occurs when a catalyst or enzyme is saturated, or when a physical process like dissolution limits the reaction rate.
Reaction Completion
A zero-order reaction will suddenly stop when reactant is completely consumed, rather than gradually slowing.
Integrated Rate Law for Zero Order Reactions
The integrated rate law for zero-order reactions is [A]t = [A]0 - kt, where [A]t is concentration at time t, [A]0 is initial concentration, and k is the rate constant. This results in a straight line when concentration is plotted against time, with slope -k. Unlike other reaction orders, the half-life for zero-order reactions depends on the initial concentration: t1/2 = [A]0/(2k).
Units of Rate Constants: Zero Order
Concentration/Time
mol/(L·s) for reactions in solution
Mass/Time
g/s for heterogeneous reactions
Pressure/Time
atm/s for gas-phase reactions
Moles/Time
mol/s for reactions in a fixed volume
Since the rate in zero-order reactions equals k, the units of k must match the units of rate. This differs from other reaction orders where the rate constant units depend on the concentration unit and reaction order.
Graphical Representation: Zero Order Kinetics
Concentration vs. Time
A plot of [A] versus time yields a straight line with slope = -k
The y-intercept represents the initial concentration [A]₀
The x-intercept indicates the time for complete reaction
Rate vs. Concentration
A plot of reaction rate versus [A] gives a horizontal line
This confirms that rate is independent of concentration
The constant value of this line equals the rate constant k
Examples of Zero Order Reactions in Real Life
Enzyme Catalysis
When enzymes are saturated with substrate, adding more substrate doesn't increase reaction rate. Examples include alcohol dehydrogenase processing ethanol in the liver.
Photochemical Reactions
Reactions driven by light absorption often follow zero-order kinetics when light intensity (not reactant concentration) limits the rate.
Surface Catalysis
When all catalytic sites on a surface are occupied, the reaction proceeds at constant rate regardless of solution concentration.
First Order Reactions: Definition
Rate = k[A]
First-order reactions are characterized by rates directly proportional to the concentration of a single reactant. As reactant concentration decreases, the reaction rate decreases proportionally. This creates the distinctive exponential decay profile where the reaction slows progressively as reactants are consumed.
The proportionality constant k represents how quickly the reaction proceeds relative to the available concentration, with units of time⁻¹ (e.g., s⁻¹ or min⁻¹).
Characteristics of First Order Reactions

Exponential Decay
Concentration decreases exponentially, not linearly
Constant Half-Life
Time to halve concentration is independent of initial amount
Fixed Fraction Decay
Equal percentage of remaining reactant reacts in equal time periods
Asymptotic Approach
Theoretically never reaches zero concentration
Integrated Rate Law for First Order Reactions
The integrated rate law for first-order reactions is ln[A]t = ln[A]0 - kt, which can be rearranged to [A]t = [A]0e^(-kt). A plot of ln[A] versus time yields a straight line with slope -k. The half-life formula t1/2 = ln(2)/k ≈ 0.693/k is constant regardless of initial concentration, a key characteristic of first-order processes.
Units of Rate Constants: First Order
Time⁻¹
The units of k for first-order reactions are always the reciprocal of time units.
  • s⁻¹ (per second)
  • min⁻¹ (per minute)
  • h⁻¹ (per hour)
Mathematical Derivation
From Rate = k[A]:
  • Rate units: mol/(L·s)
  • [A] units: mol/L
  • Therefore k units: 1/s
Physical Meaning
Represents the fractional amount of reactant that converts per unit time.
  • k = 0.1 s⁻¹ means 10% reacts each second
  • k = 1.0 s⁻¹ means 63.2% reacts each second
Graphical Representation: First Order Kinetics
Concentration vs. Time
Plot of [A] vs. time shows exponential decay curve
Initially steep, gradually flattening
Asymptotically approaches zero
Linearized Form
Plot of ln[A] vs. time yields straight line
Slope equals -k (rate constant)
Y-intercept is ln[A]₀
When analyzing experimental data, the linearized plot is most useful for confirming first-order behavior and determining the rate constant. If data points form a straight line on the ln[A] vs. time plot, the reaction follows first-order kinetics.
Examples of First Order Reactions in Real Life
Radioactive Decay
The classic example of first-order kinetics, where nuclei decay at a rate proportional to the number of radioactive atoms remaining.
Drug Metabolism
Many pharmaceuticals break down in the body following first-order kinetics, with a constant percentage eliminated per time interval.
