Reaction Rates and Rate Laws

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

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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):

Rate=−1aΔ[A]Δt=−1bΔ[B]Δt\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t}

Rate of appearance (products):

Rate=+1cΔ[C]Δt=+1dΔ[D]Δt\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:

Rate=−12Δ[N2O5]Δt=+14Δ[NO2]Δt=+Δ[O2]Δt\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:

Rate=+Δ[O2]Δt=0.10 M/s\text{Rate} = +\frac{\Delta[O_2]}{\Delta t} = 0.10 \text{ M/s}

Then:

Δ[NO2]Δt=4×0.10=0.40 M/s\frac{\Delta[NO_2]}{\Delta t} = 4 \times 0.10 = 0.40 \text{ M/s}

Δ[N2O5]Δt=−2×0.10=−0.20 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:

Average rate=Δ[A]Δt=[A]2−[A]1t2−t1\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:

Average rate=0.6−1.020−0=−0.420=−0.02 M/s\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:

Instantaneous rate=d[A]dt\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:

Initial rate=(d[A]dt)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:

Rate=k[A]m[B]n\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:

rate2rate1=([A]2[A]1)m=fm\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^m = f^m

Solve for m:

m=log⁡(rate2/rate1)log⁡([A]2/[A]1)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[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\ln[A]_t = \ln[A]_0 - kt

Or:

[A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}

Half-life (constant):

t1/2=0.693k=ln⁡2kt_{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:

1[A]t=1[A]0+kt\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt

Half-life (not constant):

t1/2=1k[A]0t_{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):

rate2rate1=1.0×10−42.5×10−5=4\frac{\text{rate}_2}{\text{rate}_1} = \frac{1.0 \times 10^{-4}}{2.5 \times 10^{-5}} = 4

[NO]2[NO]1=0.0200.010=2\frac{[NO]_2}{[NO]_1} = \frac{0.020}{0.010} = 2

Rate quadrupled when [NO] doubled:

2m=4  ⟹  m=22^m = 4 \implies m = 2

Second order in NO

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

rate3rate1=5.0×10−52.5×10−5=2\frac{\text{rate}_3}{\text{rate}_1} = \frac{5.0 \times 10^{-5}}{2.5 \times 10^{-5}} = 2

[O2]3[O2]1=0.0200.010=2\frac{[O_2]_3}{[O_2]_1} = \frac{0.020}{0.010} = 2

Rate doubled when [O₂] doubled:

2n=2  ⟹  n=12^n = 2 \implies n = 1

First order in O₂

Rate law:

Rate=k[NO]2[O2]\boxed{\text{Rate} = k[NO]^2[O_2]}

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

Calculate k from Exp 1:

k=rate[NO]2[O2]=2.5×10−5(0.010)2(0.010)k = \frac{\text{rate}}{[NO]^2[O_2]} = \frac{2.5 \times 10^{-5}}{(0.010)^2(0.010)}

k=2.5×10−51.0×10−6=2.5×101=25 M−2s−1k = \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

📚 Practice Problems

1Problem 1easy

❓ 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:

\ce4NH3(g)+5O2(g)−>4NO(g)+6H2O(g)\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

Rate=−1aΔ[A]Δt=−1bΔ[B]Δt=+1cΔ[C]Δt=+1dΔ[D]Δt\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:

Rate=−14Δ[NH3]Δt=−15Δ[O2]Δt=+14Δ[NO]Δt=+16Δ[H2O]Δt\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

Δ[O2]Δt=−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:

Rate=−15Δ[O2]Δt\text{Rate} = -\frac{1}{5}\frac{\Delta[O_2]}{\Delta t}

Rate=−15(−2.0×10−3)\text{Rate} = -\frac{1}{5}(-2.0 \times 10^{-3})

Rate=2.0×10−35\text{Rate} = \frac{2.0 \times 10^{-3}}{5}

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

Answer (a):

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

(b) Rate of NO production

Using rate expression:

Rate=+14Δ[NO]Δt\text{Rate} = +\frac{1}{4}\frac{\Delta[NO]}{\Delta t}

Solve for Δ[NO]/Δt:

Δ[NO]Δt=4×Rate\frac{\Delta[NO]}{\Delta t} = 4 \times \text{Rate}

Δ[NO]Δt=4×(4.0×10−4)\frac{\Delta[NO]}{\Delta t} = 4 \times (4.0 \times 10^{-4})

Δ[NO]Δt=16×10−4\frac{\Delta[NO]}{\Delta t} = 16 \times 10^{-4}

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

Answer (b):

