Workshop 2: Chemical Kinetics

CHM152 ~ General Chemistry II

⚛ TEHRANI ⚛

⚛ What is Chemical Kinetics?

Kinetics is the study of the rate of chemical reactions — how fast reactants are consumed and products are formed.

This matters in the real world:

⚛ The Big Idea

For the reaction: C₂H₄ + HCl → C₂H₅Cl

The rate is defined as the change in concentration over change in time, with units of M·s⁻¹ (molarity per second).

Rate = −Δ[reactant] / Δt = +Δ[product] / Δt
○ Key insight: The rate is the negative of the change in reactant concentration (because reactants decrease) and the positive of the change in product concentration (because products increase).

⚛ What You Will Learn

  1. Write rate expressions with stoichiometric coefficients
  2. Use the method of initial rates to find rate laws
  3. Determine reaction order (0th, 1st, 2nd)
  4. Apply integrated rate laws with graphical analysis
  5. Use the Arrhenius equation for temperature dependence

⚛ Important Vocabulary

TermDefinition
RateChange in concentration per unit time (M·s⁻¹)
Rate LawMathematical relationship: rate = k[A]a[B]b
Rate Constant (k)Proportionality constant specific to each reaction & temperature
OrderExponent on concentration in the rate law (determined experimentally)
Activation Energy (Ea)Minimum energy required for a reaction to occur
Arrhenius Factor (A)Pre-exponential factor relating to collision frequency/orientation

⚛ Rate Expressions & Stoichiometry

When a reaction has different stoichiometric coefficients, the rate of change of each species must be divided by its coefficient to give the same overall rate.

⚛ The Rule

For a general reaction: aA + bB → cC + dD

Rate = −(1/a)·Δ[A]/Δt = −(1/b)·Δ[B]/Δt = +(1/c)·Δ[C]/Δt = +(1/d)·Δ[D]/Δt

⚛ Worked Example

3H₂ + N₂ → 2NH₃

Rate = −(1/3)·Δ[H₂]/Δt = −Δ[N₂]/Δt = +(1/2)·Δ[NH₃]/Δt

Notice: H₂ disappears 3 times faster than N₂, so we divide its rate by 3 to get a single unified rate.

⚛ Another Example

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Rate = −Δ[CH₄]/Δt = −(1/2)·Δ[O₂]/Δt = +Δ[CO₂]/Δt = +(1/2)·Δ[H₂O]/Δt

⚛ Try It: Write the Rate Expression

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

What coefficient goes in front of Δ[O₂]/Δt?

Rate = −Δ[C₃H₈]/Δt = −(1/)·Δ[O₂]/Δt = +(1/)·Δ[CO₂]/Δt = +(1/)·Δ[H₂O]/Δt

Answer: 1/5 for O₂, 1/3 for CO₂, 1/4 for H₂O. The coefficients in the balanced equation become the denominators.

⚛ Try It: Another One

2Fe(s) + 6HCl(aq) → 2FeCl₃(aq) + 3H₂(g)

Rate = −(1/)·Δ[Fe]/Δt = −(1/)·Δ[HCl]/Δt = +(1/)·Δ[FeCl₃]/Δt = +(1/)·Δ[H₂]/Δt

Answer: 1/2 for Fe, 1/6 for HCl, 1/2 for FeCl₃, 1/3 for H₂.

⚛ The Method of Initial Rates

This is the most common experimental method to determine the rate law: rate = k[A]a[B]b

The exponents a and b are the orders of the reaction with respect to each reactant. They must be determined experimentally — they are not the stoichiometric coefficients!