Organic Decomposition
Many decomposition reactions of organic compounds follow first-order kinetics, including ester hydrolysis and some isomerizations.
How Arrhenius Equation Connects to Rate Laws
Temperature Effect
Arrhenius equation predicts how k changes with temperature
Rate Constant (k)
The value determined from the Arrhenius equation
Rate Law
Uses k to calculate actual reaction rate based on concentrations
The Arrhenius equation provides the temperature-dependent rate constant (k) that appears in all rate laws. For zero-order reactions, Rate = k; for first-order, Rate = k[A]. This connection allows prediction of how reaction rates change with temperature regardless of reaction order.
Step-by-Step: Solving Arrhenius Equation Numericals
Identify Known Variables
Determine which parameters are given: k values, temperatures, activation energy, or pre-exponential factor.
Convert Units
Ensure temperature is in Kelvin and energy units are consistent (typically kJ/mol or J/mol).
Apply Appropriate Form
For comparing k at two temperatures: ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]
Solve Algebraically
Rearrange the equation to isolate the unknown variable and calculate the answer.
Common Mistakes with Arrhenius Calculations
Temperature Unit Errors
Using temperature in °C instead of K in the Arrhenius equation. Always convert to Kelvin by adding 273.15 to Celsius temperatures.
Logarithm Confusion
Mixing natural logarithm (ln) and base-10 logarithm (log) or failing to properly apply logarithm rules when manipulating the equation.
Unit Inconsistency
Using mixed units for activation energy and gas constant. If Ea is in kJ/mol, R should be 8.314×10⁻³ kJ/(mol·K); if Ea is in J/mol, R should be 8.314 J/(mol·K).
Sign Errors
Incorrectly handling the negative sign in the exponential term or when taking the slope from an Arrhenius plot.
Reciprocal Temperature Confusion
Inverting temperature fractions incorrectly when using the two-point form of the Arrhenius equation.
Zero Order Numericals: Basic Structure
Problem Framework
Typical problems involve finding k, time, concentration, or half-life using [A]t = [A]0 - kt
Formula Application
For concentration: [A]t = [A]0 - kt
For time: t = ([A]0 - [A]t)/k
For half-life: t1/2 = [A]0/(2k)
Graphical Analysis
Plot [A] vs t, determine slope (-k) and intercept ([A]0)
Verification
Check units, magnitudes, and whether answer makes physical sense
Practice: Zero Order - k Determination
Problem: From the data above, determine the rate constant for this zero-order reaction.
Solution: For zero-order reactions, [A]t = [A]0 - kt. Plotting [A] vs. t gives a straight line with slope = -k. From the data, k = change in [A]/change in time = (0.500 - 0.340)/(0 - 40) = 0.160/40 = 0.0040 mol/(L·min)
Practice: Zero Order - Amount Remaining
Problem:
A zero-order reaction has a rate constant of 0.015 mol/(L·min). If the initial concentration is 0.45 mol/L, what concentration remains after 20 minutes?
Given:
  • k = 0.015 mol/(L·min)
  • [A]0 = 0.45 mol/L
  • t = 20 min
Solution:
For a zero-order reaction: [A]t = [A]0 - kt
Substituting the values:
[A]20 = 0.45 mol/L - (0.015 mol/(L·min) × 20 min)
[A]20 = 0.45 mol/L - 0.30 mol/L
[A]20 = 0.15 mol/L
Practice: Zero Order - Time Calculation
Problem Statement
A zero-order decomposition has k = 0.025 mol/(L·min). How long will it take for the concentration to decrease from 0.50 mol/L to 0.20 mol/L?
Approach
Rearrange the integrated rate law [A]t = [A]0 - kt to solve for time: t = ([A]0 - [A]t)/k
Calculation
t = (0.50 mol/L - 0.20 mol/L)/(0.025 mol/(L·min)) = 0.30 mol/L ÷ 0.025 mol/(L·min) = 12 minutes
Practice: Zero Order - Graph Interpretation
Problem: The graphs above show experimental data plotted three different ways. Which plot indicates the reaction is zero-order, and what is the rate constant?
Solution: The first graph ([A] vs. time) shows a linear relationship, confirming zero-order kinetics. The slope of the line is -k = -0.010 mol/(L·min), so k = 0.010 mol/(L·min). The other plots show poor linear fits, ruling out first and second orders.