Δ[NO]Δt=1.6×10−3 M/s\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:

Rate=−14Δ[NH3]Δt\text{Rate} = -\frac{1}{4}\frac{\Delta[NH_3]}{\Delta t}

Solve for Δ[NH₃]/Δt:

Δ[NH3]Δt=−4×Rate\frac{\Delta[NH_3]}{\Delta t} = -4 \times \text{Rate}

Δ[NH3]Δt=−4×(4.0×10−4)\frac{\Delta[NH_3]}{\Delta t} = -4 \times (4.0 \times 10^{-4})

Δ[NH3]Δt=−16×10−4\frac{\Delta[NH_3]}{\Delta t} = -16 \times 10^{-4}

Δ[NH3]Δt=−1.6×10−3 M/s\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}

Answer (c):

Δ[NH3]Δt=−1.6×10−3 M/s\boxed{\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}}

Or, stating as consumption rate (magnitude):

NH3 consumed at 1.6×10−3 M/s\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:

∣Δ[NH3]/Δt∣4=1.6×10−34=4.0×10−4\frac{|\Delta[NH_3]/\Delta t|}{4} = \frac{1.6 \times 10^{-3}}{4} = 4.0 \times 10^{-4} ✓

∣Δ[O2]/Δt∣5=2.0×10−35=4.0×10−4\frac{|\Delta[O_2]/\Delta t|}{5} = \frac{2.0 \times 10^{-3}}{5} = 4.0 \times 10^{-4} ✓

∣Δ[NO]/Δt∣4=1.6×10−34=4.0×10−4\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:

Δ[NH3]/ΔtΔ[O2]/Δt=−1.6×10−3−2.0×10−3=1.62.0=45\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

Δ[NH3]/ΔtΔ[O2]/Δt=45\frac{\Delta[NH_3]/\Delta t}{\Delta[O_2]/\Delta t} = \frac{4}{5}

Δ[NH3]Δt=45×Δ[O2]Δt\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times \frac{\Delta[O_2]}{\Delta t}

Δ[NH3]Δt=45×(−2.0×10−3)\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times (-2.0 \times 10^{-3})

Δ[NH3]Δt=−1.6×10−3 M/s\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

2Problem 2medium

❓ 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:

\ce2NO(g)+Cl2(g)−>2NOCl(g)\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:

Rate=k[NO]m[Cl2]n\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:

rate2rate1=k[NO]2m[Cl2]2nk[NO]1m[Cl2]1n\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:

rate2rate1=[Cl2]2n[Cl2]1n=([Cl2]2[Cl2]1)n\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:

0.360.18=(0.200.10)n\frac{0.36}{0.18} = \left(\frac{0.20}{0.10}\right)^n

2=2n2 = 2^n

n=1n = 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:

rate3rate2=k[NO]3m[Cl2]3nk[NO]2m[Cl2]2n\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:

rate3rate2=[NO]3m[NO]2m=([NO]3[NO]2)m\frac{\text{rate}_3}{\text{rate}_2} = \frac{[NO]_3^m}{[NO]_2^m} = \left(\frac{[NO]_3}{[NO]_2}\right)^m

Substitute values:

1.440.36=(0.200.10)m\frac{1.44}{0.36} = \left(\frac{0.20}{0.10}\right)^m

4=2m4 = 2^m

22=2m2^2 = 2^m

m=2m = 2

Second order in NO

Rate law:

Rate=k[NO]2[Cl2]\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:

Rate=k[NO]2[Cl2]\text{Rate} = k[NO]^2[Cl_2]

Solve for k:

k=Rate[NO]2[Cl2]k = \frac{\text{Rate}}{[NO]^2[Cl_2]}

Substitute values from Exp 1:

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

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

k=0.180.001k = \frac{0.18}{0.001}

k=180k = 180

Determine units of k:

From rate equation:

Rate (M/s)=k[NO]2[Cl2]\text{Rate (M/s)} = k[NO]^2[Cl_2]

M/s=k×M2×M\text{M/s} = k \times \text{M}^2 \times \text{M}

M/s=k×M3\text{M/s} = k \times \text{M}^3

Solve for units of k:

k=M/sM3=Ms⋅M3=1M2⋅s=M−2s−1k = \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):

k=180 M−2s−1\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=0.36(0.10)2(0.20)=0.360.002=180k = \frac{0.36}{(0.10)^2(0.20)} = \frac{0.36}{0.002} = 180 ✓

Check with Exp 3:

k=1.44(0.20)2(0.20)=1.440.008=180k = \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:

Rate=k[NO]2[Cl2]\text{Rate} = k[NO]^2[Cl_2]

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

Calculate (0.050)²:

(0.050)2=0.0025(0.050)^2 = 0.0025

Calculate rate:

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

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

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

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

Answer (c):

Rate=9.0×10−3 M/s or 0.0090 M/s\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):

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

Step 2 (slow, rate-determining):

\ceNOCl2+NO−>2NOCl\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:

  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!