⚛ The Strategy (3 Steps)

  1. Hold one reactant constant — compare two experiments where only one concentration changes
  2. Find the ratio — divide the rates and concentrations:
    Rate₂ / Rate₁ = ([A]₂ / [A]₁)a
  3. Solve for the exponent — if the ratio equals 2a, 3a, etc., find a

⚛ Worked Example 1

C₄H₉Br + OH⁻ → C₄H₉OH + Br⁻

Exp[C₄H₉Br] M[OH⁻] MRate (M·s⁻¹)
10.100.100.0010
20.200.100.0020
30.300.100.0030
40.100.200.0010
50.100.300.0010

Finding the order with respect to C₄H₉Br:

Compare Exp 1 & 2 (OH⁻ is constant at 0.10 M):

Rate₂/Rate₁ = 0.0020/0.0010 = 2     [A]₂/[A]₁ = 0.20/0.10 = 2
2 = 2a → a = 1

Finding the order with respect to OH⁻:

Compare Exp 1 & 4 (C₄H₉Br is constant at 0.10 M):

Rate₄/Rate₁ = 0.0010/0.0010 = 1     [OH⁻]₄/[OH⁻]₁ = 0.20/0.10 = 2
1 = 2b → b = 0
○ Rate law: Rate = k[C₄H₉Br]¹[OH⁻]⁰ = k[C₄H₉Br]
1st order in C₄H₉Br, zero order in OH⁻, 1st order overall
k = 0.0010 / 0.10 = 0.010 s⁻¹

⚛ Worked Example 2

2NO(g) + Br₂(g) → 2NOBr(g)

Exp[NO] M[Br₂] MRate (M·s⁻¹)
10.100.1012
20.100.2024
30.100.3036
40.200.1048
50.300.10108

Order in NO: Compare Exp 1 & 4 (Br₂ constant):

48/12 = 4    and    (0.20/0.10)a = 2a = 4 → a = 2

Order in Br₂: Compare Exp 1 & 2 (NO constant):

24/12 = 2    and    (0.20/0.10)b = 2b = 2 → b = 1
○ Rate law: Rate = k[NO]²[Br₂]
2nd order in NO, 1st order in Br₂, 3rd order overall
k = 12 / (0.10² × 0.10) = 1.2 × 10⁴ M⁻²s⁻¹

⚛ Quick Reference: Determining the Exponent

Conc. RatioRate RatioOrder
doubles (×2)stays same (×1)0
doubles (×2)doubles (×2)1
doubles (×2)quadruples (×4)2
doubles (×2)octuples (×8)3
triples (×3)stays same (×1)0
triples (×3)triples (×3)1
triples (×3)×92

⚛ Practice: Find the Rate Law

NO₂(g) + O₃(g) → NO₃(g) + O₂(g)

Exp[NO₂] M[O₃] MRate (M·s⁻¹)
15.0 × 10⁻⁵1.0 × 10⁻⁵0.022
25.0 × 10⁻⁵2.0 × 10⁻⁵0.044
32.5 × 10⁻⁵2.0 × 10⁻⁵0.022

Order in NO₂:   Order in O₃:

Solution:
Order in O₃: Compare Exp 1 & 2 (NO₂ constant): 0.044/0.022 = 2, conc doubles → b = 1
Order in NO₂: Compare Exp 2 & 3 (O₃ constant): 0.022/0.044 = 0.5, conc halves (0.5a = 0.5) → a = 1
Rate = k[NO₂][O₃], both first order, second order overall
k = 0.022 / (5.0×10⁻⁵ × 1.0×10⁻⁵) = 4.4 × 10⁷ M⁻¹s⁻¹

⚛ Integrated Rate Laws

Another way to determine reaction order: use concentration vs. time data and see which plot gives a straight line.

⚛ The Three Integrated Rate Laws

Zero Order

[A]t = −kt + [A]0

Plot: [A] vs. t

Slope = −k

Linear → zero order

First Order

ln[A]t = −kt + ln[A]0

Plot: ln[A] vs. t

Slope = −k

Linear → first order

Second Order

1/[A]t = kt + 1/[A]0

Plot: 1/[A] vs. t

Slope = k

Linear → second order

⚛ Worked Example: 2H₂O₂ → 2H₂O + O₂

Time (s)[H₂O₂] Mln[H₂O₂]1/[H₂O₂]
01.0000.0001.00
1200.910−0.0941.10
3000.780−0.2481.28
6000.590−0.5281.69
12000.370−0.9942.70
18000.220−1.5144.55
24000.130−2.0407.69
30000.083−2.48912.05
36000.050−2.99620.00
○ The plot of ln[H₂O₂] vs. time gives a straight line with R² = 0.9997.
→ The reaction is first order.
→ k = 0.000834 s⁻¹ (from the slope).

