AP Chemistry 5.3 Concentration Changes Over Time Study Notes - New Syllabus Effective fall 2024

AP Chemistry 5.3 Concentration Changes Over Time Study Notes- New syllabus

AP Chemistry 5.3 Concentration Changes Over Time Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.

LEARNING OBJECTIVE

Identify the rate law expression of a chemical reaction using data that show how the concentrations of reaction species change over time.

Key Concepts:

Concentration-Time Graphs
Manipulating Concentration-Time Graphs
Concentration-Time Graphs & Rate Constants
Concentration-Time Graphs & Half-Life
Catalysts & Reaction Rate
Catalysts in Action

AP Chemistry-Concise Summary Notes- All Topics

5.3.A.1 Inferring Reaction Order from Concentration vs. Time Graph:

1. Reaction Order Overview:

Chemical reaction order is the dependence of the reaction rate on reactant concentration. It is used to determine how changing the concentration of a reactant influences the rate of a chemical reaction. Various reaction orders—zero, first, and second—each explain how the reaction rate varies with reactant concentration.

i. Zero-Order Reactions:

– Rate Law: Rate = k
– Concentration Dependence: The rate is independent of reactant concentration.
– Meaning: Independent of the change of concentration of the reactant, the rate of reaction is constant.
– Integrated Rate Law

$[A] = [A_0] – kt$

– Graph: Plotting [A] versus time will give you a slope of (-k) (straight line).

ii. First-Order Reactions:

– Rate Law: Rate = k[A]
– Concentration Dependence: The rate is proportional to the concentration of a reactant. Doubling the concentration of the reactant will double the rate.
– Meaning: The reaction rate is linearly dependent upon the concentration of the reactant.
– Integrated Rate Law: ln[A] = ln[A0] – kt
– Graph: If you graph ln[A] as a function of time, you will get a straight line with slope -k .

iii. Second-Order Reactions:

– Rate Law:Rate equation:

$\text{rate} = k[A]^2$

– Concentration Dependence: The rate is a function of the square of the concentration of the reactant. Increasing the concentration of the reactant to twice its value will increase the rate fourfold.
– Meaning: The reaction rate is a linear function of the concentration of the reactant in quadratic form.
– Integrated Rate Law: $\frac{1}{[A]} = \frac{1}{[A_0]} + kt$

– Graph: If you plot $\frac{1}{[A]}$ vs. time, you will get a straight line with slope $k$

iv. Key Differences:
– Zero-Order: Rate is independent of concentration.
– First-Order: Rate is proportional to the concentration.
– Second-Order: Rate is proportional to the square of the concentration.

2. Graph Characteristics:

You are absolutely right! The pattern of concentration with time for zero, first, and second-order reactions can be observed from their respective graphs. Here is a more detailed description of the graph characteristics for each reaction order:

i. Zero-Order Reaction:
– Concentration vs. Time Graph: The reactant rate drops linearly with time.
– Equation:[A] = [A0] – kt
– Shape: A straight line with a negative slope, which represents a constant rate of reduction in concentration.
– Interpretation: The rate of reaction is independent of concentration, so the rate of reduction is constant.

ii. Graph Example:
– The x-axis is employed for time (t), and the y-axis for concentration [A].
– The slope of the line is represented by ( -k ), and the concentration decreases linearly.

iii. First-Order Reaction:
– Concentration vs. Time Graph: The concentration of the reactant drops off exponentially with time.
– Equation: ln[A] = ln[A0] – kt
– Shape: An exponential decline curve, approaching zero as time increases.
– Interpretation: Rate is proportional to concentration, and it drops off very rapidly initially, then more slowly as concentration drops.

iv. Graph Example:
– If you plot [A] vs. time, the curve will be exponential.
– If you plot ln[A] vs. time, you get a straight line with slope ( -k ).

v. Second-Order Reaction:
– Concentration vs. Time Graph: The concentration decreases in a hyperbolic fashion, but faster than for first-order reactions.
– Equation: $\frac{1}{[A]} = \frac{1}{[A_0]} + kt$

– Shape: Hyperbolic curve, steep at first and then leveling off with time. The fall is steeper initially but slows down gradually.
– Interpretation: The rate of reaction is directly proportional to the square of the concentration, so the reaction slows down even slower as the concentration of the reactant falls.

vi. Graph Example: If you plot $\frac{1}{[A]}$ vs. time, the curve will be a straight line, and the slope will be $k$ .

