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

Given: O₂ consumed at 2.0 × 10⁻³ M/s

Note: "Consumed" means disappearance, so Δ[O₂]/Δt is negative

(a) Rate of reaction

General rate expression:

For reaction: aA + bB → cC + dD

$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

For our reaction:

$\text{Rate} = -\frac{1}{4}\frac{\Delta[NH_3]}{\Delta t} = -\frac{1}{5}\frac{\Delta[O_2]}{\Delta t} = +\frac{1}{4}\frac{\Delta[NO]}{\Delta t} = +\frac{1}{6}\frac{\Delta[H_2O]}{\Delta t}$

Using O₂ term:

Given: O₂ consumed at 2.0 × 10⁻³ M/s

$\frac{\Delta[O_2]}{\Delta t} = -2.0 \times 10^{-3} \text{ M/s}$

(Negative because O₂ is disappearing)

Calculate rate:

$\text{Rate} = -\frac{1}{5}\frac{\Delta[O_2]}{\Delta t}$

$\text{Rate} = -\frac{1}{5}(-2.0 \times 10^{-3})$

$\text{Rate} = \frac{2.0 \times 10^{-3}}{5}$

$\text{Rate} = 0.40 \times 10^{-3} = 4.0 \times 10^{-4} \text{ M/s}$

Answer (a):

$\boxed{\text{Rate} = 4.0 \times 10^{-4} \text{ M/s}}$

(b) Rate of NO production

Using rate expression:

$\text{Rate} = +\frac{1}{4}\frac{\Delta[NO]}{\Delta t}$

Solve for Δ[NO]/Δt:

$\frac{\Delta[NO]}{\Delta t} = 4 \times \text{Rate}$

$\frac{\Delta[NO]}{\Delta t} = 4 \times (4.0 \times 10^{-4})$

$\frac{\Delta[NO]}{\Delta t} = 16 \times 10^{-4}$

$\frac{\Delta[NO]}{\Delta t} = 1.6 \times 10^{-3} \text{ M/s}$

Answer (b):

$\boxed{\frac{\Delta[NO]}{\Delta t} = 1.6 \times 10^{-3} \text{ M/s}}$

Interpretation: NO produced at 1.6 × 10⁻³ M/s (positive = production)

(c) Rate of NH₃ consumption

Using rate expression:

$\text{Rate} = -\frac{1}{4}\frac{\Delta[NH_3]}{\Delta t}$

Solve for Δ[NH₃]/Δt:

$\frac{\Delta[NH_3]}{\Delta t} = -4 \times \text{Rate}$

$\frac{\Delta[NH_3]}{\Delta t} = -4 \times (4.0 \times 10^{-4})$

$\frac{\Delta[NH_3]}{\Delta t} = -16 \times 10^{-4}$

$\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}$

Answer (c):

$\boxed{\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}}$

Or, stating as consumption rate (magnitude):

$\boxed{\text{NH}_3 \text{ consumed at } 1.6 \times 10^{-3} \text{ M/s}}$

Summary of Results

Verification

Check stoichiometric relationships:

From balanced equation: 4 NH₃ : 5 O₂ : 4 NO : 6 H₂O

From our rates:

$\frac{|\Delta[NH_3]/\Delta t|}{4} = \frac{1.6 \times 10^{-3}}{4} = 4.0 \times 10^{-4}$ ✓

$\frac{|\Delta[O_2]/\Delta t|}{5} = \frac{2.0 \times 10^{-3}}{5} = 4.0 \times 10^{-4}$ ✓

$\frac{|\Delta[NO]/\Delta t|}{4} = \frac{1.6 \times 10^{-3}}{4} = 4.0 \times 10^{-4}$ ✓

All equal to the rate! ✓

Key Insights

1. Stoichiometric relationships:

• NH₃ and NO have same coefficient (4)
• Therefore consumed/produced at same rate
• Both: 1.6 × 10⁻³ M/s

2. Ratio comparison:

$\frac{\Delta[NH_3]/\Delta t}{\Delta[O_2]/\Delta t} = \frac{-1.6 \times 10^{-3}}{-2.0 \times 10^{-3}} = \frac{1.6}{2.0} = \frac{4}{5}$

Matches stoichiometry: 4 NH₃ : 5 O₂ ✓

3. Why divide by coefficients?

• Makes "rate of reaction" unique
• Same value regardless of which species you measure
• Accounts for stoichiometric ratios

4. Sign conventions:

• Reactants: negative Δ[]/Δt (decreasing)
• Products: positive Δ[]/Δt (increasing)
• Rate of reaction: always positive

Alternative approach:

Could also use ratios directly:

Given: Δ[O₂]/Δt = -2.0 × 10⁻³ M/s

Want: Δ[NH₃]/Δt

From stoichiometry: 4 NH₃ consumed per 5 O₂

$\frac{\Delta[NH_3]/\Delta t}{\Delta[O_2]/\Delta t} = \frac{4}{5}$

$\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times \frac{\Delta[O_2]}{\Delta t}$

$\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times (-2.0 \times 10^{-3})$

$\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}$ ✓

Same answer!

This is the Ostwald process:

Industrial production of nitric acid:

Step 1: 4NH₃ + 5O₂ → 4NO + 6H₂O (this reaction)

Step 2: 2NO + O₂ → 2NO₂

Step 3: 3NO₂ + H₂O → 2HNO₃ + NO

Used to make:

• Fertilizers
• Explosives
• Various chemicals

Conditions:

• High temperature (850-900°C)
• Platinum-rhodium catalyst
• Important industrial process