Problem 23 - Calculus with Concepts in Calculus

Question

The Unimolecular Reaction Theory states that if a substance is dissolved in a large container of solvent, then the rate of reaction is proportional to the amount of the remaining substance. Suppose a sugar cube 1 cubic inch in volume is dropped into a jug of iced tea. If there is \(\frac{3}{4}\) of a cubic inch of the cube left after 1 minute, then when is there \(\frac{1}{2}\) cubic inch left?

Step-by-Step Solution

Answer

It takes approximately 2.41 minutes for the sugar cube to reduce to \( \frac{1}{2} \) cubic inch.

Step 1: Set Up the Differential Equation

Given that the rate of reaction is proportional to the amount of remaining substance, we can model this with a differential equation. Let \( V(t) \) be the volume of the sugar cube at time \( t \). Then the rate of reaction can be expressed as \( \frac{dV}{dt} = -kV \), where \( k \) is the rate constant.

Step 2: Solve the Differential Equation

This differential equation \( \frac{dV}{dt} = -kV \) is separable. We can solve it by separating variables and integrating: \[ \int \frac{1}{V} \, dV = -\int k \, dt. \] Integrating both sides, we get \( \ln V = -kt + C \), where \( C \) is the integration constant.

Step 3: Apply Initial Conditions

Initially, the volume of the sugar cube is 1 cubic inch; therefore, \( V(0) = 1 \). Using this initial condition, we determine the constant: \( \ln 1 = -k \cdot 0 + C \Rightarrow C = 0 \). This simplifies our equation to \( \ln V = -kt \).

Step 4: Determine the Rate Constant

We know that after 1 minute, \( V(1) = \frac{3}{4} \). Substituting into our equation gives \( \ln \frac{3}{4} = -k \cdot 1 \). Solving for \( k \), we find \( k = -\ln \frac{3}{4} \).

Step 5: Solve for the Desired Time

Now we want to find the time \( t \) when the volume \( V(t) = \frac{1}{2} \). Substitute \( V(t) = \frac{1}{2} \) and \( k = -\ln \frac{3}{4} \) into the equation \( \ln V = -kt \). We get: \( \ln \frac{1}{2} = \ln \frac{3}{4} \cdot t \). Solving for \( t \), we find \( t = \frac{\ln \frac{1}{2}}{\ln \frac{3}{4}} \).

Step 6: Calculate the Time

Calculate the numerical value of \( t \). This is \( t = \frac{\ln \frac{1}{2}}{\ln \frac{3}{4}} \approx 2.41 \) minutes.

Key Concepts

Unimolecular Reaction Theory

The unimolecular reaction theory describes how the rate of reaction for a substance that is dissolved in a solvent depends on the amount of the substance that remains, rather than the concentration of any reactants. In these scenarios, it's assumed that each molecule reacts independently, without the need for a second molecule to trigger the reaction.

For example, when you dissolve a sugar cube in iced tea, the speed at which it dissolves is related to the amount that's left. This theory simplifies understanding of reactions that don't require two reactive entities to meet in order to proceed.

This principle is particularly useful in predicting how long reactions will take and how much of the reactant will be left after a certain period. The unimolecular theory essentially puts forth a model where the reaction's progression can be mathematically represented by simple equations.

Reaction Rate

The reaction rate is a crucial concept that describes the speed at which reactants are turned into products in a chemical reaction. In unimolecular reactions, the rate is proportional to the remaining amount of the reactant.

Mathematically, this is expressed as \( \frac{dV}{dt} = -kV \), where \( V \) represents the volume or amount of the reacting substance, and \( k \) is a constant known as the rate constant. This equation tells us that the change in volume over time is directly dependent on the current volume, illustrating exponential decay as the reaction proceeds.

Understanding the reaction rate allows chemists and scientists to predict how quickly a reaction will occur, which is critical in both industrial applications and academic research where reaction time plays a vital role.

Integration

Integration is a fundamental concept in calculus used to solve differential equations by finding antiderivatives. In the case of unimolecular reactions, the differential equation \( \frac{dV}{dt} = -kV \) can be solved through integration.

• First, separate the variables by rearranging: \( \int \frac{1}{V} \, dV = -\int k \, dt \).
• Integrating both sides gives: \( \ln V = -kt + C \), where \( C \) is the integration constant.

Integration here helps in deriving an expression that models how the volume of the reactant decreases over time. It's a critical step in solving problems that involve changing rates.

Separable Differential Equations

Separable differential equations are a subset of differential equations that can be solved by separating variables. This means you can isolate all terms involving one variable on one side of the equation and those involving the second variable on the other side.

In the case of unimolecular reactions, the equation \( \frac{dV}{dt} = -kV \) is separable because it can be rearranged to allow integration: \( \int \frac{1}{V} \, dV = -\int k \, dt \). This separation is vital as it converts a complex differential equation into easier integrals that are solvable.

• Allows solving analytically by integration.
• Helps in understanding natural processes like decay or growth which follow a consistent rate.

Understanding how to separate and solve these equations is an essential skill in calculus and has practical applications in various scientific fields.