#Chemical Kinetics: Elementary Reactions and Rate Laws

Let's dive into the heart of reaction speeds! This guide will help you master rate laws and elementary reactions, crucial for acing your AP Chemistry exam.

#What are Rate Laws?

At its core, a rate law is a mathematical expression that shows how the rate of a chemical reaction depends on the concentration of reactants. For a simple reaction like A → B, the rate law is given by:

$$R = k[A]^n$$

Where:

• R is the reaction rate (how fast reactants turn into products).
• k is the rate constant (a proportionality constant specific to the reaction at a given temperature).
• [A] is the concentration of reactant A.
• n is the order of the reaction with respect to reactant A (experimentally determined).

#Experimental Determination of Rate Laws

#Why Experiments are Essential 🧪

• You cannot determine the rate law by simply looking at the balanced chemical equation. The stoichiometric coefficients do not necessarily equal the reaction orders.
• Rate laws must be found through experimentation.

#The Experimental Process

1. Run multiple trials of the reaction.
2. Vary the initial concentrations of reactants, typically doubling one reactant concentration at a time while keeping others constant.
3. Measure the initial rate of the reaction for each trial.
4. Analyze how changes in concentration affect the rate to determine the reaction order for each reactant.

#How Concentration Changes Affect Rate

Let's see how doubling concentration affects rate:

• If doubling the concentration of a reactant doubles the rate, the reaction is first order with respect to that reactant (n=1).
• If doubling the concentration of a reactant quadruples the rate, the reaction is second order with respect to that reactant (n=2).
• If doubling the concentration of a reactant octuples the rate, the reaction is third order with respect to that reactant (n=3).

#GIf Courtesy of GIPHY

#Example Problem: Finding a Rate Law

Let's find the rate law for the reaction: 2NO + 2H₂ → N₂ + 2H₂O using the data below:

#1. Finding the Order with Respect to NO

• Compare experiments 1 and 2: [NO] doubles, [H₂] is constant.
• The rate increases by a factor of 5.0 x 10⁻⁵ / 1.25 x 10⁻⁵ = 4. * Since doubling [NO] quadruples the rate, the reaction is second order with respect to NO.

#2. Finding the Order with Respect to H₂

• Compare experiments 2 and 3: [H₂] doubles, [NO] is constant.
• The rate increases by a factor of 1.0 x 10⁻⁴ / 5.0 x 10⁻⁵ = 2. * Since doubling [H₂] doubles the rate, the reaction is first order with respect to H₂.

#3. Putting it all Together

The rate law is: R = k[NO]²[H₂]

• To find k, plug in the values from any experiment and solve. For example, using experiment 1:

$$1.25 \times 10^{-5} = k(0.10)^2(0.10)$$

$$k = 0.0125 M^{-2}s^{-1}$$

#Elementary Reactions

#What are they?

• An elementary reaction is a reaction that occurs in a single step.
• It involves either a single molecule or a group of molecules colliding in a single event.
• Elementary reactions are the building blocks of more complex reactions.

#Types of Elementary Reactions

• Elementary reactions can be first-order or second-order, depending on the number of molecules involved in the single step.
• Examples include:
  • The reaction of hydrogen and oxygen to form water.
  • The decomposition of ozone.
  • The ionization of a gas.

Understanding elementary reactions is crucial for grasping more complex reaction mechanisms.

#AP Practice Question - 2017 #2

Let's tackle a real AP question! Here’s a breakdown of the 2017 AP Chemistry Free Response Question #2, focusing on parts (e) and (f), which relate directly to rate laws and experimental design.

#The Scenario

The decomposition of urea, CO(NH₂)₂ (aq) ⇌ NH₄⁺ (aq) + OCH⁻ (aq), is studied at 90°C. The following data is collected:

#Part (e)

(i) Explain how the data supports the rate law: rate = k[CO(NH₂)₂].

(ii) Using the proposed rate law and the data, determine the value of the rate constant, k, including units.

#Part (f)

Describe an experiment to determine whether the rate of the reaction depends on the concentration of OH⁻(aq).

#Answers and Explanations

#Part (e)(i) - Identifying First Order Reactions

There are two ways to determine if the reaction is first order:

1. Graph ln[CO(NH₂)₂] vs. time: If the graph is linear, the reaction is first order.
2. Check for a constant half-life: If the time it takes for the concentration to halve is constant, the reaction is first order.

