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

The time required for the concentration of substance A to become equal to 4 times the concentration of substance B is approximately \(t\approx 406.52\,\text{s}\).

Step-by-step solution

Write down the first-order kinetics formula for the concentrations of A and B

The general formula for the concentration of a species after t seconds in a first-order reaction is given by: \[ C(t) = C_0 e^{-kt} \] We are given that the initial concentrations of A and B are equal, and we can denote this as \(C_0\). So, we will apply this formula for A and B: \[ [\mathrm{A}](t) = C_0 e^{-k_\mathrm{A}t} \] \[ [\mathrm{B}](t) = C_0 e^{-k_\mathrm{B}t} \]

Write down the condition for the required concentrations of A and B

Given that we need to find a time at which the concentration of substance A becomes equal to 4 times the concentration of substance B, we can write our equation as follows: \[ [\mathrm{A}](t) = 4 [\mathrm{B}](t) \]

Substitute the formulas from Step 1 into the condition from Step 2

Substitute the expressions for the concentrations of A and B from Step 1 into the condition equation from Step 2: \[ C_0 e^{-k_\mathrm{A}t} = 4 (C_0 e^{-k_\mathrm{B}t}) \]

Solve for t

We now have a single equation with only one variable, t, and we can solve for it. First, divide both sides by \(C_0\): \[ e^{-k_\mathrm{A}t} = 4 e^{-k_\mathrm{B}t} \] Now take the natural logarithm of both sides: \[ \ln (e^{-k_\mathrm{A}t}) = \ln (4 e^{-k_\mathrm{B}t}) \] Using the properties of natural logarithms, we can simplify the equation: \[ -k_\mathrm{A}t = \ln(4) - k_\mathrm{B}t \] Now, isolate t on one side of the equation: \[ t(k_\mathrm{B} + k_\mathrm{A}) = \ln(4) \] Finally, divide by the sum of the rate constants to find the value of t: \[ t = \frac{\ln(4)}{k_\mathrm{B} + k_\mathrm{A}} \] Now, plug in the given values for \(k_\mathrm{A}\) and \(k_\mathrm{B}\): \[ t = \frac{\ln(4)}{(4.50 \times 10^{-4}\,\text{s}^{-1}) + (3.70 \times 10^{-3}\,\text{s}^{-1})} \]

Calculate t

Perform the calculation to find the value of t: \[ t \approx \frac{\ln(4)}{4.150\times10^{-3}\,\text{s}^{-1}} \approx 406.52\,\text{s} \] Therefore, approximately 406.52 seconds must pass for the concentration of substance A to become equal to 4 times the concentration of substance B.

Key concepts

Rate Constant

When studying chemical kinetics, the rate constant is a pivotal component, symbolized by the letter 'k' in mathematical expressions. It's a quantifier of how quickly a reaction proceeds under specific conditions, including temperature and, if present, catalysts. First-order kinetics, in particular, are characterized by a reaction rate that is directly proportional to the concentration of one reactant.
For a reaction like the decomposition of substances A and B in our example, the rate constants, denoted by k_A and k_B respectively, play a fundamental role in determining the time it takes for the concentrations to change. These constants, intrinsic to each substance, allow us to calculate not just the reaction rate at any given moment, but also predict future concentrations using the formula C(t) = C₀ e^-kt, where C₀ is the initial concentration.
Having different rate constants, such as those given for A and B, indicates that they decompose at different rates, with B decomposing faster due to a higher rate constant. This difference allows us to compare and predict how their concentrations relate over time.

Half-Life

The concept of half-life is particularly important when discussing first-order kinetics. It represents the time it takes for the concentration of a reactant to decrease by half. In mathematical terms, the half-life, usually denoted by t_1/2, is given for a first-order reaction as t_1/2 = ln(2) / k, where 'k' is the rate constant.
This value is constant for first-order reactions and does not depend on the initial concentration. As such, it is a useful measure for comparing how quickly different substances react over time. For instance, if we were given the task of determining the half-life of substances A and B from our exercise based on their rate constants, we'd find substance B has a quicker half-life implying it reacts and thus depletes more rapidly.
Understanding half-life can aid in analyzing various processes from radioactive decay in physics to medication metabolism in pharmacology. It serves also as a tool for making predictions about reaction progress and is integral in fields like geochronology and environmental science for its role in decay and degradation processes.

Reaction Rate

Reaction rate is a descriptor of the speed at which reactants convert into products in a chemical reaction. It is affected by various factors including concentration, temperature, and the presence of catalysts. In the context of first-order kinetics, the reaction rate is directly proportional to the concentration of a single reactant.
To find this rate, we can use the formula Rate = k * [Reactant], where 'k' is the rate constant, and '[Reactant]' denotes the concentration of the reactant at that moment. As seen in the textbook example involving substances A and B, though they both decompose via first-order kinetics, their rates of reaction vary due to different rate constants.
Additionally, the rate at which the concentration of a reactant decreases is critical for calculating how long it takes to reach a certain concentration, such a scenario was illustrated in the original exercise where we determined the time required for the concentration of A to become four times greater than that of B. Understanding reaction rates is vital in areas like the design of chemical reactors and the pharmaceutical industry where reaction speed is crucial for efficiency and safety.