– The graph will be hyperbolic if you plot [A] vs. time.

vi. Summary of Graph Shapes:

– Zero-Order: Concentration drops linearly with time (hyperbolic graph when concentration is plotted vs. time).
– First-Order: Decrease in concentration exponentially (curved when concentration vs. time is plotted; linear when ln[A] vs. time is plotted).
– Second-Order: Hyperbolic decrease in concentration (s-shaped when concentration vs. time is plotted; linear when $\frac{1}{[A]}$ vs. time is plotted).

3. Half-Life:

Reaction Order	Half-Life Expression	Impact of Initial Concentration
Zero-Order	$t_{1 / 2} = \frac{[A_{0}]}{2 k}$	Half-life decreases as initial concentration decreases.
First-Order	$t_{1 / 2} = \frac{0.693}{k}$	Half-life is constant, independent of initial concentration.
Second-Order	$t_{1 / 2} = \frac{1}{k [A_{0}]}$	Half-life increases as initial concentration decreases.

Zero-order: The half-life is directly proportional to the initial concentration.
First-order: The half-life is constant, independent of the initial concentration.
Second-order: The half-life is inversely proportional to the initial concentration.

5.3.A.2 First-Order Reaction: ln[Concentration] vs. Time:

1. First-Order Reactions:

– Rate Law: Rate = k[A] (proportional to reactant concentration).
– Integrated Rate Law: ln[A] = ln[A0] – kt or

$[A] = [A_0] e^{-kt}$

where:
– [A] is the concentration at time ( t ),
– [A0] is the initial concentration,
– ( k ) is rate constant,
– ( t ) is time.

– Half-Life:

$t_{1/2} = \frac{0.693}{k}$

(not depending on initial concentration, a constant).
– Graph: A plot of ln[A] vs. time is a straight line with slope ( -k ).

2. Graphical Interpretation:

For a first-order reaction, the integrated rate law can be expressed as: ln[A] = ln[A0] – kt
This equation illustrates that the natural logarithm of the reactant’s concentration ln[A] has a linear relationship with time ( t ).

i. Why ln[A] vs. Time is Linear:
– The equation ln[A] = ln[A0] – kt resembles a straight-line equation in the form y = mx + b , where:
– ( y ) represents ln[A],
– ( x ) denotes time ( t ),
– ( m ) (the slope) is ( -k ) (the rate constant),
– ( b ) is the y-intercept, which is ln[A0] (the natural logarithm of the initial concentration).

– This linear relationship indicates that if the reaction adheres to first-order kinetics, plotting ln[A] (on the y-axis) against time (on the x-axis) will produce a straight line.

ii. Significance of the Slope:
– The slope of the graph depicting ln[A] versus time corresponds to ( -k ), the rate constant of the reaction.
– A negative slope signifies that the concentration of the reactant diminishes over time, which aligns with the behavior of a reaction where the reactant is being consumed.
– The rate constant ( k ) provides information about the reaction’s speed: a higher ( k ) indicates a quicker reaction (meaning a shorter duration for the concentration to decrease).

3. Half-Life:
For first-order reactions, the half-life remains constant, which means it does not vary with the initial concentration of the reactant. This characteristic is distinctive to first-order reactions.

i. Half-Life Formula for First-Order Reactions:
The half-life for a first-order reaction can be calculated using the formula:

$t_{1/2} = \frac{0.693}{k}$

ii. Significance of Constant Half-Life:
– In a first-order reaction, the duration needed to reduce the concentration of the reactant by half remains consistent, regardless of the initial concentration A0] .
– This indicates that each subsequent half-life (the time taken for the concentration to halve) will be identical. For instance, if the initial concentration is halved in the first half-life, it will take the same duration to halve again in the next half-life, and this pattern continues.

iii. Why Does the Half-Life Stay Constant?:
The reason the half-life is constant for first-order reactions is that the reaction rate is directly proportional to the concentration of the reactant. As the concentration diminishes, the reaction naturally slows down, but due to the linear relationship in terms of ln[A] , the time required for half of the reactant to be consumed remains unchanged.

