Section 1

General Chemistry II: Kinetics and Equilibrium Midterm/Final - Cheatsheet

STUDY GUIDE

🎓 General Chemistry II: Kinetics and Equilibrium - Study Guide

📋 Course Structure


code
📚 General Chemistry II
├── 📖 Chapter 1: Foundations of Chemical Kinetics
├── 📖 Chapter 2: Rate Laws and the Method of Initial Rates
├── 📖 Chapter 3: Integrated Rate Laws and Half-Life
├── 📖 Chapter 4: Temperature, Collision Theory, and Activation Energy
├── 📖 Chapter 5: Reaction Mechanisms and Catalysis
├── 📖 Chapter 6: Principles of Chemical Equilibrium
├── 📖 Chapter 7: Equilibrium Calculations and the Reaction Quotient
└── 📖 Chapter 8: Le Chatelier’s Principle

Section 2

📖 Chapter 1: Foundations of Chemical Kinetics

What this chapter covers: Introduction to reaction rates and their measurement. It explores the relationship between stoichiometry and the rate of disappearance/appearance of species, alongside the qualitative factors that influence reaction speed.

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	Key Variable/Unit	When to Use
Reaction Rate	$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = \frac{1}{d}\frac{\Delta[D]}{\Delta t}$	$M/s$ (Molarity/sec)	Relating rates of different species
Surface Area	Smaller particles = Higher Rate	$m^2/g$	Solid reactants in heterogeneous mix
Instantaneous Rate	Slope of tangent at time $t$	$\frac{d[A]}{dt}$	Finding rate at a specific moment
Average Rate	Rate over a time interval $\Delta t$	$\frac{\Delta[A]}{\Delta t}$	General speed over a long duration

🛠️ Problem Types

Type A: Rate Stoichiometry

Setup: Given the rate of disappearance of one reactant, find the rate of appearance of a product.

Method: Use the balanced equation coefficients: $\text{Rate}_A / \text{coeff}_A = \text{Rate}_B / \text{coeff}_B$ .

Type B: Qualitative Rate Factors

Setup: Predict how changing conditions (temperature, concentration, surface area) affects speed.

Method: Identify the factor; e.g., crushing a solid increases surface area, thus increasing collision frequency and rate.

🧮 Solved Example

Problem: In the reaction $2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)$ , the rate of disappearance of $N_2O_5$ is $6.0 \times 10^{-4} M/s$ . Find the rate of appearance of $NO_2$ .

Given: $\text{Rate}(N_2O_5) = 6.0 \times 10^{-4} M/s$ ; Coefficients: $N_2O_5 = 2$ , $NO_2 = 4$ .

Steps:

Set up the equality: $-\frac{1}{2}\frac{\Delta[N_2O_5]}{\Delta t} = \frac{1}{4}\frac{\Delta[NO_2]}{\Delta t}$ .
Substitute known value: $\frac{1}{2}(6.0 \times 10^{-4}) = \frac{1}{4}(\text{Rate of } NO_2)$ .
Solve for $NO_2$ : $\text{Rate} = 4 \times (3.0 \times 10^{-4}) = 1.2 \times 10^{-3} M/s$ .

"

✅
Answer: $1.2 \times 10^{-3} M/s$

⚠️ Common Mistakes

❌ Mistake: Forgetting the negative sign for reactants.

✅ How to avoid: Remember reactants are consumed (negative change), but "Rate" is always reported as a positive value.

📖 Chapter 2: Rate Laws and the Method of Initial Rates

What this chapter covers: The mathematical dependency of reaction rate on reactant concentrations. It focuses on determining reaction orders ( $n, m$ ) and the rate constant ( $k$ ) using experimental data.

🔑 Essential Concepts & Formulas

Concept/Formula	Definition/Equation	Key Variable/Unit	When to Use
Rate Law	$\text{Rate} = k[A]^n[B]^m$	$n, m$ : Reaction Orders	Calculating rate from concentration
Overall Order	Sum of exponents ( $n + m + \dots$ )	Unitless integer	Determining units of $k$
0th Order $k$	Rate is independent of $[A]$	$M \cdot s^{-1}$	When doubling $[A]$ does nothing
1st Order $k$	Rate is proportional to $[A]$	$s^{-1}$	When doubling $[A]$ doubles rate
2nd Order $k$	Rate proportional to $[A]^2$	$M^{-1} \cdot s^{-1}$	When doubling $[A]$ quadruples rate

🛠️ Problem Types

Type A: Determining Rate Law from Initial Rates Table

Setup: Given a table with multiple trials of initial concentrations and rates.

