Problem 63 - Calculus Single Variable

Question

When sucrose is dissolved in water, glucose and fructose are formed. The reaction rate satisfies $$ \frac{d\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]}{d t}=-5.7 \times 10^{-5} \cdot\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] $$ For what value of \(t\) is the concentration of sucrose. \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\), equal to \(1 / 3\) its initial value?

Step-by-Step Solution

Answer

The time \(t\) is \(19,251\) seconds.

Step 1: Understanding the Rate Law Equation

The given rate of reaction is expressed as \(\frac{d\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]}{dt} = -5.7 \times 10^{-5} \cdot \left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\). This indicates a first-order reaction, where the rate of change of sucrose concentration is proportional to its current concentration.

Step 2: Integrating the Rate Equation

For a first-order reaction \(\frac{d[A]}{dt} = -k[A]\), the integrated rate law is \([A] = [A]_0 e^{-kt}\). Here, \(k = 5.7 \times 10^{-5}\). This equation will allow us to relate the concentration of sucrose to time.

Step 3: Setting Up the Problem

We want the concentration of sucrose \([\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}](t)\) to be \(\frac{1}{3}\) of its initial concentration \([\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0\). Thus, set \(\frac{1}{3} [A]_0 = [A]_0 e^{-kt}\).

Step 4: Solving for Time \(t\)

Cancel \([A]_0\) from both sides to get \(\frac{1}{3} = e^{-kt}\). Take the natural logarithm of both sides to solve for \(t\): \(\ln\left(\frac{1}{3}\right) = -kt\).

Step 5: Isolating \(t\) and Calculating the Value

We find \(t\) by rearranging: \(t = \frac{\ln\left(\frac{1}{3}\right)}{-k}\). Substitute \(k = 5.7 \times 10^{-5}\): \(t = \frac{\ln\left(\frac{1}{3}\right)}{-5.7 \times 10^{-5}}\).

Key Concepts

Understanding Reaction Rate

In chemical reactions, the reaction rate measures how fast a reactant is consumed or a product is formed. For a first-order reaction like sucrose decomposition into glucose and fructose, the rate of reaction is directly proportional to the concentration of the reactant. This means if the concentration of sucrose decreases, the reaction rate decreases as well. The rate of reaction can be mathematically expressed using a differential equation, such as \( \frac{d[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]}{dt} = -k[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}] \), where

• \(k\) is the rate constant specific to the reaction, indicating how fast the reaction proceeds.
• The negative sign shows that sucrose is being consumed.

This proportional relationship helps chemists predict how the reaction will proceed over time by knowing the current concentration.

Integrated Rate Law for First-Order Reactions

When dealing with first-order reactions, the integrated rate law becomes an invaluable tool. For such reactions, the integrated rate law is given by \( [A] = [A]_0 e^{-kt} \), where

• \([A]\) is the concentration of the reactant at time \(t\).
• \([A]_0\) is the initial concentration of the reactant.
• \(k\) is the rate constant.
• \(t\) is the time elapsed.

This equation stems from integrating the differential rate law. It provides a direct relationship between the concentration of sucrose at any time point and the elapsed time. By rearranging this equation, chemists can solve for any variable of interest, whether it be time, concentration, or the rate constant, as demonstrated in the textbook solution.

Exponential Decay in Chemistry

Exponential decay is a concept frequently used to describe processes where quantities decrease over time at a rate proportional to their current value. In the context of chemical reactions, such as the first-order decomposition of sucrose, exponential decay describes how the concentration of the reactant decreases over time. This decay can be expressed mathematically by the equation \( [\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}] = [\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}]_0 e^{-kt} \).

• The term "exponential" signifies that the decrease is not linear but rather becomes slower over time.
• As the concentration of sucrose decreases, the rate of the reaction slows down, following the exponential pattern.

This principle is crucial in understanding not only chemical reactions but also various other natural processes, such as radioactive decay and population dynamics.

Calculating with Natural Logarithms

Natural logarithms (ln) are a mathematical tool that simplifies calculations involving exponential decay, especially in first-order reactions. When the ratio of concentrations in an integrated rate law leads to an equation such as \( \frac{1}{3} = e^{-kt} \),

• The natural logarithm helps by transforming the exponential form into a solvable linear equation.
• By taking the natural logarithm of both sides, we convert the problem to \( \ln\left(\frac{1}{3}\right) = -kt \).
• This step is crucial as it simplifies the process of isolating and solving for time \(t\).

In the context of our exercise, the natural logarithm allows for the calculation of how long it takes for the sucrose concentration to reduce to one-third of its original value. It provides a bridge between the exponential decay form and a linear equation that is easier to manipulate.