Question

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?

Short answer

To find the time when the concentration of A is 4 times the concentration of B, we can use the first-order reaction equation and the given rate constants. The equation is \(e^{-k_At} = 4 \times e^{-k_Bt}\). Solving for \(t\) gives \(t = \frac{\ln{4}}{k_B - k_A}\). Substituting the given values of \(k_A\) and \(k_B\) and calculating, we find that it takes approximately \(538.8\) seconds for the condition to be met.

Step-by-step solution

Identify the formula for the first-order reaction

The general equation to describe first-order reactions is given by: $$ [A]_t = [A]_0 \times e^{-k_At} $$ Similarly, for substance B, it would be: $$ [B]_t = [B]_0 \times e^{-k_Bt} $$ Where: - \([A]_t\) and \([B]_t\) are the concentrations of substances A and B, respectively, at a given time \(t\), - \([A]_0\) and \([B]_0\) are the initial concentrations at time \(t=0\), - \(k_A\) and \(k_B\) are the rate constants of substances A and B, respectively, - \(t\) is the time elapsed.

Express the given condition \([A] = 4[B]\) using the formulas

To find the time when the condition \([A] = 4[B]\) is met, we can substitute the concentration expressions from the previous step into this equation. It would look like this: $$ [A]_0 \times e^{-k_At} = 4 \times ([B]_0 \times e^{-k_Bt}) $$Since \([A]_0 = [B]_0\) (given that both substances have equal concentrations initially), we can simplify the equation to: $$ e^{-k_At} = 4 \times e^{-k_Bt} $$

Solve for time t

In order to solve for time \(t\), we can first take the natural logarithm of both sides to get rid of the exponentials: $$ \ln{e^{-k_At}} = \ln{(4 \times e^{-k_Bt})} $$This simplifies to: $$ -k_At = \ln{4} - k_Bt $$Now we can solve for \(t\): $$ t = \frac{\ln{4}}{k_B - k_A} $$Plug in the given values of \(k_A\) and \(k_B\): $$ t = \frac{\ln{4}}{3.70 \times 10^{-3} s^{-1} - 4.50 \times 10^{-4} s^{-1}} $$

Calculate the time

Finally, evaluate the expression to find the time: $$ t = \frac{\ln{4}}{3.25 \times 10^{-3} s^{-1}} \approx 538.8 \,\text{seconds} $$ So, it takes approximately 538.8 seconds for the concentration of A to be 4 times the concentration of B.

Key concepts

Reaction Rate Constants

In the study of chemical reactions, understanding reaction rate constants is fundamental. These constants, typically represented by the symbol \(k\), provide insight into how fast a reaction proceeds.

The rate constant is unique to each substance and specific to the conditions of the reaction, such as temperature and pressure.

• For first-order reactions, the rate of reaction is directly proportional to the concentration of the single reactant.
• The units of the rate constant for a first-order reaction are \(s^{-1}\).

In the exercise, both substances A and B decompose following first-order kinetics. With their rate constants given as \(k_A = 4.50 \times 10^{-4} \ \text{s}^{-1}\) and \(k_B = 3.70 \times 10^{-3} \ \text{s}^{-1}\), we can determine that substance B decomposes at a faster rate than substance A due to its higher rate constant.

Exponential Decay

Exponential decay is a fundamental concept in first-order kinetics, where the quantity of a substance decreases exponentially over time. This means the rate of decline of the substance is proportional to its current amount.

In practice, exponential decay is represented mathematically by the formula:\[[X]_t = [X]_0 \times e^{-kt}\]where:

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

By applying this formula, you describe how the concentration of substances A and B decreases with time. This predictable nature of exponential decay helps in calculating the time required for a substance to reach a certain concentration, as seen when determining how long it takes for \([A]\) to be 4 times \([B]\).

Chemical Decomposition Reactions

Chemical decomposition reactions involve breaking down a chemical compound into two or more simpler substances. These reactions are pivotal in various biological, industrial, and environmental processes.

In the context of the exercise, both substances A and B undergo decomposition following first-order kinetics. This indicates that:

• The rate of decomposition is solely dependent on the concentration of the substance.
• Each substance breaks down into simpler products at a rate characterized by its specific rate constant.

Understanding the kinetics of decomposition reactions is crucial because it helps scientists and engineers predict how long it might take for a substance to break down under certain conditions.

By applying our knowledge of first-order kinetics, the time can be calculated for scenarios like when one substance's concentration becomes a multiple of another, as explained in the solution.