Question

A flask contains a mixture of compounds \(A\) and \(B\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(A\) and 18.0 min for \(B\). If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, how long will it take for the concentration of \(\mathrm{A}\) to be four times that of \(\mathrm{B} ?\)

Short answer

It will take 56.25 minutes for the concentration of A to be four times that of B.

Step-by-step solution

Formulate the Problem

The problem asks for the time it takes for the concentration of A to be four times that of B. It needs to be solved as follows: \(C_{A} = 4C_{B}\)\nLet the initial concentrations of A and B be \(C_{0}\), then the concentrations at time \(t\) are given by: \n\(C_{A} = C_0 \times \frac{1}{2}^{(t/50)}\) for A and \(C_{B} = C_0 \times \frac{1}{2}^{(t/18)}\) for B.

Solve the Equality

In order to find the time when the concentration of A is four times that of B, we can set up the following equality and solve for \(t\). Set \(C_{A} = 4C_{B}\) and solve for \(t\). Plug in the specific expressions for \(C_{A}\) and \(C_{B}\) from step 1 into this equality, giving: \(C_0 \times \frac{1}{2}^{(t/50)} = 4 \times (C_0 \times \frac{1}{2}^{(t/18)})\). Since the initial concentration of A and B are the same, \(C_0\) cancels on both sides of the equation.

Solve for Time

The equation simplifies to \(\frac{1}{2}^{(t/50)} = 4 \times \frac{1}{2}^{(t/18)}\). Raise each side to the base 2, giving \(2^{t/50} = 2^{2 + t/18}\). Since the bases are the same, the exponents must be equal, giving \(t/50 = 2 + t/18\). Now solve for \(t\): Multiply through by 900 to clean up the fractions, providing 18t = 1800 + 50t. Subtract 18t from both sides: 1800 = 32t. Divide by 32 to isolate \(t\), giving \(t = 56.25\) minutes.

Key concepts

Half-Life Calculations

The half-life of a substance is the time it takes for its concentration to decrease by half through decay. It's a fundamental concept in first-order kinetics, where the rate of reaction depends on the concentration of one reactant. In the given problem, compounds \(A\) and \(B\) decompose with half-lives of 50.0 and 18.0 minutes, respectively, indicating how quickly each compound reduces.

To tackle problems involving half-life, it's crucial to recognize that the concentration at any time \(t\) can be expressed as a fraction of the initial concentration. For first-order reactions, this is calculated using \( C = C_0 \times \left(\frac{1}{2}\right)^{(t/\text{half-life})} \).

This setup allows us to determine how long it takes for concentrations to reach a certain multiple or fraction of their initial values. In the exercise, given equal initial concentrations, we need to find when \(A\) reaches four times the concentration of \(B\). This involves setting up an equation that considers their respective half-lives and solving for \(t\).

Concentration of Reactants

The concentration of reactants decreases over time in first-order reactions. It can be described using an exponential decay model where a single reactant's concentration diminishes based on time and the reaction's rate constant.

For compounds \(A\) and \(B\), both following first-order kinetics, the concentration at a given time \(t\) can be defined using the formula \( C = C_0 \times \left(\frac{1}{2}\right)^{(t/\text{half-life})} \). Here, \( C_0 \) is the initial concentration, and \( \text{half-life} \) is different for each compound, allowing us to compare their rates of decay.

In the problem, both compounds start with the same initial concentration \(C_0\). As they decay, the aim is to find when \(C_A = 4C_B\). By substituting the formulas of \(C_A\) and \(C_B\), we equate and solve the logarithmic equation, which helps in understanding the quantitative comparison of rates for two substances under the same initial conditions.

Exponential Decay

Exponential decay describes how quantities decrease at a consistent rate over time. It's observed in first-order kinetics where the rate of reaction is proportional to the concentration of a single reactant.

In the exercise, the reduction in concentration of both compounds \(A\) and \(B\) follows an exponential decay process. This type of decay can be expressed mathematically as \(C = C_0 \times e^{-kt}\), where \(k\) is the rate constant, and \(t\) is the time. However, when the half-life is known, it can be more intuitively presented using the half-life formula.

By recognizing the role of exponential decay, we gain insights into how fast \(A\) and \(B\) diminish in the flask. Understanding this concept helps in setting up the necessary equations to predict when one concentration surpasses another, as required in the problem when \(C_A = 4C_B\). This is achieved by equating and solving the expressions for the exponential decay of each compound, adjusted by their specific half-lives.