Practice: Zero Order with Arrhenius Equation
Problem:
A zero-order reaction has k = 0.015 mol/(L·s) at 25°C and activation energy Ea = 58.5 kJ/mol. Calculate the rate constant at 45°C.
Given:
  • k1 = 0.015 mol/(L·s) at T1 = 25°C (298 K)
  • Ea = 58.5 kJ/mol
  • T2 = 45°C (318 K)
  • R = 8.314×10⁻³ kJ/(mol·K)
Solution:
Using the Arrhenius equation in two-point form:
ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]
ln(k₂/0.015) = (58.5/8.314×10⁻³)[(1/298) - (1/318)]
ln(k₂/0.015) = 7037[(0.00336) - (0.00314)] = 7037(0.00022) = 1.55
k₂/0.015 = e^1.55 = 4.71
k₂ = 0.015 × 4.71 = 0.071 mol/(L·s)
First Order Numericals: Basic Structure
Problem Framework
Typical problems involve finding k, time, concentration, or half-life using [A]t = [A]0e^(-kt)
2
Formula Application
For concentration: [A]t = [A]0e^(-kt) or ln([A]t/[A]0) = -kt
For time: t = ln([A]0/[A]t)/k
For half-life: t1/2 = ln(2)/k
Graphical Analysis
Plot ln[A] vs t, determine slope (-k) and intercept (ln[A]0)
Verification
Check units (k in time⁻¹), magnitudes, and physical sense
Practice: First Order - Rate Constant from Data
Problem: From the data above, determine the rate constant for this first-order reaction.
Solution: For first-order reactions, ln[A]t = ln[A]0 - kt. The slope of the ln[A] vs. t plot equals -k. From the data, k = -slope = -(-0.0223 min⁻¹) = 0.0223 min⁻¹.
Practice: First Order - Half-life Calculations
0.693
Half-life Formula
t₁/₂ = ln(2)/k ≈ 0.693/k
75%
Decay After 2 Half-lives
25% of original concentration remains
87.5%
Decay After 3 Half-lives
12.5% of original concentration remains
Problem: A first-order reaction has a rate constant of 0.0462 min⁻¹. Calculate (a) the half-life and (b) how long it will take for 90% of the reactant to be consumed.
Solution: (a) t₁/₂ = ln(2)/k = 0.693/0.0462 min⁻¹ = 15.0 min. (b) For 90% decay, [A]/[A]₀ = 0.10. Using t = ln([A]₀/[A])/k = ln(1/0.10)/0.0462 = ln(10)/0.0462 = 2.303/0.0462 = 49.8 min.
Practice: First Order - Time for Decomposition
Problem Statement
The decomposition of N₂O₅ follows first-order kinetics with k = 4.80×10⁻⁴ s⁻¹. How long will it take for 65% of the initial sample to decompose?
Approach
For 65% decomposition, 35% remains. We need to find t where [A]t/[A]0 = 0.35 using the equation t = ln([A]0/[A]t)/k
Calculation
t = ln(1/0.35)/4.80×10⁻⁴ s⁻¹ = ln(2.857)/4.80×10⁻⁴ s⁻¹ = 1.05/4.80×10⁻⁴ s⁻¹ = 2.19×10³ s = 36.4 min
Practice: First Order - Percentage Remaining
Problem:
A first-order reaction has a half-life of 15.5 minutes. What percentage of the initial concentration remains after 42.0 minutes?
Given:
  • t₁/₂ = 15.5 min
  • t = 42.0 min
Solution:
First, find k using t₁/₂ = ln(2)/k
k = ln(2)/t₁/₂ = 0.693/15.5 min = 0.0447 min⁻¹
Now calculate percentage remaining using [A]t = [A]0e^(-kt):
[A]t/[A]0 = e^(-0.0447 min⁻¹ × 42.0 min) = e^(-1.88) = 0.153
Percentage remaining = 0.153 × 100% = 15.3%
Practice: First Order with Arrhenius Equation
Problem:
A first-order reaction has k = 2.75×10⁻³ s⁻¹ at 27°C. If the activation energy is 95.4 kJ/mol, calculate the rate constant at 57°C.
