12.2: Integrated Rate Laws and Collision Theory

Integrated Rate Laws and Collision Theory

Context: Understanding how reaction rate depends on time, concentration, temperature, and molecular factors.

12.4 Integrated Rate Laws

Simple rate laws can be rearranged to solve for the concentration at any time t, provided you know the rate constant k and the initial concentration [A]0.
Three common rate laws and their integrated forms:
- Zero order:
- Rate = k
- Integrated: $[A]t = -k t + [A]0$
- First order:
- Rate = k[A]
- Integrated: $\ln [A]t = -k t + \ln [A]0$
- Second order:
- Rate = k[A]^2
- Integrated: $\frac{1}{[A]t} = k t + \frac{1}{[A]0}$
How to identify the order from integrated laws:
- Look for patterns in plots: 1/[A] vs t is linear for second order; ln[A] vs t is linear for first order; [A] vs t is linear for zero order.
- You must know the reaction order before applying an integrated rate law.
- Note: [A]_t always decreases with time for a reacting system (concentration fall as time increases).

How to determine order and use integrated laws

Pattern recognition: if 1/[A] is linear with t → second order; if ln[A] is linear with t → first order; if [A] is linear with t → zero order.
Units of k often indicate the reaction order:
- Zero order: k has units of concentration per time, e.g., M s^-1.
- First order: k has units of s^-1.
- Second order: k has units of M^-1 s^-1.
Practical steps:
- Determine the linear plot from experimental [A] vs t data.
- Use the appropriate integrated rate law to extract k (slope) and [A]0 (intercept).

Worked examples and problems (key ideas and solutions)

Example problem (page 5):
- Question 1: What is [A] at t = 25 s if [A]0 = 0.75 M and k = 0.15 s^-1?
- Question 2: What is [A]0 if [A]_{15} = 0.89 and k = 0.014 L mol^-1 s^-1?
- Note: The units of k reveal the reaction order.
Example (page 6): Zero-order reaction
- Given: [A]t = -k t + [A]0 with [A]0 = 0.85 M, [A]t = 0.65 M after t = 50 s.
- Calculation: k = ([A]0 - [A]t) / t = (0.85 - 0.65) / 50 = 0.20 / 50 = 0.004 M s^-1
- Correct choice: A. 0.004 mol/L-s
Pattern-check activity (page 7-11):
- A dataset shows time versus [A], with [A] decreasing over time.
- Students assess which plot yields a straight line ([A] vs t, ln[A] vs t, or 1/[A] vs t) to identify the order.
- For a given dataset, the corresponding plot is checked for linearity to decide the order.
Data interpretation (pages 9-11):
- Example dataset shows:
- Time (s): 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
- [A]: values decreasing from ~1.750 to ~0.644 M
- ln[A]: ln of those concentrations
- 1/[A]: reciprocals of concentrations
- The data are arranged so that one of the plots (either [A] vs t, ln[A] vs t, or 1/[A] vs t) is linear; this identifies the order.
Summary from the dataset (page 12):
- For the provided data, ln[A]t vs t is a straight line, so the reaction is first-order:
- $\ln [A]t = -k t + \ln [A]0$

Plotting rules and table (Integrated rate laws)

Zero-order: plot of [A]t vs t is a straight line.
- Slope: -k
- y-intercept: [A]0
First-order: plot of ln([A]t) vs t is a straight line.
- Slope: -k
- y-intercept: ln([A]0)
Second-order: plot of 1/[A]t vs t is a straight line.
- Slope: k
- y-intercept: 1/[A]0

Half-life concepts (12.4)

Definition: t1/2 is the time required for [A] to drop to half of its initial value.
How to obtain t1/2: set [A]t = 1/2 [A]0 and solve using the integrated rate law.
Formulas by order:
- Zero-order: $t{1/2} = \frac{[A]0}{2k}$
- First-order: $t_{1/2} = \frac{\ln 2}{k}$
- Second-order: $t{1/2} = \frac{1}{k [A]0}$

Example problem (page 15) – Second-order half-life scenario

Given: The half-life t1/2 = 25 s for a 0.80 M solution; determine [A] after t = 100 s for a second-order reaction.
Use second-order integrated law: $\frac{1}{[A]t} = k t + \frac{1}{[A]0}$
First find k from t1/2 relation: $t{1/2} = \frac{1}{k [A]0} \Rightarrow k = \frac{1}{t{1/2} [A]0} = \frac{1}{25 \times 0.80} = 0.05\ \text{L}\,\text{mol}^{-1}\,\text{s}^{-1}$
Then compute: $\frac{1}{[A]t} = (0.05)(100) + \frac{1}{0.80} = 5 + 1.25 = 6.25 \Rightarrow [A]t = \frac{1}{6.25} = 0.16\ \text{M}$
Answer: C) 0.16 M

