Kinetics: Study of reaction rates and mechanisms

Reaction rate: How fast reactants convert to products

• Measured by change in concentration over time
• Units: M/s (molarity per second) or similar

Why study kinetics?

• Control reaction speed (industrial processes)
• Understand reaction mechanisms
• Optimize conditions (temperature, catalysts)
• Predict product formation rates

Key distinction:

• Thermodynamics: Will a reaction occur? (ΔG)
• Kinetics: How fast will it occur? (rate)

Example: Diamond → Graphite

• Thermodynamically favorable (ΔG < 0)
• Kinetically very slow (rate ≈ 0)
• That's why diamonds last "forever"

Expressing Reaction Rate

For General Reaction: aA + bB → cC + dD

Rate definitions:

Rate of disappearance (reactants):

$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t}$

Rate of appearance (products):

$\text{Rate} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

Why negative for reactants?

• [A] decreases, so Δ[A] < 0
• Negative sign makes rate positive

Why divide by coefficients?

• Makes rate same regardless of which species measured
• Accounts for stoichiometry

Example: 2N₂O₅ → 4NO₂ + O₂

All these express the same rate:

$\text{Rate} = -\frac{1}{2}\frac{\Delta[N_2O_5]}{\Delta t} = +\frac{1}{4}\frac{\Delta[NO_2]}{\Delta t} = +\frac{\Delta[O_2]}{\Delta t}$

If O₂ appears at 0.10 M/s:

$\text{Rate} = +\frac{\Delta[O_2]}{\Delta t} = 0.10 \text{ M/s}$

Then:

$\frac{\Delta[NO_2]}{\Delta t} = 4 \times 0.10 = 0.40 \text{ M/s}$

$\frac{\Delta[N_2O_5]}{\Delta t} = -2 \times 0.10 = -0.20 \text{ M/s}$

Note: NO₂ forms 4 times faster than O₂ (stoichiometry)

Instantaneous vs Average Rate

Average Rate

Over time interval Δt:

$\text{Average rate} = \frac{\Delta[A]}{\Delta t} = \frac{[A]_2 - [A]_1}{t_2 - t_1}$

Example: If [A] goes from 1.0 M to 0.6 M in 20 seconds:

$\text{Average rate} = \frac{0.6 - 1.0}{20 - 0} = \frac{-0.4}{20} = -0.02 \text{ M/s}$

Instantaneous Rate

Rate at specific moment:

$\text{Instantaneous rate} = \frac{d[A]}{dt}$

Found by:

• Tangent to concentration vs time curve
• Slope at that instant
• Derivative (calculus)

Rate typically decreases over time:

• Fast at start (high [reactants])
• Slows as reactants consumed
• Eventually approaches zero

Initial Rate

Rate at t = 0:

$\text{Initial rate} = \left(\frac{d[A]}{dt}\right)_{t=0}$

Why useful?

• Before significant reverse reaction
• Simplifies analysis
• Used to determine rate law experimentally

Rate Law (Rate Equation)

Rate law: Equation relating rate to concentrations

General form:

$\text{Rate} = k[A]^m[B]^n$

Where:

• k = rate constant (specific to reaction and temperature)
• [A], [B] = molar concentrations
• m, n = reaction orders (exponents)

Key points:

• Rate law must be determined experimentally
• Cannot be predicted from balanced equation
• Exponents usually 0, 1, or 2 (can be fractional)

Rate Constant (k)

Units depend on overall order:

Zero order: M/s or M·s⁻¹

First order: s⁻¹ or 1/s

Second order: M⁻¹·s⁻¹ or 1/(M·s)

Third order: M⁻²·s⁻¹

Temperature dependence:

• k increases with temperature
• Related by Arrhenius equation (later)

Reaction Order

Order with respect to reactant:

• Exponent in rate law
• How rate depends on that concentration

Example: Rate = k[A]²[B]

• Order with respect to A: 2 (second order in A)
• Order with respect to B: 1 (first order in B)

Overall order:

• Sum of all exponents
• Example above: 2 + 1 = 3 (third order overall)

Determining Reaction Order

Cannot determine from balanced equation!