Method: Compare two trials where only one concentration changes. Use $\frac{\text{Rate}_2}{\text{Rate}_1} = (\frac{[A]_2}{[A]_1})^n$ to solve for $n$ .

Type B: Calculating the Rate Constant $k$

Setup: Once orders are found, solve for the constant.

Method: Plug values from any single trial into the determined rate law: $k = \frac{\text{Rate}}{[A]^n[B]^m}$ .

🧮 Solved Example

Problem: Find the rate law for $A + B \rightarrow C$ given: Trial 1 ( $[A]=0.1, [B]=0.1, \text{Rate}=2 \times 10^{-3}$ ), Trial 2 ( $[A]=0.2, [B]=0.1, \text{Rate}=8 \times 10^{-3}$ ).

Given: Trial 1 vs 2: $[A]$ doubles, $[B]$ is constant, Rate quadruples ( $8/2 = 4$ ).

Steps:

Use ratio: $4 = (2)^n$ .
Solve for $n$ : $n = 2$ (Second order in $A$ ).
Repeat for $B$ (if data provided). If $B$ was constant, order is determined by other trials.

"

✅
Answer: $\text{Rate} = k[A]^2$

⚠️ Common Mistakes

❌ Mistake: Using stoichiometric coefficients as reaction orders.

✅ How to avoid: Orders must be determined experimentally (from data tables), not from the balanced equation.

📖 Chapter 3: Integrated Rate Laws and Half-Life

What this chapter covers: Relating concentration to time. This chapter provides the tools to calculate how much reactant remains after a specific duration and defines the concept of half-life.

🔑 Essential Concepts & Formulas

Order	Integrated Rate Law	Linear Plot	Half-Life ( $t_{1/2}$ )
0th	$[A]_t = -kt + [A]_0$	$[A]$ vs $t$	$\frac{[A]_0}{2k}$
1st	$\ln[A]_t = -kt + \ln[A]_0$	$\ln[A]$ vs $t$	$\frac{0.693}{k}$
2nd	$\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$	$\frac{1}{[A]}$ vs $t$	$\frac{1}{k[A]_0}$

🛠️ Problem Types

Type A: Concentration at Time $t$

Setup: Given $k$ , initial concentration, and time, find remaining amount.

Method: Identify the order, select the correct integrated equation, and solve for $[A]_t$ .

Type B: Identifying Order Graphically

Setup: Given three graphs ( $[A], \ln[A], 1/[A]$ vs $t$ ).

Method: The graph that is a straight line indicates the order. Slope absolute value $= k$ .

🧮 Solved Example

Problem: A 1st-order reaction has $k = 0.0231 \text{ min}^{-1}$ . How long until 75% of the reactant is consumed?

Given: $k = 0.0231$ ; 75% consumed means 25% remains ( $[A]_t = 0.25[A]_0$ ).

Steps:

Use 1st order law: $\ln(\frac{[A]_t}{[A]_0}) = -kt$ .
Substitute: $\ln(0.25) = -(0.0231)t$ .
Solve: $-1.386 = -0.0231t \rightarrow t = 60 \text{ min}$ .

"

✅
Answer: $60 \text{ minutes}$

⚠️ Common Mistakes

❌ Mistake: Using the wrong units for time (e.g., mixing seconds and minutes).

✅ How to avoid: Ensure the time unit in $t$ matches the time unit in the rate constant $k$ .

7 more sections

Create a free account to import and read the full study notes — all 9 sections.

No credit card · 2 free imports included

General Chemistry II: Kinetics and Equilibrium Midterm/Final - Cheatsheet

Study Notes Preview

General Chemistry II: Kinetics and Equilibrium Midterm/Final - Cheatsheet

🎓 General Chemistry II: Kinetics and Equilibrium - Study Guide

📋 Course Structure

📖 Chapter 1: Foundations of Chemical Kinetics

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

📖 Chapter 2: Rate Laws and the Method of Initial Rates

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

📖 Chapter 3: Integrated Rate Laws and Half-Life

🔑 Essential Concepts & Formulas

🛠️ Problem Types

🧮 Solved Example

⚠️ Common Mistakes

7 more sections