Temperature effects and Arrhenius equation (12.5)

Key idea: Temperature changes rate constant k; higher T typically increases k because reactant molecules collide more energetically.
Activation energy (Ea): energy barrier that must be overcome for a reaction to occur.
Frequency factor (A): accounts for the number of collisions that could lead to a reaction.
Arrhenius equations:
- Original form: $k = A \, e^{-\frac{E_a}{R T}}$
- Linear form: $\ln k = \ln A - \frac{E_a}{R T}$
Temperature dependence in practice:
- As T increases, k increases, making the reaction faster.
- Ea is the energy barrier; A reflects how often productive collisions occur.

Common exercise (12.5) – Temperature example (page 18)

Given: Ea = 12{,}500 J/(mol·K); R = 8.314 J/(mol·K).
Relationship to compare two temperatures T1 and T2:
- $\ln \left( \frac{k2}{k1} \right) = -\frac{Ea}{R} \left( \frac{1}{T2} - \frac{1}{T_1} \right)$
Example values (qualitative): If k at 25°C is known, you can compute k at 125°C using the above formula by converting temperatures to Kelvin:
- T1 = 25°C = 298.15 K
- T2 = 125°C = 398.15 K
- With Ea = 12{,}500 J/mol, compute the ratio and then k2.
Note: This demonstrates how rate constants respond to temperature changes via Ea and R.

Collision Theory (12.5)

Core idea: A chemical reaction A + B → products occurs only when A and B collide in the correct way.
Factors affecting collision frequency and outcome:
- Concentration: Higher concentrations increase the number of collisions, raising the rate.
- Temperature: Higher temperature speeds molecules, increasing the fraction of collisions with enough energy to overcome Ea.
- Activation energy: Collisions must have energy at least Ea to lead to reaction; energy distribution matters.
- Orientation: Collisions must occur with correct relative orientation for bonds to form; not all collisions are productive.

Energy and transition concepts (pages 22-27)

Energy diagram concepts:
- Reactants → Activation energy Ea → Transition state → products
- The transition state (activated complex) is a high-energy, unstable species halfway between reactants and products.
- The energy difference between reactants and products determines endothermic vs exothermic character.
Transition state details:
- During a reaction, partial bonds exist where bonds are being formed and broken.
- The transition state is higher in energy than either reactants or products.
Orientation and hindrance:
- The top side of a molecule can be hindered, limiting productive collisions.
- More hindrance lowers the frequency factor A; less hindrance raises A.

Example orientation-factor question (page 25)

Question: Which reaction would have the highest orientation factor? (H2 adding to blue carbons in various substrates.)
Concept: Orientation factor reflects how likely a collision aligns reactants correctly to form products; variations depend on molecular geometry.

Quick connections to broader ideas

Integrated rate laws connect kinetics with observable concentration-time data, enabling determination of reaction order from experimental plots.
Arrhenius kinetics links microscopic molecular properties (Ea, A) to macroscopic rate constants and their temperature dependence.
Collision theory provides a molecular-level rationale for why temperature, concentration, and orientation affect reaction rates.
Transition state theory builds on collision ideas to describe the high-energy peak that must be crossed during a reaction.

Quick reference formulas (LaTeX)

Zero-order integrated law: $[A]t = -k t + [A]0$
First-order integrated law: $\ln [A]t = -k t + \ln [A]0$
Second-order integrated law: $\frac{1}{[A]t} = k t + \frac{1}{[A]0}$
Half-lives:
- $t{1/2} = \frac{[A]0}{2k}$ (zero-order)
- $t_{1/2} = \frac{\ln 2}{k}$ (first-order)
- $t{1/2} = \frac{1}{k [A]0}$ (second-order)
Arrhenius equation:
- $k = A \, e^{-\frac{E_a}{R T}}$
- $\ln k = \ln A - \frac{E_a}{R T}$
Temperature comparison (two temperatures):
- $\ln \left( \frac{k2}{k1} \right) = -\frac{Ea}{R} \left( \frac{1}{T2} - \frac{1}{T_1} \right)$ where T1, T2 are in Kelvin

Summary takeaways

To analyze a reaction, identify the appropriate integrated rate law by recognizing patterns in data plots or unit analysis of k.
The order of the reaction determines how concentration changes with time and what the half-life looks like.
Temperature and catalysts affect rate; Arrhenius equation provides a quantitative link between k and T via Ea and A.
Collision theory emphasizes the importance of sufficient energy, correct orientation, and frequent collisions in driving reaction rates.
The concept of the transition state helps explain why not all collisions lead to products and how energy barriers govern reaction rates.