Must use experimental data:

Method 1: Initial Rates Method

Procedure:

1. Run multiple experiments
2. Vary one reactant concentration at a time
3. Measure initial rate for each
4. Compare how rate changes

Analysis:

If [A] doubles and rate doubles:

• Rate ∝ [A]¹
• First order in A

If [A] doubles and rate quadruples (×4):

• Rate ∝ [A]²
• Second order in A

If [A] doubles and rate unchanged:

• Rate ∝ [A]⁰
• Zero order in A

General rule:

If [A] changes by factor f, rate changes by factor f^m:

$\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m = f^m$

Solve for m:

$m = \frac{\log(\text{rate}_2/\text{rate}_1)}{\log([A]_2/[A]_1)}$

Method 2: Graphical Method (Integrated Rate Laws)

Use integrated rate laws (covered in next topic):

• Zero order: [A] vs t is linear
• First order: ln[A] vs t is linear
• Second order: 1/[A] vs t is linear

Plot data, see which is linear

Reaction Order Types

Zero Order in Reactant

Rate law: Rate = k[A]⁰ = k

Characteristics:

• Rate independent of [A]
• Rate constant over time (until A runs out)
• Graph [A] vs t: straight line, negative slope

Example:

• Surface-catalyzed reactions (surface saturated)
• Enzyme reactions (enzyme saturated)

Integrated form:

$[A]_t = [A]_0 - kt$

First Order in Reactant

Rate law: Rate = k[A]¹ = k[A]

Characteristics:

• Rate proportional to [A]
• If [A] doubles, rate doubles
• Graph ln[A] vs t: straight line, slope = -k

Example:

• Radioactive decay
• Many decomposition reactions

Integrated form:

$\ln[A]_t = \ln[A]_0 - kt$

Or:

$[A]_t = [A]_0 e^{-kt}$

Half-life (constant):

$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$

Second Order in Reactant

Rate law: Rate = k[A]²

Characteristics:

• Rate proportional to [A]²
• If [A] doubles, rate quadruples
• Graph 1/[A] vs t: straight line, slope = k

Example:

• Some gas-phase reactions
• Dimerization reactions

Integrated form:

$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$

Half-life (not constant):

$t_{1/2} = \frac{1}{k[A]_0}$

Depends on initial concentration!

Mixed Orders

Different orders for different reactants:

Example: Rate = k[A]²[B]

• Second order in A
• First order in B
• Third order overall

To determine each order:

• Vary [A] while keeping [B] constant
• Vary [B] while keeping [A] constant
• Use initial rates method

Summary Table: Reaction Orders

Example: Initial Rates Data

Reaction: 2NO + O₂ → 2NO₂

Experimental data:

Determine rate law:

Compare Exp 1 and 2 ([O₂] constant):

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{1.0 \times 10^{-4}}{2.5 \times 10^{-5}} = 4$

$\frac{[NO]_2}{[NO]_1} = \frac{0.020}{0.010} = 2$

Rate quadrupled when [NO] doubled:

$2^m = 4 \implies m = 2$

Second order in NO

Compare Exp 1 and 3 ([NO] constant):

$\frac{\text{rate}_3}{\text{rate}_1} = \frac{5.0 \times 10^{-5}}{2.5 \times 10^{-5}} = 2$

$\frac{[O_2]_3}{[O_2]_1} = \frac{0.020}{0.010} = 2$

Rate doubled when [O₂] doubled:

$2^n = 2 \implies n = 1$

First order in O₂

Rate law:

$\boxed{\text{Rate} = k[NO]^2[O_2]}$

Overall order: 2 + 1 = 3 (third order)

Calculate k from Exp 1:

