Concentration Changes Over Time

Ethan Taylor
7 min read
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Study Guide Overview
This study guide covers chemical kinetics, focusing on rate laws and integrated rate laws. It explains reaction orders (0, 1, 2), rate law expressions, and the derivation and application of integrated rate laws for different orders. Graphical analysis techniques for determining reaction order and half-life calculations for first-order reactions are also discussed. Practice questions and exam tips are provided.
Chemical Kinetics: Rate Laws and Integrated Rate Laws
Introduction to Reaction Orders and Rate Laws
This section is crucial for understanding how reaction rates change with reactant concentrations. Expect to see these concepts in both multiple-choice and free-response questions.
Reaction Order (n)
- The reaction order, denoted by n, describes how the concentration of a reactant affects the reaction rate.
- While n is often an integer (0, 1, or 2), it can be a fraction in complex reactions.
- For the AP exam, focus on integer orders (0, 1, and 2).
Rate Law
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A rate law expresses the relationship between the rate of a reaction and the concentrations of the reactants.
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General form: Rate = k[A]n
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k is the rate constant (temperature-dependent)
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[A] is the concentration of reactant A
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n is the order of the reaction with respect to reactant A
Rate laws are determined experimentally and cannot be predicted from the balanced chemical equation.
Analogies to Understand Rate Laws
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Think of population dynamics:
- In a crisis, population decay is exponential.
- The rate of decay is proportional to the current population.
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Similarly, in chemical reactions: - As reactants are consumed, the reaction rate typically decreases. - The rate depends on the concentration of reactants.
Deriving Integrated Rate Laws
The Math Behind Rate Laws
Don't get bogged down in the calculus if it's not your strength. Focus on applying the integrated rate law equations rather than deriving them.
- For a simple reaction A → B, the rate of reaction can be expressed as:
- Rate = k[A]n
- Using calculus, we can derive the integrated rate laws, which relate reactant concentration to time.
General Derivation (Calculus-Based)
- Differential rate law:
- Separating variables and integrating (if you're comfortable with calculus):
- Solving this integral gives us the integrated rate laws.
Integrated Rate Laws for Different Orders
Remember the acronym ZIF to recall the order of reactions when matching graphs to rate laws. ZIF stands for Zero, First, and Second order reactions.
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Zeroth-Order (n = 0):
- Rate = k
- Integrated rate law: [A] - [A]₀ = -kt
- Graph of [A] vs. time is linear with a slope of -k.
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First-Order (n = 1):
- Rate = k[A]
- Integrated rate law: ln[A] - ln[A]₀ = -kt
- Graph of ln[A] vs. time is linear with a slope of -k.
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Second-Order (n = 2):
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Rate = k[A]2
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Integrated rate law: 1/[A] - 1/[A]₀ = kt
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Graph of 1/[A] vs. time is linear with a slope of k.
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The integrated rate laws and their corresponding linear plots are essential for determining reaction order from experimental data.
Graphical Analysis of Integrated Rate Laws
Linear Relationships
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The integrated rate laws can be rearranged to fit the form of a linear equation: y = mx + b.
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By plotting the appropriate function of concentration versus time, you can determine the reaction order from the linear relationship.
Summary of Linear Plots
- Zeroth-Order: [A] vs. time (slope = -k)
- First-Order: ln[A] vs. time (slope = -k)
- Second-Order: 1/[A] vs. time (slope = k)
Example of Determining Reaction Order from Graphs
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If the plot of 1/[A] vs. time is linear, the reaction is second order.
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The slope of the line gives you the rate constant, k.
Half-Life in First-Order Reactions
Definition of Half-Life
- Half-life (t1/2) is the time it takes for the concentration of a reactant to decrease to half of its initial value.
Half-Life Equation for First-Order Reactions
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t1/2 = 0.693 / k
This equation is provided on the AP Chemistry formula sheet. Remember that half-life is constant for first-order reactions.
Importance of Half-Life
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Half-life is particularly important in first-order processes like radioactive decay.
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It helps determine the rate of decay and the amount of substance remaining after a given time.
Don't confuse half-life with the time it takes for a reaction to reach completion. Half-life is specifically about halving the concentration.
Practice Question
Practice Questions
Multiple Choice Questions
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The decomposition of a certain substance is a first-order reaction. After 80 minutes, 75% of the substance has decomposed. What is the half-life of this reaction? (A) 20 min (B) 40 min (C) 60 min (D) 80 min
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For the reaction 2A + B → C, the following data were collected:
Experiment | [A] (M) | [B] (M) | Initial Rate (M/s) |
---|---|---|---|
1 | 0.1 | 0.1 | 2.0 x 10-3 |
2 | 0.2 | 0.1 | 8.0 x 10-3 |
3 | 0.1 | 0.2 | 2.0 x 10-3 |
What is the rate law for this reaction? (A) Rate = k[A][B] (B) Rate = k[A]2[B] (C) Rate = k[A]2 (D) Rate = k[B]
Free Response Question
The decomposition of dinitrogen pentoxide (N2O5) in the gas phase is represented by the following equation:
2 N2O5(g) → 4 NO2(g) + O2(g)
The rate of the reaction was studied at 330 K, and the following data were obtained:
Time (s) | [N2O5] (M) |
---|---|
0 | 0.100 |
100 | 0.0703 |
200 | 0.0497 |
300 | 0.0351 |
a) Determine the experimental rate law for the decomposition of N2O5. Justify your answer. b) Calculate the rate constant for the reaction, including units. c) Calculate the half-life of the reaction. d) What is the concentration of N2O5 after 600 seconds?
Answer Key
Multiple Choice
- (B)
- (C)
Free Response
a) The reaction is first order in N2O5. This can be determined by plotting ln[N2O5] vs. time, which results in a linear graph. (1 point for correct order, 1 point for justification)
b) The rate constant (k) is the negative of the slope of the linear plot. Using two points from the ln[N2O5] vs. time graph, we get:
k = - (ln(0.0351) - ln(0.100)) / (300 - 0) = 0.0105 s-1 (1 point for correct setup, 1 point for correct answer with units)
c) The half-life (t1/2) is calculated using the formula t1/2 = 0.693/k:
t1/2 = 0.693 / 0.0105 s-1 = 66 s (1 point for correct setup, 1 point for correct answer with units)
d) Using the integrated rate law for a first-order reaction: ln[A] - ln[A]₀ = -kt
ln[A] = ln(0.100) - (0.0105 s-1)(600 s) = -7.28
[A] = e-7.28 = 0.00069 M (1 point for correct setup, 1 point for correct answer with units)
Final Exam Focus
Prioritize understanding integrated rate laws, determining reaction orders, and calculating half-lives. These concepts frequently appear in both MCQs and FRQs.
Key Topics to Review
- Rate Laws: How to write and interpret them.
- Integrated Rate Laws: Equations for zero, first, and second-order reactions.
- Graphical Analysis: Identifying reaction order from linear plots.
- Half-Life: Calculating half-life for first-order reactions.
Common Question Types
- Determining reaction order from experimental data.
- Calculating rate constants from graphs or data tables.
- Using integrated rate laws to predict concentrations at different times.
- Applying half-life concepts to first-order reactions.
Last-Minute Tips
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Time Management: Quickly identify the type of question and apply the appropriate formula.
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Common Pitfalls: Pay attention to units and ensure correct graph interpretation.
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Strategies: If stuck on a problem, move on and come back to it later.
Good luck, you've got this! 💪

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Question 1 of 11
In the rate law, Rate = , what does 'n' represent? 🤔
The rate constant
The concentration of reactant A
The order of the reaction with respect to reactant A
The overall rate of the reaction