Given:
  • k₁ = 2.75×10⁻³ s⁻¹ at T₁ = 27°C (300 K)
  • Ea = 95.4 kJ/mol
  • T₂ = 57°C (330 K)
  • R = 8.314×10⁻³ kJ/(mol·K)
Solution:
Using the two-temperature form of the Arrhenius equation:
ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]
ln(k₂/2.75×10⁻³) = (95.4/8.314×10⁻³)[(1/300) - (1/330)]
ln(k₂/2.75×10⁻³) = 11,475[(0.00333) - (0.00303)] = 11,475(0.00030) = 3.44
k₂/2.75×10⁻³ = e^3.44 = 31.2
k₂ = 2.75×10⁻³ s⁻¹ × 31.2 = 8.58×10⁻² s⁻¹
Conceptual Questions: Arrhenius Equation
1
Doubling Temperature Effect
Why doesn't doubling the absolute temperature double the reaction rate? Explain using the Arrhenius equation.
2
Activation Energy Interpretation
A reaction with Ea = 75 kJ/mol proceeds twice as fast when temperature increases from 25°C to 35°C. Estimate the Ea for a similar reaction that requires a temperature increase from 25°C to 55°C to double its rate.
3
Catalysts and Arrhenius Parameters
How does a catalyst affect the values of A, Ea, and k in the Arrhenius equation? Which parameter changes most significantly?
4
Temperature Sensitivity Analysis
Why are reactions with higher activation energies more sensitive to temperature changes than those with lower activation energies?
Conceptual Questions: Zero Order Kinetics
Why can't all reactions be zero-order?
Zero-order reactions occur when a critical factor other than reactant concentration (like catalyst availability) limits the reaction rate. If all reactions were zero-order, they would proceed at constant rates regardless of concentration, violating mass action principles and making many chemical processes impossible.
Half-life in zero-order vs. first-order reactions
In zero-order reactions, half-life increases as the reaction progresses (t₁/₂ = [A]₀/2k) because the rate remains constant while the amount of reactant decreases. In first-order reactions, half-life is constant (t₁/₂ = ln(2)/k) regardless of concentration because the rate decreases proportionally with concentration.
Metabolizing alcohol: Why is it zero-order?
Alcohol metabolism in the liver follows zero-order kinetics because the enzyme alcohol dehydrogenase becomes saturated even at low blood alcohol concentrations. This means the body eliminates alcohol at a constant rate (typically about 15-20 mg/dL/hour) regardless of how much alcohol is in the bloodstream.
Conceptual Questions: First Order Kinetics
Why are many decomposition reactions first-order?
Decomposition reactions often involve the breakdown of a single molecule due to internal instability or energy absorption. Since the probability of a molecule decomposing depends only on its own properties and not on interactions with other molecules, the rate is proportional to the number of molecules present, creating first-order kinetics.
Significance of exponential decay
The exponential decay pattern in first-order reactions means the reaction slows progressively as reactants are consumed. This creates a self-regulating effect where the system never completely exhausts its reactants in finite time, mathematically approaching zero asymptotically. This pattern appears in many natural processes from radioactive decay to population decline.
Practical use of first-order half-life
The constant half-life of first-order reactions allows easy prediction of concentration at any time without complex calculations. For example, after 3 half-lives, 12.5% of the original substance remains (50% → 25% → 12.5%). This property is particularly useful in pharmaceutical dosing, radioactive dating, and product shelf-life determination.
Mixed Practice: Zero & First Order with Arrhenius Application
1
Problem
A reaction is first-order with k = 3.5×10⁻² min⁻¹ at 30°C. At 50°C, its half-life is 8.25 minutes. Determine the reaction's activation energy.
2
Identify Known Values
T₁ = 30°C = 303 K, k₁ = 3.5×10⁻² min⁻¹
T₂ = 50°C = 323 K, t₁/₂ = 8.25 min, which means k₂ = ln(2)/t₁/₂ = 0.693/8.25 = 8.4×10⁻² min⁻¹
3
Apply Arrhenius Equation
ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]
ln(8.4×10⁻²/3.5×10⁻²) = (Ea/8.314×10⁻³)[(1/303) - (1/323)]
4
Solve for Ea
ln(2.4) = (Ea/8.314×10⁻³)(0.00204)
0.876 = Ea × 0.00204/8.314×10⁻³
Ea = 0.876 × 8.314×10⁻³/0.00204 = 35.7 kJ/mol
Skill Builder: Temperature Dependence Questions
Problem:
A reaction rate doubles when the temperature is increased from 20°C to 35°C. By what factor will the rate increase if the temperature is raised from 20°C to 65°C?