$k = \frac{\text{rate}}{[NO]^2[O_2]} = \frac{2.5 \times 10^{-5}}{(0.010)^2(0.010)}$

$k = \frac{2.5 \times 10^{-5}}{1.0 \times 10^{-6}} = 2.5 \times 10^1 = 25 \text{ M}^{-2}\text{s}^{-1}$

Factors Affecting Reaction Rate

1. Concentration

Higher concentration → faster rate

• More molecules
• More collisions
• Rate law quantifies this

2. Temperature

Higher temperature → faster rate

• Molecules move faster
• More energetic collisions
• More exceed activation energy
• Arrhenius equation quantifies

3. Surface Area

Greater surface area → faster rate

• For heterogeneous reactions
• More contact between phases
• Example: powder vs chunk

4. Catalysts

Catalyst → faster rate

• Lower activation energy
• Provides alternate pathway
• Not consumed
• Doesn't change equilibrium position

5. Nature of Reactants

Ionic reactions: Very fast (no bonds to break)

Covalent reactions: Slower (bonds must break)

Example:

• Ag⁺ + Cl⁻ → AgCl: instantaneous
• Organic reactions: may take hours/days

Applications

Industrial Processes

Haber process (NH₃ synthesis):

• High temperature (increases k)
• High pressure (increases [N₂], [H₂])
• Catalyst (Fe/Fe₃O₄)

Pharmaceuticals

Drug stability:

• Rate of decomposition
• Shelf life predictions
• Storage temperature

Environmental

Atmospheric reactions:

• Ozone depletion rates
• Pollutant degradation
• Climate modeling

Food Chemistry

Spoilage rates:

• Temperature dependence (refrigeration)
• Preservatives (inhibitors)
• Oxidation rates

Key Concepts Summary

1. Rate = Δ[concentration]/Δt

   • Negative for reactants
   • Positive for products
   • Divide by stoichiometric coefficient

2. Rate law: Rate = k[A]^m[B]^n

   • Determined experimentally
   • Exponents ≠ stoichiometric coefficients
   • k increases with temperature

3. Reaction order:

   • Zero: rate independent of [A]
   • First: rate ∝ [A]
   • Second: rate ∝ [A]²

4. Initial rates method:

   • Vary one reactant at a time
   • Compare rate changes
   • Determine orders

5. Rate depends on:

   • Concentration (rate law)
   • Temperature (Arrhenius)
   • Catalysts, surface area

(a) Determine the rate law. (b) Calculate the rate constant k with units. (c) What is the overall order of the reaction?

Data table:

General rate law form:

$\text{Rate} = k[NO]^m[Cl_2]^n$

Need to find: m, n, k

(a) Determine rate law

Find order with respect to Cl₂ (n):

Compare Exp 1 and 2 (keep [NO] constant, vary [Cl₂])

Set up ratio:

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{k[NO]_2^m[Cl_2]_2^n}{k[NO]_1^m[Cl_2]_1^n}$

Since [NO] is same in both:

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{[Cl_2]_2^n}{[Cl_2]_1^n} = \left(\frac{[Cl_2]_2}{[Cl_2]_1}\right)^n$

Substitute values:

$\frac{0.36}{0.18} = \left(\frac{0.20}{0.10}\right)^n$

$2 = 2^n$

$n = 1$

First order in Cl₂

Find order with respect to NO (m):

Compare Exp 2 and 3 (keep [Cl₂] constant, vary [NO])

Set up ratio:

$\frac{\text{rate}_3}{\text{rate}_2} = \frac{k[NO]_3^m[Cl_2]_3^n}{k[NO]_2^m[Cl_2]_2^n}$

Since [Cl₂] is same in both:

$\frac{\text{rate}_3}{\text{rate}_2} = \frac{[NO]_3^m}{[NO]_2^m} = \left(\frac{[NO]_3}{[NO]_2}\right)^m$

Substitute values:

$\frac{1.44}{0.36} = \left(\frac{0.20}{0.10}\right)^m$

$4 = 2^m$

$2^2 = 2^m$

$m = 2$

Second order in NO

Rate law:

$\boxed{\text{Rate} = k[NO]^2[Cl_2]}$

Orders:

• With respect to NO: 2 (second order)
• With respect to Cl₂: 1 (first order)
• Overall: 2 + 1 = 3 (third order)

(b) Calculate k with units

Use any experiment; let's use Exp 1:

$\text{Rate} = k[NO]^2[Cl_2]$

Solve for k:

$k = \frac{\text{Rate}}{[NO]^2[Cl_2]}$

Substitute values from Exp 1:

$k = \frac{0.18}{(0.10)^2(0.10)}$

$k = \frac{0.18}{(0.01)(0.10)}$

$k = \frac{0.18}{0.001}$

$k = 180$

Determine units of k:

From rate equation:

$\text{Rate (M/s)} = k[NO]^2[Cl_2]$

$\text{M/s} = k \times \text{M}^2 \times \text{M}$

$\text{M/s} = k \times \text{M}^3$

Solve for units of k:

$k = \frac{\text{M/s}}{\text{M}^3} = \frac{\text{M}}{\text{s} \cdot \text{M}^3} = \frac{1}{\text{M}^2 \cdot \text{s}} = \text{M}^{-2}\text{s}^{-1}$

Answer (b):

$\boxed{k = 180 \text{ M}^{-2}\text{s}^{-1}}$

Or equivalently: k = 180 L²·mol⁻²·s⁻¹

Verify with other experiments:

Check with Exp 2:

$k = \frac{0.36}{(0.10)^2(0.20)} = \frac{0.36}{0.002} = 180$ ✓

Check with Exp 3:

$k = \frac{1.44}{(0.20)^2(0.20)} = \frac{1.44}{0.008} = 180$ ✓

Consistent! ✓

(c) Calculate rate at [NO] = 0.050 M, [Cl₂] = 0.020 M

Use rate law with k = 180 M⁻²s⁻¹:

$\text{Rate} = k[NO]^2[Cl_2]$

$\text{Rate} = 180 \times (0.050)^2 \times (0.020)$

Calculate (0.050)²:

$(0.050)^2 = 0.0025$

Calculate rate:

$\text{Rate} = 180 \times 0.0025 \times 0.020$

$\text{Rate} = 180 \times 0.000050$

$\text{Rate} = 0.0090 \text{ M/s}$

$\text{Rate} = 9.0 \times 10^{-3} \text{ M/s}$

Answer (c):

$\boxed{\text{Rate} = 9.0 \times 10^{-3} \text{ M/s} \text{ or } 0.0090 \text{ M/s}}$

Summary of Results

Key Observations

1. Rate law vs stoichiometry:

Balanced equation: 2NO + Cl₂ → 2NOCl

Rate law: Rate = k[NO]²[Cl₂]

Notice:

• Exponent for NO: 2 (matches coefficient)
• Exponent for Cl₂: 1 (matches coefficient)

This is coincidental!

• Rate law determined experimentally
• Sometimes matches, sometimes doesn't
• Never assume exponents = coefficients

2. Reaction order interpretation:

Second order in NO:

• If [NO] doubles, rate quadruples
• NO concentration has large effect
• Suggests 2 NO molecules in rate-determining step

First order in Cl₂:

• If [Cl₂] doubles, rate doubles
• Linear dependence
• Suggests 1 Cl₂ molecule in rate-determining step

Third order overall:

• Complex reaction
• Likely multi-step mechanism
• Rate-determining step involves 2 NO + 1 Cl₂

3. Initial rates method steps:

Step 1: Identify pairs where only one concentration varies

Step 2: Set up rate ratio equation

Step 3: Concentrations constant cancel out

Step 4: Solve for exponent

Step 5: Repeat for each reactant

Step 6: Calculate k from any experiment

Step 7: Verify k consistent across all experiments

Units of k for different orders:

Our reaction: Order = 3, so k has units M⁻²·s⁻¹ ✓

Real-world context:

This reaction:

• Gas-phase radical reaction
• Studied extensively for mechanism
• Shows typical behavior for NO reactions

Proposed mechanism:

Step 1 (fast equilibrium):

$NO + Cl2 \rightleftharpoons NOCl2$

Step 2 (slow, rate-determining):

$NOCl2 + NO \rightarrow 2NOCl$

Overall: 2NO + Cl₂ → 2NOCl ✓

Rate-determining step involves:

• 1 NOCl₂ (which comes from 1 NO + 1 Cl₂)
• 1 NO
• Total: 2 NO + 1 Cl₂

This explains the rate law: Rate = k[NO]²[Cl₂] ✓

Practice tip:

When doing initial rates problems:

1. ✓ Organize data in table
2. ✓ Find pairs where only ONE concentration changes
3. ✓ Set up ratio (things that don't change cancel)
4. ✓ Solve for exponent
5. ✓ Calculate k from any experiment
6. ✓ Check k is same for all experiments
7. ✓ Pay attention to units!

Or:

$[A]_t = [A]_0 e^{-kt}$

(a) Concentration after 1.0 hour

Given:

• t = 1.0 hour
• [H₂O₂]₀ = 0.30 M

Convert time to seconds:

$t = 1.0 \text{ hour} \times \frac{60 \text{ min}}{1 \text{ hour}} \times \frac{60 \text{ s}}{1 \text{ min}} = 3600 \text{ s}$

Use integrated rate law:

$\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt$

Substitute values:

$\ln[H_2O_2]_t = \ln(0.30) - (1.8 \times 10^{-5})(3600)$

Calculate each term:

$\ln(0.30) = -1.204$

$(1.8 \times 10^{-5})(3600) = 0.0648$

Substitute:

$\ln[H_2O_2]_t = -1.204 - 0.0648$

$\ln[H_2O_2]_t = -1.269$

Take exponential:

$[H_2O_2]_t = e^{-1.269}$

$[H_2O_2]_t = 0.281 \text{ M}$

Answer (a):

$\boxed{[H_2O_2]_t = 0.28 \text{ M}}$

Alternative method using exponential form:

$[H_2O_2]_t = [H_2O_2]_0 e^{-kt}$

$[H_2O_2]_t = 0.30 \times e^{-(1.8 \times 10^{-5})(3600)}$

$[H_2O_2]_t = 0.30 \times e^{-0.0648}$

$e^{-0.0648} = 0.937$

$[H_2O_2]_t = 0.30 \times 0.937 = 0.281 \text{ M}$

Same answer! ✓

Interpretation:

• Initial: 0.30 M
• After 1 hour: 0.28 M
• Decrease: 0.02 M (about 6.3% decomposed)
• Very slow reaction at 20°C without catalyst

(b) Time to drop from 0.30 M to 0.10 M

Given:

• [H₂O₂]₀ = 0.30 M
• [H₂O₂]_t = 0.10 M
• Find: t

Use integrated rate law:

$\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt$

Rearrange to solve for t:

$kt = \ln[H_2O_2]_0 - \ln[H_2O_2]_t$

$kt = \ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)$

$t = \frac{1}{k}\ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)$

Substitute values:

$t = \frac{1}{1.8 \times 10^{-5}}\ln\left(\frac{0.30}{0.10}\right)$

$t = \frac{1}{1.8 \times 10^{-5}}\ln(3.0)$

Calculate ln(3.0):

$\ln(3.0) = 1.099$

Calculate t:

$t = \frac{1.099}{1.8 \times 10^{-5}}$

$t = 6.106 \times 10^4 \text{ s}$

$t = 61,060 \text{ s}$

Convert to hours:

$t = 61,060 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{1 \text{ hour}}{60 \text{ min}}$

$t = \frac{61,060}{3600} \text{ hours}$

$t = 16.96 \text{ hours} \approx 17 \text{ hours}$

Answer (b):

