Integrated Rate Law: Videos & Practice Problems

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.

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 and how does it relate reactant concentration to time?

The integrated rate law describes the relationship between the concentration of a reactant and time during a chemical reaction. It helps determine how long it takes for a certain amount of reactant to be consumed. The form of the integrated rate law depends on the order of the reaction (zero, first, or second order). For example, in a zero order reaction, the concentration decreases linearly with time, while in a first order reaction, the natural logarithm of concentration decreases linearly with time. In a second order reaction, the inverse of the concentration increases linearly with time. Understanding these relationships allows us to predict reactant concentrations at any time and identify the reaction order from experimental data.

How do you determine the order of a reaction using integrated rate laws?

To determine the order of a reaction using integrated rate laws, you analyze how the concentration of the reactant changes over time by plotting data in different ways. For a zero order reaction, a plot of concentration ($\mathrm{[A]}$) versus time ($t$) is linear. For a first order reaction, a plot of the natural logarithm of concentration ($\mathrm{ln[A]}$) versus time is linear. For a second order reaction, a plot of the inverse of concentration ($\frac{1}{\mathrm{[A]}}$) versus time is linear. The plot that yields a straight line indicates the reaction order. This method is essential for identifying the kinetics of a reaction and calculating the rate constant.

What are the integrated rate law equations for zero, first, and second order reactions?

The integrated rate law equations differ based on the reaction order:

Zero order: ${\mathrm{[A]}}_t={\mathrm{[A]}}_0-kt$, where ${\mathrm{[A]}}_t$ is the concentration at time $t$, ${\mathrm{[A]}}_0$ is the initial concentration, and $k$ is the rate constant.

First order: $\ln {\mathrm{[A]}}_t=\ln {\mathrm{[A]}}_0-kt$.

Second order: $\frac{1}{{\mathrm{[A]}}_t}=\frac{1}{{\mathrm{[A]}}_0}+kt$.

Each equation corresponds to a linear plot that helps identify the reaction order and calculate the rate constant.

What are the units of the rate constant (k) for zero, first, and second order reactions?

The units of the rate constant $k$ depend on the reaction order:

For zero order reactions, $k$ has units of concentration per time, typically $mol\; \backslash ,\; L^\{-1\}\; \backslash ,\; s^\{-1\}$.

For first order reactions, $k$ has units of inverse time, such as $s^\{-1\}$.

For second order reactions, $k$ has units of inverse concentration per time, for example $L\; \backslash ,\; mol^\{-1\}\; \backslash ,\; s^\{-1\}$.

These units arise from the integrated rate law equations and help confirm the reaction order experimentally.

How can you identify a first order reaction from experimental data?

A first order reaction can be identified by plotting the natural logarithm of the reactant concentration ($ln[A]$) versus time ($t$). If the plot is a straight line with a negative slope, the reaction follows first order kinetics. The slope of this line equals the negative rate constant ($-k$). Additionally, many radioactive decay processes are first order, so the presence of radioactive isotopes in a problem is a strong hint. This method is widely used because it simplifies the analysis and allows easy calculation of the rate constant.

What is the significance of the slope in integrated rate law plots?

The slope in integrated rate law plots represents the rate constant $k$ and provides insight into the reaction kinetics. For zero and first order reactions, the slope is negative and equals $-k$, indicating that the reactant concentration decreases over time. In zero order, the plot of concentration versus time has slope $-k$. In first order, the plot of $ln[A]$ versus time has slope $-k$. For second order reactions, the slope is positive and equals $k$ in the plot of the inverse concentration versus time, showing that $1/[A]$ increases over time. Understanding the slope helps determine the rate constant and confirms the reaction order.