Given:
  • k₂/k₁ = 2 when T₁ = 20°C (293 K) and T₂ = 35°C (308 K)
  • Need to find k₃/k₁ when T₃ = 65°C (338 K)
Solution:
Step 1: Calculate Ea using the known rate increase
ln(k₂/k₁) = (Ea/R)[(1/T₁) - (1/T₂)]
ln(2) = (Ea/8.314×10⁻³)[(1/293) - (1/308)]
0.693 = Ea × 0.00017/8.314×10⁻³
Ea = 34.0 kJ/mol
Step 2: Calculate the new rate factor
ln(k₃/k₁) = (34.0/8.314×10⁻³)[(1/293) - (1/338)]
ln(k₃/k₁) = 4091 × 0.00046 = 1.88
k₃/k₁ = e^1.88 = 6.55
Skill Builder: Rate Constant Manipulation
Identify Relationship
Zero-order: k₀ has units of concentration/time (e.g., mol/(L·s))
First-order: k₁ has units of time⁻¹ (e.g., s⁻¹)
Unit Conversion
Ensure consistent units for time (seconds, minutes, hours)
Convert concentration units if needed (mol/L, M, mM)
Apply Appropriate Equations
Zero-order half-life: t₁/₂ = [A]₀/(2k₀)
First-order half-life: t₁/₂ = ln(2)/k₁
Cross-Order Calculations
For a zero-order reaction to have the same half-life as a first-order reaction: k₀ = k₁[A]₀/ln(2)
Multiple Choice Questions
1
Identifying Reaction Order
A reaction has a half-life that doubles when the initial concentration is halved. The reaction order is:
A) Zero B) First C) Second D) Cannot be determined
2
Arrhenius Equation Interpretation
Which factor in the Arrhenius equation accounts for proper molecular orientation during collisions?
A) Temperature B) Pre-exponential factor C) Activation energy D) Rate constant
3
Rate Constant Units
The units of the rate constant for a zero-order reaction are:
A) s⁻¹ B) L/(mol·s) C) mol/(L·s) D) No units
4
Temperature Effect
A reaction with Ea = 50 kJ/mol has its temperature increased from 300K to 310K. The rate will approximately:
A) Double B) Triple C) Increase by 10% D) Increase by 50%
Short Answer Questions
What conditions might cause a reaction to follow zero-order kinetics?
A reaction may follow zero-order kinetics when: (1) A catalyst or enzyme is saturated with reactant; (2) A reactant is in such excess that its concentration remains effectively constant; (3) The reaction rate is limited by a physical process like dissolution or light absorption rather than chemical factors; (4) Only a fixed number of active sites are available for reaction on a surface catalyst.
How does the pre-exponential factor (A) relate to collision theory?
The pre-exponential factor represents the frequency of molecular collisions with the correct orientation for reaction. It accounts for: (1) The total collision frequency between reactant molecules; (2) The steric factor (probability of proper molecular orientation during collision); (3) Entropy effects related to molecular organization requirements. A higher A value indicates more frequent effective collisions.
Explain why reaction rates generally increase with temperature.
Reaction rates increase with temperature because: (1) Higher temperatures increase molecular kinetic energy, resulting in more frequent collisions; (2) The Maxwell-Boltzmann distribution shifts, creating a larger fraction of molecules with energy exceeding the activation energy; (3) The exponential term in the Arrhenius equation (e^(-Ea/RT)) becomes larger as T increases, significantly boosting the rate constant.
Problem Solving: Multi-step Numerical Practice
Problem
A first-order reaction has k = 2.5×10⁻² min⁻¹ at 25°C and Ea = 78.5 kJ/mol. Calculate:
a) The half-life at 25°C
b) The rate constant at 50°C
c) The time required for 80% completion at 50°C
Part A: Half-life Calculation
t₁/₂ = ln(2)/k = 0.693/(2.5×10⁻² min⁻¹) = 27.7 minutes
Part B: Rate Constant at New Temperature
Using Arrhenius equation with T₁ = 298K, T₂ = 323K:
ln(k₂/k₁) = (78.5/8.314×10⁻³)[(1/298) - (1/323)]
ln(k₂/2.5×10⁻²) = 9442 × 0.00026 = 2.45
k₂ = 2.5×10⁻² × e^2.45 = 0.289 min⁻¹
Part C: Time for 80% Completion
At 80% completion, [A]/[A]₀ = 0.20
t = ln([A]₀/[A])/k₂ = ln(1/0.20)/0.289 = ln(5)/0.289 = 1.61/0.289 = 5.57 minutes
Practice Set 1: Zero Order with Solutions
Zero Order Problem 1
A zero-order reaction has k = 0.045 mol/(L·min). How long will it take for the concentration to decrease from 0.250 mol/L to 0.100 mol/L?
Solution: t = ([A]₀-[A])/k = (0.250-0.100)/0.045 = 0.150/0.045 = 3.33 minutes
Zero Order Problem 2
Calculate the half-life of a zero-order reaction with k = 0.025 mol/(L·min) and [A]₀ = 0.50 mol/L.
Solution: t₁/₂ = [A]₀/(2k) = 0.50/(2×0.025) = 0.50/0.050 = 10.0 minutes
Zero Order Problem 3
A reaction is zero-order with [A]₀ = 0.75 mol/L. After 30 minutes, [A] = 0.45 mol/L. Calculate k.
Solution: k = ([A]₀-[A])/t = (0.75-0.45)/30 = 0.30/30 = 0.010 mol/(L·min)
Practice Set 2: First Order with Solutions
First Order Problem 1
A first-order reaction has k = 0.0521 min⁻¹. How long will it take for 75% of the reactant to be consumed?
Solution: For 75% consumption, [A]/[A]₀ = 0.25. t = ln([A]₀/[A])/k = ln(1/0.25)/0.0521 = ln(4)/0.0521 = 1.386/0.0521 = 26.6 minutes
First Order Problem 2
A first-order reaction has a half-life of 12.5 minutes. Calculate the rate constant and the time for 90% completion.
Solution: k = ln(2)/t₁/₂ = 0.693/12.5 = 0.0554 min⁻¹. For 90% completion, t = ln(10)/0.0554 = 2.303/0.0554 = 41.6 minutes
First Order Problem 3
A reactant concentration decreases from 0.85 mol/L to 0.35 mol/L in 15 minutes. Calculate the first-order rate constant and half-life.
Solution: k = ln([A]₀/[A])/t = ln(0.85/0.35)/15 = ln(2.43)/15 = 0.888/15 = 0.0592 min⁻¹. t₁/₂ = ln(2)/k = 0.693/0.0592 = 11.7 minutes
Practice Set 3: Arrhenius Focus
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Temperature Effect
A reaction has k = 3.45×10⁻³ s⁻¹ at 25°C and k = 1.03×10⁻² s⁻¹ at 40°C. Calculate the activation energy.
Solution: Ea = R·ln(k₂/k₁)/[(1/T₁)-(1/T₂)] = 8.314×10⁻³·ln(1.03×10⁻²/3.45×10⁻³)/[(1/298)-(1/313)] = 8.314×10⁻³·ln(2.986)/0.000161 = 8.314×10⁻³·1.094/0.000161 = 56.4 kJ/mol
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Rate at New Temperature
A reaction with Ea = 85.5 kJ/mol has k = 4.85×10⁻² min⁻¹ at 30°C. Calculate k at 55°C.
Solution: ln(k₂/k₁) = (Ea/R)[(1/T₁)-(1/T₂)] = (85.5/8.314×10⁻³)[(1/303)-(1/328)] = 10,284×0.000251 = 2.58. Therefore, k₂ = 4.85×10⁻²×e^2.58 = 4.85×10⁻²×13.2 = 0.640 min⁻¹
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Pre-exponential Factor
A reaction has Ea = 95.8 kJ/mol and k = 2.35×10⁻³ s⁻¹ at 35°C. Calculate the pre-exponential factor A.
Solution: k = Ae^(-Ea/RT), so A = k/e^(-Ea/RT) = 2.35×10⁻³/e^(-95800/(8.314×308)) = 2.35×10⁻³/e^(-37.3) = 2.35×10⁻³/(6.03×10⁻¹⁷) = 3.90×10¹³ s⁻¹
Challenge Problems: Arrhenius Equation
Challenge 1: Comparing Reaction Rates
Two reactions have activation energies of 55 kJ/mol and 95 kJ/mol. At 300K, both have the same rate constant. At what temperature will the reaction with the higher Ea proceed twice as fast as the reaction with the lower Ea?
Solution Approach:
Let reactions be X (Ea = 55 kJ/mol) and Y (Ea = 95 kJ/mol).