$\boxed{t = 6.1 \times 10^4 \text{ s} \text{ or } 17 \text{ hours}}$

Interpretation:

• To drop from 0.30 M to 0.10 M takes 17 hours
• Concentration reduced to 1/3 of original
• Much longer than 1 hour (part a)
• Shows exponential decay nature

(c) Half-life

For first-order reactions, half-life is constant:

$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$

Substitute k:

$t_{1/2} = \frac{0.693}{1.8 \times 10^{-5}}$

$t_{1/2} = 3.85 \times 10^4 \text{ s}$

Convert to hours:

$t_{1/2} = \frac{3.85 \times 10^4}{3600} \text{ hours}$

$t_{1/2} = 10.7 \text{ hours}$

Answer (c):

$\boxed{t_{1/2} = 3.85 \times 10^4 \text{ s} \text{ or } 10.7 \text{ hours}}$

Verification using half-life:

After one half-life (10.7 hours):

• [H₂O₂] drops from 0.30 M to 0.15 M

After two half-lives (21.4 hours):

• [H₂O₂] drops from 0.15 M to 0.075 M

To go from 0.30 M to 0.10 M:

$\frac{0.30}{0.10} = 3 = 2^{1.585}$

So time needed:

$t = 1.585 \times t_{1/2} = 1.585 \times 10.7 = 17.0 \text{ hours}$

Matches part (b)! ✓

Summary Table

Key Concepts Illustrated

1. First-order characteristics:

Constant half-life:

• t₁/₂ = 10.7 hours (independent of concentration)
• After each 10.7 hours, concentration halves
• Exponential decay pattern

Linear ln[A] vs t:

• If plotted, would give straight line
• Slope = -k
• Intercept = ln[A]₀

2. Time conversions:

Always check units!

• k given in s⁻¹
• Time must be in seconds
• Convert hours → seconds for calculations
• Can convert back to hours for answer

3. Integrated rate law applications:

Forward calculation (given t, find [A]):

• Use: ln[A]_t = ln[A]₀ - kt
• Or: [A]_t = [A]₀e^(-kt)

Reverse calculation (given [A], find t):

• Rearrange: t = (1/k)ln([A]₀/[A]_t)

Half-life:

• Special case: [A]_t = [A]₀/2
• Gives: t₁/₂ = ln(2)/k = 0.693/k

Real-world context:

H₂O₂ decomposition:

Storage:

• Brown bottles (blocks light)
• Light catalyzes decomposition
• Cool, dark place preferred

Catalysts:

• MnO₂: very fast decomposition
• Enzymes (catalase): extremely fast
• Used to demonstrate catalysis

Applications:

• Bleaching agent
• Disinfectant
• Rocket propellant (concentrated)
• Must monitor concentration over time

Medical use:

• 3% solution (drugstore)
• Slow decomposition allows storage
• Eventually loses potency (O₂ escapes)

Graphical representation:

Concentration vs time:

• Exponential decay curve
• Starts at 0.30 M
• Asymptotically approaches 0
• After 10.7 h: 0.15 M
• After 21.4 h: 0.075 M

ln[H₂O₂] vs time:

• Straight line
• Slope = -k = -1.8 × 10⁻⁵ s⁻¹
• Useful for determining if reaction is first order

Practice tips:

For first-order problems:

1. ✓ Identify it's first order (given or determined)
2. ✓ Use ln[A]_t = ln[A]₀ - kt
3. ✓ Check time units match k units
4. ✓ For half-life: t₁/₂ = 0.693/k (constant!)
5. ✓ Remember ln(a/b) = ln(a) - ln(b)
6. ✓ Calculator: ln is natural log (base e)

Common mistakes:

• ✗ Using log instead of ln
• ✗ Mixing time units (hours vs seconds)
• ✗ Forgetting negative sign in rate law
• ✗ Using wrong integrated rate law for order