⚛ How to Decide: The Graphical Method

  1. Make three plots from your data: [A] vs t, ln[A] vs t, and 1/[A] vs t
  2. The plot that gives a straight line tells you the order
  3. The slope of that line gives you k (note: slope = −k for zero and first order, slope = +k for second order)

⚛ Interactive: Which Plot is Linear?

Given this data for the reaction A → B:

Time[A]ln[A]1/[A]
00.200−1.6095.00
400.153−1.8776.54
800.124−2.0888.06
1200.104−2.2639.62
1600.090−2.40811.11
2000.079−2.53812.66

Look at the columns. Which changes by roughly equal increments?

Answer: Second Order. The 1/[A] column increases by roughly 1.5–1.6 each time interval — that's approximately constant. The reaction is second order.
k = slope ≈ (12.66 − 5.00) / (200 − 0) = 0.0383 M⁻¹s⁻¹

⚛ Units of k Depend on Order

OrderUnits of k
ZeroM·s⁻¹
Firsts⁻¹
SecondM⁻¹·s⁻¹
○ Tip: The units of k always work out so that the rate law gives M·s⁻¹.

⚛ Temperature Dependence of Reaction Rates

Increasing temperature typically increases the rate because molecules have more energy and collide more frequently with enough energy to react.

⚛ The Arrhenius Equation

k = A·e−Ea/RT

Where:

⚛ The Linearized Form

Taking the natural log of both sides:

ln k = ln A − Ea/(RT) = −(Ea/R)·(1/T) + ln A

This is in the form y = mx + b:

⚛ Worked Example

H₂(g) + I₂(g) → 2HI(g)

T (°C)T (K)k (M⁻¹s⁻¹)1/T (K⁻¹)ln k
2835561.20×10⁻⁴0.001799−9.028
3025753.50×10⁻⁴0.001739−7.958
3556286.80×10⁻³0.001592−4.991
3936663.80×10⁻²0.001502−3.270
4307031.70×10⁻¹0.001422−1.772
○ From the plot of ln k vs. 1/T:
Slope = −19,432 K = −Ea/R
Ea = 19,432 K × 8.314 J/(mol·K) = 162 kJ/mol
A = e25.9 = 1.8 × 10¹¹ M⁻¹s⁻¹

⚛ Two-Point Arrhenius Form

If you only have two temperatures, use:

ln(k₂/k₁) = −(Ea/R)·(1/T₂ − 1/T₁)

This lets you find Ea from just two data points, or predict k at a new temperature if you know Ea.

⚛ Quick Check

If the slope of an Arrhenius plot (ln k vs 1/T) is −12,500 K, what is Ea?

Ea = kJ/mol

Answer: Ea = 12,500 K × 8.314 J/(mol·K) = 103,925 J/mol = 104 kJ/mol

⚛ Kinetics Calculator

Use these tools to check your work on practice problems.

⚛ Rate Law Calculator (Method of Initial Rates)

Enter two experiments to find the order:



⚛ Arrhenius Ea Calculator

Enter slope from ln k vs. 1/T plot:

⚛ Two-Point Arrhenius Calculator

Find Ea from two (T, k) pairs:



⚛ Self-Check Quiz

Test your understanding before tackling the worksheet. No grade — just practice!

Question 1

The rate of a reaction has units of:




Question 2

For 2A + 3B → C, how is the rate expressed for reactant B?




Question 3

In the rate law rate = k[A]²[B], if [A] is doubled while [B] stays the same, the rate:




Question 4

If a plot of ln[A] vs. time gives a straight line, the reaction is:




Question 5

In an Arrhenius plot of ln k vs. 1/T, the slope equals:




Question 6

A first-order reaction has a rate constant with units of: