Reaction Rates and Rate Laws

Learn to measure reaction rates, determine rate laws experimentally, understand reaction order, and calculate rate constants.

Reaction Rates and Rate Laws

Introduction to Reaction Rates

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:

Run multiple experiments
Vary one reactant concentration at a time
Measure initial rate for each
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

| Order | Rate Law | Linear Plot | Half-life | Units of k | |-------|----------|-------------|-----------|------------| | 0 | Rate = k | [A] vs t | t₁/₂ = [A]₀/(2k) | M·s⁻¹ | | 1 | Rate = k[A] | ln[A] vs t | t₁/₂ = 0.693/k | s⁻¹ | | 2 | Rate = k[A]² | 1/[A] vs t | t₁/₂ = 1/(k[A]₀) | M⁻¹·s⁻¹ |

Example: Initial Rates Data

Reaction: 2NO + O₂ → 2NO₂

Experimental data:

| Exp | [NO]₀ (M) | [O₂]₀ (M) | Initial Rate (M/s) | |-----|-----------|-----------|-------------------| | 1 | 0.010 | 0.010 | 2.5 × 10⁻⁵ | | 2 | 0.020 | 0.010 | 1.0 × 10⁻⁴ | | 3 | 0.010 | 0.020 | 5.0 × 10⁻⁵ |

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

Rate = Δ[concentration]/Δt
- Negative for reactants
- Positive for products
- Divide by stoichiometric coefficient
Rate law: Rate = k[A]^m[B]^n
- Determined experimentally
- Exponents ≠ stoichiometric coefficients
- k increases with temperature
Reaction order:
- Zero: rate independent of [A]
- First: rate ∝ [A]
- Second: rate ∝ [A]²
Initial rates method:
- Vary one reactant at a time
- Compare rate changes
- Determine orders
Rate depends on:
- Concentration (rate law)
- Temperature (Arrhenius)
- Catalysts, surface area

📚 Practice Problems

1Problem 1medium

❓ Question:

For the reaction A + 2B → C, the following initial rate data were obtained:

| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) | |------------|---------|---------|-------------------| | 1 | 0.10 | 0.10 | 0.015 | | 2 | 0.20 | 0.10 | 0.030 | | 3 | 0.10 | 0.20 | 0.060 |

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

💡 Show Solution

Solution:

(a) Rate law: Rate = k[A]^m[B]^n

Find m (order with respect to A): Compare Experiments 1 and 2 ([B] constant):

[A] doubles: 0.10 → 0.20
Rate doubles: 0.015 → 0.030
Therefore: 2^m = 2, so m = 1

Find n (order with respect to B): Compare Experiments 1 and 3 ([A] constant):

[B] doubles: 0.10 → 0.20
Rate quadruples: 0.015 → 0.060
Therefore: 2^n = 4, so n = 2

Rate law: Rate = k[A][B]²

(b) Rate constant: Using Experiment 1: 0.015 = k(0.10)(0.10)² 0.015 = k(0.0010) k = 15 M⁻²s⁻¹

2Problem 2easy

❓ Question:

For the reaction 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), oxygen is consumed at a rate of 2.0 × 10⁻³ M/s. (a) What is the rate of the reaction? (b) At what rate is NO produced? (c) At what rate is NH₃ consumed?

💡 Show Solution

Solution:

Given reaction:

$\ce{4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)}$

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

| Species | Stoichiometric Coefficient | Rate of Change (M/s) | Type | |---------|---------------------------|---------------------|------| | NH₃ | 4 | -1.6 × 10⁻³ | Consumed | | O₂ | 5 | -2.0 × 10⁻³ | Consumed (given) | | NO | 4 | +1.6 × 10⁻³ | Produced | | H₂O | 6 | +2.4 × 10⁻³ | Produced | | Reaction | — | 4.0 × 10⁻⁴ | Rate |

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

3Problem 3medium

❓ Question:

For the reaction A + 2B → C, the following initial rate data were obtained:

| Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) | |------------|---------|---------|-------------------| | 1 | 0.10 | 0.10 | 0.015 | | 2 | 0.20 | 0.10 | 0.030 | | 3 | 0.10 | 0.20 | 0.060 |

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

💡 Show Solution

Solution:

(a) Rate law: Rate = k[A]^m[B]^n

Find m (order with respect to A): Compare Experiments 1 and 2 ([B] constant):

[A] doubles: 0.10 → 0.20
Rate doubles: 0.015 → 0.030
Therefore: 2^m = 2, so m = 1

Find n (order with respect to B): Compare Experiments 1 and 3 ([A] constant):

[B] doubles: 0.10 → 0.20
Rate quadruples: 0.015 → 0.060
Therefore: 2^n = 4, so n = 2

Rate law: Rate = k[A][B]²

(b) Rate constant: Using Experiment 1: 0.015 = k(0.10)(0.10)² 0.015 = k(0.0010) k = 15 M⁻²s⁻¹

4Problem 4hard

❓ Question:

A reaction is found to be first order in A and second order in B. (a) Write the rate law. (b) By what factor does the rate increase if [A] is tripled and [B] is doubled? (c) What are the units of the rate constant?

