Integrated Rate Law: Videos & Practice Problems

The integrated rate law is crucial for understanding the relationship between reactant concentrations and time, revealing how long it takes for a certain amount of reactant to be consumed. It varies according to the reaction order: zero, first, or second. For zero order reactions, the rate constant (K) has units of molarity per time, and the concentration decreases linearly over time. First order reactions, often seen in radioactive processes, have a rate constant with units of time inverse, and a plot of the natural logarithm of concentration versus time is linear. Second order reactions feature a rate constant with units of molarity inverse per time, and a plot of the inverse of concentration versus time shows an increasing linear relationship. Recognizing these patterns is key to identifying reaction orders and understanding the kinetics of chemical processes.

Zero-Order Reactions

The integrated rate law is a crucial concept in chemical kinetics, illustrating the relationship between the concentration of reactants and time. It allows chemists to determine how long it takes for a specific amount of a reactant to be consumed. The form of the integrated rate law varies depending on the order of the reaction, with the zero-order reaction being the first type to consider.

For zero-order reactions, the integrated rate law is expressed by the equation:

\( A_t = -kt + A_0 \)

In this equation, \( A_t \) represents the final concentration of the reactant, \( A_0 \) is the initial concentration, and \( k \) is the rate constant. The units of \( k \) can be derived from the equation, where the order of the reaction \( n \) is zero. Thus, the units for \( k \) are:

\( k = \text{M}^{1} \cdot \text{time}^{-1} \)

Here, "M" stands for molarity, and "time" can be measured in seconds, days, or years, depending on the context.

This integrated rate law can also be interpreted graphically. The equation resembles the equation of a straight line, where \( A_t \) corresponds to the y-axis, \( t \) to the x-axis, and the slope of the line is represented by \(-k\). The initial concentration \( A_0 \) serves as the y-intercept. As time progresses, the concentration of the reactant decreases, resulting in a negative slope, which indicates that the concentration diminishes over time.

Understanding the relationship between the change in concentration and time is essential, as the slope of the line can be expressed as:

\( \text{slope} = \frac{\Delta y}{\Delta x} = \frac{\Delta A}{\Delta t} \)

This relationship reinforces the concept that in a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant, making it a unique case in the study of reaction kinetics.

Zero-Order Reaction Example

In a zero-order reaction, the concentration of the reactant decreases linearly over time. The relationship can be expressed with the equation:

\[ [A]_t = -kt + [A]_0 \]

Where:

• \([A]_t\) is the concentration at time \(t\),
• \(k\) is the rate constant,
• \([A]_0\) is the initial concentration,
• \(t\) is the time.

In this scenario, the slope of the concentration versus time plot is given as 0.260, which indicates that:

\[ k = -\text{slope} = -0.260 \, \text{M/s} \]

After 35 seconds, the concentration of nitrogen trioxide is measured to be:

\[ [A]_t = 2.75 \times 10^{-2} \, \text{M} \]

Substituting the known values into the zero-order equation, we have:

\[ 2.75 \times 10^{-2} = -(-0.260)(35) + [A]_0 \]

Calculating the term \(-(-0.260)(35)\) gives:

\[ 9.1 \, \text{M} \]

Now, rearranging the equation to solve for the initial concentration \([A]_0\):

\[ [A]_0 = 2.75 \times 10^{-2} + 9.1 \]

Thus, the initial concentration is:

\[ [A]_0 = 9.1275 \, \text{M} \]

Considering significant figures, the final answer for the initial concentration of nitrogen trioxide, rounded to two significant figures, is:

\[ [A]_0 \approx 9.1 \, \text{M} \]

First-Order Reactions

In first-order reactions, the relationship between the concentration of a reactant and time can be described using the equation:

\( \ln[A_t] = -kt + \ln[A_0] \)

In this equation, \( [A_t] \) represents the concentration of the reactant at time \( t \), while \( [A_0] \) is the initial concentration. The variable \( k \) is the rate constant, which has units of time^-1 (e.g., seconds^-1) for first-order reactions. This means that the rate constant is dimensionally consistent with the reaction order, where \( n \) is 1, leading to \( k \) being simply time^-1.

When analyzing first-order reactions, it is useful to recognize that the equation can be rearranged to resemble the equation of a straight line, \( y = mx + b \). Here, \( \ln[A_t] \) serves as the dependent variable (y), \( -k \) is the slope (m), \( t \) is the independent variable (x), and \( \ln[A_0] \) is the y-intercept (b). This linear relationship indicates that plotting \( \ln[A_t] \) against time \( t \) will yield a straight line, confirming the first-order nature of the reaction.

Additionally, it is important to note that all first-order processes, including radioactive decay, follow this first-order rate law. While not all first-order processes are radioactive, any mention of a radioactive isotope in a problem typically indicates that the process is first-order. This recognition can be a helpful strategy when determining the order of a reaction in various scenarios.

First-Order Reaction Example

In a first-order reaction, the rate constant (k) is expressed in units of time inverse, indicating that the reaction rate depends on the concentration of the reactant. Given a rate constant of 0.289 s^-1, we can determine the time required for the concentration of reactant A to decrease from 1.43 M to 0.850 M using the first-order rate equation:

ln([A]_t / [A]₀) = -kt

Here, [A]₀ is the initial concentration (1.43 M), and [A]_t is the final concentration (0.850 M). We start by calculating the natural logarithm of the initial and final concentrations:

ln(1.43) ≈ 0.3567

ln(0.850) ≈ -0.1625

Substituting these values into the equation gives:

-0.1625 - 0.3567 = -0.289t

Solving for t involves rearranging the equation:

-0.5192 = -0.289t

Dividing both sides by -0.289 yields:

t ≈ 1.80 seconds

This result is consistent with significant figures, as the values used in the calculation (0.289, 1.43, and 0.850) all have three significant figures. Therefore, the time required for the concentration of reactant A to decrease from 1.43 M to 0.850 M is approximately 1.80 seconds.

Second-Order Reactions

In second order reactions, the relationship between the concentration of reactants and time can be described using the integrated rate law equation:

$$\frac{1}{A_t} = k t + \frac{1}{A_0}$$

Here, A_t represents the final concentration of the reactant, A_0 is the initial concentration, and k is the rate constant. The units of the rate constant k for second order reactions are crucial to understand, as they are expressed in molarity^-1 time^-1 (M^-1 s^-1), reflecting the nature of the reaction order.

In this context, the variable t denotes time. The integrated rate law for second order reactions can be visualized as a linear equation, where:

$$y = mx + b$$

In this analogy, y corresponds to 1/A_t, m is the slope represented by k, x is time t, and b is 1/A_0. When plotting 1/A_t against time t, the resulting graph will yield a straight line with a positive slope, indicating that the reaction is second order. This is a distinctive feature of second order kinetics, as the slope increases over time due to the positive value of k.

It is important to note that this behavior contrasts with zero and first order reactions, which exhibit different characteristics in their respective plots. Recognizing these patterns can help in identifying the order of a reaction based on experimental data.

Second-Order Reaction Example

In a second-order reaction, the relationship between the concentration of reactants and time can be described by the equation:

\(\frac{1}{[A_t]} = k t + \frac{1}{[A_0]}\)

In this scenario, we are given the initial concentration \([A_0]\) as 0.7670 M after 300 seconds and the final concentration \([A_t]\) as \(7.3 \times 10^{-2}\) M after 750 seconds. To find the rate constant \(k\), we first need to determine the elapsed time:

Elapsed time = 750 seconds - 300 seconds = 450 seconds.

Now, substituting the known values into the equation:

\(\frac{1}{7.3 \times 10^{-2}} = k(450) + \frac{1}{0.7670}\)

Calculating the left side:

\(\frac{1}{7.3 \times 10^{-2}} \approx 13.6986\)

And the right side becomes:

\(13.6986 = k(450) + 1.3035\)

Next, we isolate \(k\) by subtracting \(1.3035\) from both sides:

\(12.3951 = k(450)\)

Now, divide both sides by 450 to solve for \(k\):

\(k = \frac{12.3951}{450} \approx 0.0272\) M^-1s^-1

Thus, the rate constant \(k\) for this second-order reaction is approximately \(2.72 \times 10^{-2}\) M^-1s^-1. The units for \(k\) in a second-order reaction are derived from the formula, where \(n\) is the order of the reaction. For second-order reactions, the units are M^-1s^-1, indicating that the rate constant depends on the concentration of the reactants and the time elapsed.

For the reaction A →  B, the rate constant is 0.0837 M^–1•sec^–1. How long would it take for [A] to decrease by 85%?

• A. 0.0211 s
• B. 1.94 s
• C. 179 s
• D. 0.677 s

Answer: (D)

The following reaction has a rate constant of 3.7 × 10^–3 M•s^–1 at 25°C:
A → B + C
Calculate the concentration of C after 2.7 × 10^–3 sec where [A]₀ was 0.750 M at 25°C; assume [C]₀ = 0 M.

• A. 0.75 M
• B. 9.99×10^–6 M
• C. 3.7×10^–3 M
• D. 7.5×10^–6 M

Answer: (B)

For the decomposition of urea, NH₂CONH₂ (aq) + H⁺(aq) + 2 H₂O (l) → 2 NH₄⁺ (aq) + HCO₃^– (aq), the rate constant is 3.24 × 10^–4 s^–1 at 35°C. The initial concentration of urea is 2.89 mol/L. What fraction of urea has decomposed after 3.5 minutes?

• A. 0.934
• B. 0.0658
• C. 1.13×10^–3
• D. 0.0235

Answer: (B)

Iodine-123 is used to study thyroid gland function. As this radioactive isotope breaks down, after 5.7 hrs the concentration of iodine-123 is 56.3% complete. Find the rate constant of this reaction.

• A. 0.359 hr^–1
• B. 0.15 hr^–1
• C. 1.36×10^–3 M^–1•hr^–1
• D. 2.26×10^–3 M^–1•hr^–1

Answer: (B)

Here’s what students ask on this topic:

What is the integrated rate law?

The integrated rate law is an equation that describes the concentration of a reactant in a chemical reaction as a function of time. It's derived from the rate law, which expresses the rate of a reaction in terms of the concentration of reactants. For a reaction of order n, the rate law is:

$\text{rate =}k{\left[A\right]}^n$

where k is the rate constant, [A] is the concentration of the reactant, and n is the order of the reaction.

There are different forms of integrated rate laws for different reaction orders. For example:

1. Zero-order reactions:

$\left[A\right]=\left[A\right]0_{}-\mathrm{kt}$

where [A]₀ is the initial concentration and [A] is the concentration at time t. The plot of [A] vs. time is linear for a zero-order reaction.

2. First-order reactions:

Which of the following statements correctly describe an integrated rate law? Can you select all that apply?

An integrated rate law expresses the concentration of a reactant in a chemical reaction as a function of time. Unlike the differential rate law, which shows the rate of reaction at a particular moment (instantaneous rate), the integrated rate law provides a means to calculate the amount of reactant remaining after a certain period. It is derived by integrating the differential rate law.

The integrated rate law can be used to determine the order of a reaction, which indicates the power to which the concentration of the reactant is raised in the rate law. For example, a first-order reaction will have a linear relationship between the natural logarithm of the reactant concentration and time, while a second-order reaction will exhibit a linear relationship between the inverse of the reactant concentration and time.

Furthermore, the integrated rate law allows for the calculation of the half-life of a reaction, which is the time required for half of the reactant to be consumed. In first-order reactions, the half-life is constant and does not depend on the initial concentration, while in second-order reactions, the half-life is dependent on the initial concentration.

Which of the following represents the integrated rate law for a first-order reaction?

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

$\ln \left[A\right]=-kt+\ln \left[A\right]0_{}$

In this equation, $\left[A\right]$ represents the concentration of the reactant at time $t$, $k$ is the first-order rate constant, and $\left[A\right]0_{}$ is the initial concentration of the reactant when $t$ = 0. The negative sign indicates that the concentration of the reactant decreases over time. This equation can also be rearranged to solve for $t$ or $k$ depending on the given data. When plotted on a graph with

Which of the following represents the integrated rate law for a second-order reaction?

The integrated rate law for a second-order reaction, where the concentration of the reactant decreases over time as the reaction proceeds, is represented by the following equation:

$\frac{1}{A_t}=kt+\frac{1}{A_0}$

In this equation, $A_t$ is the concentration of the reactant A at time t, $A_0$ is the initial concentration of the reactant, k is the second-order rate constant, and t is the time. The units of the rate constant k in a second-order reaction are typically M^-1s^-1 (inverse molarity per second).

This form of the integrated rate law shows that the inverse of the concentration of the reactant at any given time is directly proportional to the time elapsed. A plot of $\frac{1}{A_t}$ versus time (t) for a second-order reaction will yield a straight line, with the slope equal to the rate constant k.