Problem 64 - Calculus Single Variable

Question

The chemical reaction $$ \mathrm{CH}_{3} \mathrm{COCH}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}+\mathrm{HI} $$ between acetone and iodine is governed by the differential equation $$ \frac{d\left[\mathrm{I}_{2}\right]}{d t}=-0.41 \times 10^{-3} \cdot\left[\mathrm{I}_{2}\right]^{2} $$ (in moles per liter per second). If the initial iodine concentration is \(c_{0}\), solve for \(\left[\mathrm{I}_{2}\right]\) as a function of time. Suppose that \(\left[\mathrm{I}_{2}\right]\) is plotted as a function of time. What is the slope of the graph when \(\left[\mathrm{I}_{2}\right]=1 / 10\) mole \(/\) liter? Suppose that \(1 /\left[\mathrm{I}_{2}\right]\) is plotted as a function of time. What is the slope of the graph when \(\left[\mathrm{I}_{2}\right]=1 / 10\) mole/liter?

Step-by-Step Solution

Answer

For [I2] vs. t, the slope is -0.41 × 10^{-5} when [I2]=0.1 M. For 1/[I2] vs. t, the slope is 0.41 × 10^{-3}.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{d[I_2]}{dt} = -0.41 \times 10^{-3} \cdot [I_2]^2 \). This is a first-order separable differential equation for the concentration \([I_2]\) with respect to time \(t\). Our goal is to find \([I_2]\) as a function of time. We start by separating variables: \( \frac{d[I_2]}{[I_2]^2} = -0.41 \times 10^{-3} dt \).

Step 2: Integrate Both Sides

Integrate both sides of the equation from Step 1. The left side integrates to \(-\frac{1}{[I_2]} + C\) and the right side integrates to \(-0.41 \times 10^{-3}t + C_1\). Equating these, we get \(-\frac{1}{[I_2]} = -0.41 \times 10^{-3}t + C\).

Step 3: Solve for Integration Constant

Using the initial condition \([I_2](0) = c_0\), substitute \(t = 0\) and \([I_2] = c_0\) into the equation from Step 2 to find \(-\frac{1}{c_0} = C\). Thus, \(C = -\frac{1}{c_0}\).

Step 4: Express [I_2] as a Function of Time

Substitute the value of \(C\) back into the equation: \(-\frac{1}{[I_2]} = -0.41 \times 10^{-3}t - \frac{1}{c_0}\). Rearrange to get \([I_2]\) as a function of time: \([I_2] = \frac{1}{0.41 \times 10^{-3}t + \frac{1}{c_0}}\).

Step 5: Find Slope of [I_2] vs. t Graph

The slope \(\frac{d[I_2]}{dt}\) of the \([I_2]\) vs. \(t\) graph is exactly the negative of the right side of the original differential equation: \(\frac{d[I_2]}{dt} = -0.41 \times 10^{-3} \cdot [I_2]^2\). When \([I_2] = \frac{1}{10} = 0.1\) mole/liter, the slope \(= -0.41 \times 10^{-3} \cdot (0.1)^2 = -0.41 \times 10^{-5}\).

Step 6: Find Slope of 1/[I_2] vs. t Graph

Starting from \([I_2] = \frac{1}{0.41 \times 10^{-3}t + \frac{1}{c_0}}\), we find \(\frac{1}{[I_2]} = 0.41 \times 10^{-3}t + \frac{1}{c_0}\). The slope \(\frac{d}{dt}\left(\frac{1}{[I_2]}\right) = 0.41 \times 10^{-3}\) is constant with respect to time, regardless of \([I_2]\) value.

Key Concepts

Separable Differential Equations

Separable differential equations play a pivotal role in understanding complex systems, such as chemical reactions. These types of differential equations can be rearranged such that all terms involving one variable, say \(y\), are on one side, while terms involving another variable, say \(x\), are on the other side. This makes them 'separable'.
For the differential equation given in the exercise, \(\frac{d[I_2]}{dt} = -0.41 \times 10^{-3} \cdot [I_2]^2\), the goal is to isolate the \( [I_2] \) terms from the \(t\) terms. This leads to reformatting it as: \[\frac{d[I_2]}{[I_2]^2} = -0.41 \times 10^{-3} \, dt.\]
By separating the variables, the functional dependence between \(t\) and \([I_2]\) can be clearly analyzed, which is crucial for solving these types of problems. After separation, the next step involves integrating both sides, allowing the solution in terms of function of \(t\). Separable differential equations provide a straightforward method to approach and solve problems in chemical kinetics among other applications.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that studies the rates of chemical reactions. It helps in understanding how reaction rates change with varying conditions and can aid in predicting the speed of a chemical process.
For the reaction between acetone and iodine, represented by \(\mathrm{CH}_3\mathrm{COCH}_3 + \mathrm{I}_2 \rightarrow \mathrm{CH}_3\mathrm{COCH}_2\mathrm{I} + \mathrm{HI}\), kinetics is key. By examining the differential equation \(\frac{d[I_2]}{dt} = -0.41 \times 10^{-3} \cdot [I_2]^2\), we can see the reaction follows a second-order rate law concerning iodine. This means the rate of reaction depends on the square of iodine concentration.

• The negative sign indicates \(\mathrm{I}_2\) concentration decreases over time.
• The proportionality constant \(-0.41 \times 10^{-3}\) gives insight into how fast the reaction proceeds.

Understanding these aspects of chemical kinetics is essential in controlling and optimizing chemical processes, especially in industrial settings.

Reaction Rate

The reaction rate is a measure of how quickly a chemical reaction proceeds, often defined as the change in concentration of a reactant or product per unit time.
In the context of the given differential equation, \(\frac{d[I_2]}{dt}\), the reaction rate tells us the rate at which iodine is being consumed. For the specific concentration value \( [I_2] = 0.1 \), the rate of change is calculated as \(-0.41 \times 10^{-5} \).
This negative value indicates a decrease in iodine over time. Moreover, when examining the graph of \([I_2]\) versus time, the slope at any point \([I_2] = 0.1\) corresponds to this rate.
The reaction rate also remains constant when considering the graph of \(\frac{1}{[I_2]}\) versus time, with the slope \(\frac{d}{dt}\left(\frac{1}{[I_2]}\right) = 0.41 \times 10^{-3}\), a key insight for kinetics analysis.