At 300K: kX = kY
At T₂: kY = 2kX
Using Arrhenius: ln(kY,T₂/kY,300) = (95/R)[(1/300)-(1/T₂)]
ln(kX,T₂/kX,300) = (55/R)[(1/300)-(1/T₂)]
Since kY,T₂ = 2kX,T₂, we can set up equations and solve for T₂ ≈ 423K (150°C)
Challenge Problems: Zero & First Order Combined
Challenge 2: Sequential Reactions
A substrate A undergoes a two-step reaction: A → B → C. The first step is zero-order with k₁ = 0.025 mol/(L·min), and the second step is first-order with k₂ = 0.050 min⁻¹. If [A]₀ = 0.50 mol/L and [B]₀ = [C]₀ = 0, calculate [B] and [C] after 15 minutes.
Solution Approach:
For A → B (zero-order): [A]t = [A]₀ - k₁t = 0.50 - 0.025(15) = 0.125 mol/L
Amount of A converted to B = 0.50 - 0.125 = 0.375 mol/L
For B → C (first-order): [B] is trickier as B is both forming and disappearing. This requires solving differential equations where d[B]/dt = k₁ - k₂[B].
The solution gives: [B] = 0.245 mol/L, [C] = 0.130 mol/L after 15 minutes.
Real-World Data Problems: Arrhenius and Kinetics
Problem: The data above shows the degradation rate constant of a pharmaceutical compound at different storage temperatures. Determine the activation energy and predict the shelf-life (time for 10% degradation) at room temperature (25°C).
Solution: Plotting ln(k) vs. 1/T gives a straight line with slope = -Ea/R = -10,600K. Thus, Ea = 88.1 kJ/mol. For 10% degradation, t = ln(1/0.9)/0.0012 = 0.105/0.0012 = 87.5 days at 25°C.
Summary Table: Formulas & Shortcuts
Common Exam Questions from Recent Years
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Identifying Reaction Order
Given a set of concentration vs. time data, determine the reaction order by testing which plot gives a straight line: [A] vs. t, ln[A] vs. t, or 1/[A] vs. t.
2
Rate Constant Calculation
Calculate the rate constant from half-life data or from concentration measurements at different times.
3
Arrhenius Parameters
Determine activation energy from rate constants measured at different temperatures.
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Time Prediction
Predict the time required for a reaction to reach a specific percentage completion.
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Temperature Effect
Calculate how much the rate will change when temperature changes by a specific amount.
Advanced Applications: Catalysts and Activation Energy

Enhanced Reaction Rate
Catalysts dramatically increase reaction speed
Lowered Energy Barrier
Catalysts create alternative reaction pathway with lower Ea
Unchanged Overall Energetics
ΔG of reaction remains the same, only barrier height changes
Industrial Applications
Catalysts make many industrial processes economically viable
Catalysts function by providing alternative reaction mechanisms with lower activation energy barriers. This dramatically increases reaction rates without being consumed in the process. The Arrhenius equation quantifies this effect through the exponential dependence of rate on activation energy.
Further Reading and References
Textbooks
  • Physical Chemistry by P. Atkins & J. de Paula
  • Chemical Kinetics by K.J. Laidler
  • Chemical Kinetics and Reaction Dynamics by P.L. Houston
  • Elements of Chemical Reaction Engineering by H.S. Fogler
Online Resources
  • Khan Academy: Chemical Kinetics
  • MIT OpenCourseWare: Chemical Kinetics
  • Journal of Chemical Education: Practical Approaches to Chemical Kinetics
  • ChemCollective Virtual Labs: Reaction Rates
Downloadable Practice Sheets (PDF)
These downloadable practice sheets contain additional problems organized by topic and difficulty level. Each worksheet includes fully worked solutions to help you check your understanding. Topics range from basic rate calculations to complex mixed-order reactions with temperature effects.
Conclusion & Key Takeaways
Arrhenius Foundation
The Arrhenius equation (k = Ae^(-Ea/RT)) links temperature to reaction rate through activation energy and provides a theoretical framework for understanding reaction kinetics.
Reaction Orders
Zero-order reactions (rate = k) proceed at constant rates regardless of concentration, while first-order reactions (rate = k[A]) show rates proportional to concentration.
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Problem-Solving Skills
Successful problem-solving requires identifying reaction order, applying the correct integrated rate law, and recognizing when to use the Arrhenius equation for temperature effects.
Real-World Applications
These kinetic principles apply across chemistry, biochemistry, medicine, and industry, helping predict everything from drug shelf-lives to industrial reactor performance.