5.3.A.3 Second-Order Reaction: 1/[Concentration] vs. Time:

1. Second-Order Reaction:

Second-order reaction: The rate of the reaction is proportional to the square of the concentration of the reactant , A $\text{Rate} = k[A]^2$
Integrated rate law: This form allows you to calculate the concentration of the reactant at any time , given the initial concentration and the rate constant k.

$[A] = \frac{1}{\frac{1}{[A_0]} + kt}$

2. Graph:

The reason why the graph of $\frac{1}{[\text{Concentration}]}$ versus time ( $t$ ) is linear for a second-order reaction is due to the mathematical form of the integrated rate law for second-order reactions.

The integrated rate law for a second-order reaction is:

$\frac{1}{[A]} = \frac{1}{[A_0]} + kt$

i. Why the graph is linear:

This equation is in the form of a straight line equation,

$y = mx + b$

, where:

The y-axis is $\frac{1}{[A]}$
The x-axis is t (time),
The slope of the line is k (the rate constant),
The y-intercept is $\frac{1}{[A_0]}$

So, when you plot $\frac{1}{[A]}$ against time $t$ , the relationship between them is linear, and the slope of the line is directly related to the rate constant $k$ . The greater the rate constant $k$ , the steeper the slope of the line.

This linearity is a unique characteristic of second-order reactions and makes it easy to determine the rate constant experimentally. If the plot of $\frac{1}{[A]}$ vs. $t$ is not linear, then the reaction is not second-order.

5.3.A.4 Determining Rate Constant from Slopes for Different Reaction Orders:

1. Reaction Orders & Rate Laws:

Reaction Order	Rate Law	Integrated Rate Law	Graph	Plot	Slope
Zero-Order	$\text{Rate} = k$	$[A] = [A_0] – kt$	$A vs. t$	Linear	$-k$
First-Order	$\text{Rate} = k[A]$	$\ln[A] = \ln[A_0] – kt$	$\ln [A] vs.$ $t$	Linear	$-k$
Second-Order	$\text{Rate} = k[A]^2$	$\frac{1}{[A]} = \frac{1}{[A_0]} + kt$	$\frac{1}{[A]}$	Linear	$k$

2. Slope and Rate Constant:

Zero-Order Reaction

Concept	Details
Integrated Rate Law	$[A] = [A_0] – kt$
Graph to Plot	Plot [ A ] (concentration) vs. time t
Slope	The slope of the graph is− $-k$
Rate Constant k	$k = -\text{slope}$
Units of k	$M/s (molarity per second)$
Example	If the slope is $- 0.03 M/s, then k = 0.03 M/s$ $k = 0.03 \, \text{M/s}$

First-Order Reaction

Concept	Details
Integrated Rate Law	$\ln[A] = \ln[A_0] – kt$
Graph to Plot	Plot ln $\ln[A]$ $t$
Slope	The slope of the graph is $- k$
Rate Constant k	$k = - slope (negative of the slope)$
Units of k	$1/s (inverse seconds)$
Example	If the slope is $-0.15 \, \text{1/s}$ , $then$ $k = 0.15 1/s$ $k = 0.15 \, \text{1/s}$

Second-Order Reaction

Concept	Details
Integrated Rate Law	$\frac{1}{[A]} = \frac{1}{[A_0]} + kt$
Graph to Plot	Plot $\frac{1}{[A]}$ $t$
Slope	The slope of the graph is k
Rate Constant k	$k = slope(slope is directly equal to k)$
Units of k	$\text{M}^{-1}\text{s}^{-1}$ (inverse molarity per second)
Example	If the slope is $0.25 \, \text{M}^{-1}\text{s}^{-1}$ , then $k = 0.25 \, \text{M}^{-1}\text{s}^{-1}$

5.3.A.5 Half-Life for First-Order Reactions: t₁/₂ = 0.693/k:

1. First-Order Reactions:

For first-order reactions, a defining characteristic is that the half-life remains constant and does not depend on the concentration of the reactant. This differs from zero-order and second-order reactions, where the half-life is influenced by the concentration of the reactant.

i. First-Order Reaction: Half-Life:

The half-life of a reaction refers to the time it takes for the concentration of the reactant to reduce to half its initial value. For first-order reactions, the half-life can be calculated using the following formula:

$t_{1 / 2} = \frac{\ln 2}{k}$

Where:
– is the half-life

– ( k ) is the rate constant (expressed in units of text1/s),
– ln 2 is the natural logarithm of 2 (approximately 0.693).

ii. Key Points about Half-Life in First-Order Reactions:

– The half-life t 1/2 is independent of the initial concentration [A0] of the reactant.
– The half-life remains constant throughout the reaction. Regardless of the initial amount of reactant, the duration for half of it to react stays the same.

iii. Why is this the case?
In a first-order reaction, the rate is directly proportional to the concentration of the reactant. As a result, the time required for half of the reactant to be consumed remains unchanged as the reaction progresses, since the rate constant ( k ) is the main factor influencing the reaction rate, rather than the concentration.

Aspect	First-Order Reactions
Half-Life Equation	$t_{1 / 2} = \frac{\ln 2}{k}$
Dependence on Concentration	Half-life is independent of the initial concentration[A0]
Units of k	$1/s (inverse seconds)$
Constant Half-Life	Half-life is constant throughout the reaction

2. Half-Life and Rate Constant:

The half-life
$t_{1 / 2}$

$is inversely proportional to the rate constant$
As the rate constant k increases (meaning the reaction happens faster), the half-life t1/2 decreases, i.e., the time it takes for half of the reactant to be consumed gets shorter.
Conversely, as k decreases, the reaction takes longer to reach half completion, so the half-life increases.

UNIT OF K:

has units of s−1 (inverse seconds), and thus the half-life t 1/2 will have units of time (seconds).

5.3.A.6 Radioactive Decay as an Example of First-Order Kinetics:

1. First-Order Rate Law:

In a first-order reaction, the reaction rate is influenced by the concentration of just one reactant. The rate law for this type of reaction is expressed as:

Rate = k[A]

Where:
– Rate indicates how fast the reaction occurs.
– ( k ) represents the rate constant.
– [A] denotes the concentration of the reactant.

A key aspect of first-order reactions is that the half-life—the duration required for half of the reactant to be used up—remains constant. This constancy means that it does not vary with the initial concentration of the reactant. The formula for calculating the half-life in a first-order reaction is:

$t_{1 / 2} = \frac{0.693}{k}$

In this equation, ( t1/2 ) stands for the half-life, while ( k ) is the rate constant.

Unlike zero or second-order reactions, where the half-life can change based on the initial concentration, the half-life in a first-order reaction stays the same regardless of how much reactant is present at the start. This characteristic is a defining trait of first-order kinetics.

2. Half-Life and Decay Constant:
For a first-order reaction, the half-life ( t1/2) is determined by the rate constant ( k ) using the formula:

$t_{1 / 2} = \frac{0.693}{k}$

Where:
– ( t1/2) represents the half-life, which is the duration required for half of the reactant to be used up.
– ( k ) denotes the decay constant (or rate constant) for the reaction.

This indicates that in a first-order reaction, the half-life remains constant and is not affected by the initial concentration. Instead, it solely relies on the value of the rate constant ( k ).

3. Applications:

i. Carbon-14 Dating:
– Used in the dating of old organic matter from the level of remaining Carbon-14, which is determined to decay with a specific rate.

ii. Medical Imaging:
– Radioisotopes with specified half-lives are used in diagnostics like PET scans for the imaging and identification of diseases.

iii. Nuclear Energy:
– Decay of radioactive materials such as uranium in nuclear reactors uses half-life principles to regulate the consumption of fuel and waste.

AP Chemistry 5.3 Concentration Changes Over Time Study Notes

AP Chemistry 5.3 Concentration Changes Over Time Study Notes - New Syllabus Effective fall 2024