💡 Show Solution

Solution:

(a) Rate law: Rate = k[A]¹[B]² or Rate = k[A][B]²

(b) Rate change factor: Initial rate: Rate₁ = k[A]₁[B]₁² New rate: Rate₂ = k[A]₂[B]₂²

Where: [A]₂ = 3[A]₁ and [B]₂ = 2[B]₁

Rate₂ = k(3[A]₁)(2[B]₁)² Rate₂ = k(3[A]₁)(4[B]₁²) Rate₂ = 12k[A]₁[B]₁² Rate₂ = 12 × Rate₁

The rate increases by a factor of 12

General pattern for units:

0th order: M/s
1st order: s⁻¹
2nd order: M⁻¹s⁻¹
3rd order: M⁻²s⁻¹

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

(a) Rate law: Rate = k[A]¹[B]² or Rate = k[A][B]²

(b) Rate change factor: Initial rate: Rate₁ = k[A]₁[B]₁² New rate: Rate₂ = k[A]₂[B]₂²

Where: [A]₂ = 3[A]₁ and [B]₂ = 2[B]₁

Rate₂ = k(3[A]₁)(2[B]₁)² Rate₂ = k(3[A]₁)(4[B]₁²) Rate₂ = 12k[A]₁[B]₁² Rate₂ = 12 × Rate₁

The rate increases by a factor of 12

General pattern for units:

0th order: M/s
1st order: s⁻¹
2nd order: M⁻¹s⁻¹
3rd order: M⁻²s⁻¹

6Problem 6medium

❓ Question:

The following data were collected for the reaction: 2NO(g) + Cl₂(g) → 2NOCl(g). Determine (a) the rate law, (b) the value of k with units, (c) the rate when [NO] = 0.050 M and [Cl₂] = 0.020 M. | Exp | [NO]₀ (M) | [Cl₂]₀ (M) | Initial Rate (M/s) | |-----|-----------|-----------|-------------------| | 1 | 0.10 | 0.10 | 0.18 | | 2 | 0.10 | 0.20 | 0.36 | | 3 | 0.20 | 0.20 | 1.44 |

💡 Show Solution

Solution:

Given reaction:

$\ce{2NO(g) + Cl2(g) -> 2NOCl(g)}$

Data table:

| Exp | [NO]₀ (M) | [Cl₂]₀ (M) | Initial Rate (M/s) | |-----|-----------|-----------|-------------------| | 1 | 0.10 | 0.10 | 0.18 | | 2 | 0.10 | 0.20 | 0.36 | | 3 | 0.20 | 0.20 | 1.44 |

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

| Question | Answer | |----------|--------| | (a) Rate law | Rate = k[NO]²[Cl₂] | | (b) Rate constant | k = 180 M⁻²s⁻¹ | | (c) Rate at given [NO], [Cl₂] | 9.0 × 10⁻³ M/s |

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:

| Overall Order | Units of k | |---------------|------------| | 0 | M·s⁻¹ | | 1 | s⁻¹ | | 2 | M⁻¹·s⁻¹ | | 3 | M⁻²·s⁻¹ | | n | M¹⁻ⁿ·s⁻¹ |

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):

$\ce{NO + Cl2 <=> NOCl2}$

Step 2 (slow, rate-determining):

$\ce{NOCl2 + NO -> 2NOCl}$

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:

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

7Problem 7hard

❓ Question:

The decomposition of hydrogen peroxide is first order: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). At 20°C, k = 1.8 × 10⁻⁵ s⁻¹. (a) If the initial concentration is 0.30 M, what will be the concentration after 1.0 hour? (b) How long will it take for the concentration to drop from 0.30 M to 0.10 M? (c) What is the half-life of this reaction?

💡 Show Solution

Solution:

Given:

Reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Order: First order
k = 1.8 × 10⁻⁵ s⁻¹ at 20°C
[H₂O₂]₀ = 0.30 M

First order integrated rate law:

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

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

| Part | Question | Answer | |------|----------|--------| | (a) | [H₂O₂] after 1.0 hour | 0.28 M | | (b) | Time: 0.30 M → 0.10 M | 6.1 × 10⁴ s (17 hours) | | (c) | Half-life | 3.85 × 10⁴ s (10.7 hours) |

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:

✓ Identify it's first order (given or determined)
✓ Use ln[A]_t = ln[A]₀ - kt
✓ Check time units match k units
✓ For half-life: t₁/₂ = 0.693/k (constant!)
✓ Remember ln(a/b) = ln(a) - ln(b)
✓ 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

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