Question

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(\mathrm{A}\) and 18.0 min for \(\mathrm{B}\). 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 takes approximately 56.85 minutes for \(A\) to be four times \(B\).

Step-by-step solution

Understanding First-order Kinetics

In first-order kinetics, the rate of decomposition for each compound is proportional to its current concentration. The concentration over time can be calculated using the formula: \[ c(t) = c_0 e^{-kt} \]where \( c(t) \) is the concentration at time \( t \), \( c_0 \) is the initial concentration, and \( k \) is the rate constant.

Finding the Rate Constants

To find the rate constants (\(k\)) for compounds \(A\) and \(B\), we use the relation between half-life and rate constant for first-order reactions:\[ k = \frac{0.693}{t_{1/2}} \]For \(A\), \(k_A = \frac{0.693}{50.0} = 0.01386\, \text{min}^{-1}\).For \(B\), \(k_B = \frac{0.693}{18.0} = 0.0385\, \text{min}^{-1}\).

Setting Up the Equation

Initially, the concentrations of \(A\) and \(B\) are equal, say \(c_0\). After time \(t\), their concentrations are:\[ c_A(t) = c_0 e^{-k_A t} \]\[ c_B(t) = c_0 e^{-k_B t} \]We want \(c_A(t) = 4 \times c_B(t)\).Thus, the equation becomes:\[ c_0 e^{-k_A t} = 4 \times c_0 e^{-k_B t} \]

Solving for Time

Cancelling \(c_0\) from both sides gives:\[ e^{-k_A t} = 4 e^{-k_B t} \]Taking the natural logarithm of both sides:\[ -k_A t = \ln 4 - k_B t \]Rearranging gives:\[ t(k_B - k_A) = \ln 4 \]Solving for \(t\):\[ t = \frac{\ln 4}{k_B - k_A} = \frac{1.3863}{0.0385 - 0.01386} \approx 56.85 \, \text{minutes} \]

Key concepts

Half-Life Calculations

The concept of half-life plays a crucial role in understanding first-order kinetics. Half-life ( t_{1/2}) is the time required for a substance's concentration to decrease by half. In first-order reactions, the half-life is consistent regardless of the concentration.

When performing half-life calculations, we often use the formula for first-order kinetics:\[ t_{1/2} = \frac{0.693}{k} \]where k is the rate constant. This simple equation highlights how the half-life is inversely proportional to the rate constant. Contrast this with other kinetics where half-life can vary with initial concentration.

• For instance, compound A with a half-life of 50.0 minutes indicates slower decomposition compared to compound B's 18.0 minutes.
• This insight allows us to predict and calculate how long a particular reactant will last under certain conditions.

Understanding half-life is pivotal as it simplifies calculations for various reactions, especially in predicting how long a compound takes to reach a desired concentration.

Rate Constant Determination

Determining the rate constant is essential for solving problems related to first-order reactions. The rate constant ( k ) quantifies the speed of the reaction at a molecular level.

The relationship between the rate constant and half-life is straightforward for first-order kinetics:\[ k = \frac{0.693}{t_{1/2}} \]For compound A, substituting its half-life into the formula gives:\[ k_A = \frac{0.693}{50.0} = 0.01386\, \text{min}^{-1} \] For compound B, it is:\[ k_B = \frac{0.693}{18.0} = 0.0385\, \text{min}^{-1} \]

• A higher rate constant, as seen with B, indicates a faster reaction.
• These calculated values help predict concentration changes over time.

This determination allows chemists to manipulate conditions to favor specific reaction rates and outcomes, making it a vital step in applying the principles of kinetics practically.

Concentration Equations

Concentration equations are pivotal in tracking how concentrations of reactants change over time in a chemical reaction. For first-order reactions, the concentration at any time ( t ) can be described by:\[ c(t) = c_0 e^{-kt} \]where c_0 is the initial concentration, and k is the rate constant.

Using these equations for compounds A and B:

• For A: \[ c_A(t) = c_0 e^{-k_A t} \]
• For B: \[ c_B(t) = c_0 e^{-k_B t} \]

These expressions allow us to set up equations to solve for specific conditions, like when the concentration of A is four times that of B. By adjusting the time t , you can manipulate concentrations to achieve desired ratios.

Chemical Decomposition Reactions

Chemical decomposition reactions involve breaking down compounds into simpler substances. When these reactions occur via first-order kinetics, they depend on the concentration of only one reactant.

For example, the exercise illustrates this concept using compounds A and B decomposing independently:

• Both A and B follow first-order kinetics, where the reaction rate depends only on their individual concentrations.
• This makes it possible to express each compound's decomposition through its own rate constant and initial concentrations.

Understanding these decomposition processes provides insights into reaction mechanisms, essential in fields like biochemistry and pharmaceuticals, where controlling reaction rates is critical for desired outcomes.