3Problem 3hard

❓ 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\ln[A]_t = \ln[A]_0 - kt

Or:

[A]t=[A]0e−kt[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 hour×60 min1 hour×60 s1 min=3600 st = 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⁡[H2O2]t=ln⁡[H2O2]0−kt\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt

Substitute values:

ln⁡[H2O2]t=ln⁡(0.30)−(1.8×10−5)(3600)\ln[H_2O_2]_t = \ln(0.30) - (1.8 \times 10^{-5})(3600)

Calculate each term:

ln⁡(0.30)=−1.204\ln(0.30) = -1.204

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

Substitute:

ln⁡[H2O2]t=−1.204−0.0648\ln[H_2O_2]_t = -1.204 - 0.0648

ln⁡[H2O2]t=−1.269\ln[H_2O_2]_t = -1.269

Take exponential:

[H2O2]t=e−1.269[H_2O_2]_t = e^{-1.269}

[H2O2]t=0.281 M[H_2O_2]_t = 0.281 \text{ M}

Answer (a):

[H2O2]t=0.28 M\boxed{[H_2O_2]_t = 0.28 \text{ M}}

Alternative method using exponential form:

[H2O2]t=[H2O2]0e−kt[H_2O_2]_t = [H_2O_2]_0 e^{-kt}

[H2O2]t=0.30×e−(1.8×10−5)(3600)[H_2O_2]_t = 0.30 \times e^{-(1.8 \times 10^{-5})(3600)}

[H2O2]t=0.30×e−0.0648[H_2O_2]_t = 0.30 \times e^{-0.0648}

e−0.0648=0.937e^{-0.0648} = 0.937

[H2O2]t=0.30×0.937=0.281 M[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⁡[H2O2]t=ln⁡[H2O2]0−kt\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt

Rearrange to solve for t:

kt=ln⁡[H2O2]0−ln⁡[H2O2]tkt = \ln[H_2O_2]_0 - \ln[H_2O_2]_t

kt=ln⁡([H2O2]0[H2O2]t)kt = \ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)

t=1kln⁡([H2O2]0[H2O2]t)t = \frac{1}{k}\ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)

Substitute values:

t=11.8×10−5ln⁡(0.300.10)t = \frac{1}{1.8 \times 10^{-5}}\ln\left(\frac{0.30}{0.10}\right)

t=11.8×10−5ln⁡(3.0)t = \frac{1}{1.8 \times 10^{-5}}\ln(3.0)

Calculate ln(3.0):

ln⁡(3.0)=1.099\ln(3.0) = 1.099

Calculate t:

t=1.0991.8×10−5t = \frac{1.099}{1.8 \times 10^{-5}}

t=6.106×104 st = 6.106 \times 10^4 \text{ s}

t=61,060 st = 61,060 \text{ s}

Convert to hours:

t=61,060 s×1 min60 s×1 hour60 mint = 61,060 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{1 \text{ hour}}{60 \text{ min}}

t=61,0603600 hourst = \frac{61,060}{3600} \text{ hours}

t=16.96 hours≈17 hourst = 16.96 \text{ hours} \approx 17 \text{ hours}

Answer (b):

t=6.1×104 s or 17 hours\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:

t1/2=0.693k=ln⁡2kt_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}

Substitute k:

t1/2=0.6931.8×10−5t_{1/2} = \frac{0.693}{1.8 \times 10^{-5}}

t1/2=3.85×104 st_{1/2} = 3.85 \times 10^4 \text{ s}

Convert to hours:

t1/2=3.85×1043600 hourst_{1/2} = \frac{3.85 \times 10^4}{3600} \text{ hours}

t1/2=10.7 hourst_{1/2} = 10.7 \text{ hours}

Answer (c):

t1/2=3.85×104 s or 10.7 hours\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:

0.300.10=3=21.585\frac{0.30}{0.10} = 3 = 2^{1.585}

So time needed:

t=1.585×t1/2=1.585×10.7=17.0 hourst = 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:

  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
Next Topic →
Integrated Rate Laws and Half-Life
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