In this case, the provided graph is not linear, so we'll use the half-life method:

• The initial concentration is 0.1000 M. After one half-life, it should be 0.0500 M, which occurs at 10 hours.
• After another half-life, the concentration should be 0.0250 M, which occurs at 20 hours.
• The half-life is constant at 10 hours, confirming the reaction is first order.

#Part (e)(ii) - Calculating the Rate Constant

• We know the reaction is first order, and we know the half-life is 10 hours.
• Use the half-life equation for first-order reactions: t₁/₂ = 0.693/k

$$10 = \frac{0.693}{k}$$

$$k = \frac{0.693}{10} = 0.0693 h^{-1}$$

#Part (f) - Designing an Experiment

To test if the reaction depends on [OH⁻], you should:

1. Change the initial concentration of OH⁻ while keeping other conditions (like temperature) constant.
2. Measure how the rate of decomposition of CO(NH₂)₂ changes with the varying [OH⁻].

#Final Exam Focus

#High-Priority Topics

• Rate Laws: How to write them, determine reaction orders experimentally, and calculate the rate constant.
• Elementary Reactions: Understanding single-step reactions and how they relate to overall reaction mechanisms.
• Half-Life: Applying the concept to first-order reactions and using it to calculate the rate constant.

#Common Question Types

• Multiple Choice: Identifying reaction orders, analyzing experimental data, and understanding the relationship between rate laws and reaction mechanisms.
• Free Response: Designing experiments to determine rate laws, calculating rate constants, and explaining reaction mechanisms.

#Last-Minute Tips

• Time Management: Quickly identify the key information in the question and don't get bogged down on one part.
• Common Pitfalls: Watch out for incorrect units, and remember that stoichiometric coefficients are not the same as reaction orders.
• Strategies for Challenging Questions: Break down complex problems into smaller parts, and always show your work for partial credit.

#Practice Questions

Practice Question

#Multiple Choice Questions

1. The rate law for the reaction 2A + B → C is given by rate = k[A][B]². If the concentration of A is doubled and the concentration of B is halved, the rate of the reaction will be: (A) unchanged (B) doubled (C) halved (D) quadrupled

2. For a reaction A → products, the following data were obtained:

Time (min)	[A] (M)
0	1.00
10	0.50
20	0.25

What is the order of the reaction with respect to A? (A) zero order (B) first order (C) second order (D) third order

#Free Response Question

The reaction between hydrogen and iodine in the gas phase is represented by the equation: H₂(g) + I₂(g) → 2HI(g)

The following data were obtained in a series of experiments:

Experiment	[H₂] (M)	[I₂] (M)	Initial Rate (M/s)
1	0.10	0.10	2.0 x 10⁻⁵
2	0.20	0.10	4.0 x 10⁻⁵
3	0.10	0.20	8.0 x 10⁻⁵

(a) Write the rate law for the reaction. (b) Calculate the rate constant, k, including units. (c) Calculate the initial rate of the reaction if [H₂] = 0.30 M and [I₂] = 0.40 M. (d) Propose a possible mechanism for this reaction involving an intermediate.

#Scoring Breakdown

(a) Rate Law (2 points)

• 1 point for correct orders with respect to H₂ and I₂
• 1 point for correct rate law expression: rate = k[H₂][I₂]

(b) Rate Constant (2 points)

• 1 point for correct calculation of k
• 1 point for correct units: M⁻¹s⁻¹ $$k = \frac{2.0 \times 10^{-5}}{(0.10)(0.10)} = 2.0 \times 10^{-3} M^{-1}s^{-1}$$

(c) Initial Rate (1 point)

• 1 point for correct calculation using the rate law and given concentrations $$rate = (2.0 \times 10^{-3})(0.30)(0.40) = 2.4 \times 10^{-4} M/s$$

(d) Mechanism (2 points)

• 1 point for a two-step mechanism with an intermediate
• 1 point for correct stoichiometry and the overall balanced equation
  • Step 1: H₂ + I₂ ⇌ 2HI (slow)
  • Step 2: 2HI → H₂ + I₂ (fast)

Good luck! You've got this! 💪