Problem 85 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The time required to reach a condition where the concentration of A is 4 times the concentration of B is approximately \(340.12\, s\).

Step 1: Write the first-order reaction equations for both substances

For both substances A and B, we can write the first-order reaction equation as: \[A = A_0e^{-k_At}\] \[B = B_0e^{-k_Bt}\] Where: - \(A\) and \(B\) are the concentrations of A and B at time \(t\), - \(A_0\) and \(B_0\) are the initial concentrations of A and B, respectively, - \(k_A\) and \(k_B\) are the rate constants, - \(t\) is the time for which we need to find.

Step 2: Set the given condition ([A] = 4[B]) and solve for t

We have the given condition that at some time 't', \([A] = 4[B]\). So, \[A_0e^{-k_At} = 4 \times B_0e^{-k_Bt}\] Since we are given that initially A and B have equal concentration (\(A_0 = B_0\)), we can rewrite the equation as: \[e^{-k_At} = 4 \times e^{-k_Bt}\] Now, we need to solve this equation for time, \(t\).

Step 3: Solve the equation to get the value of t

Take the natural log of both sides of the equation: \[-k_At \times ln(e) = ln(4) - k_Bt \times ln(e)\] Simplify and rearrange for \(t\): \[t = \frac{ln(4)}{k_B - k_A}\] Now, substitute the given values of \(k_A\) and \(k_B\) (\(k_A = 4.50 \times 10^{-4}\, \mathrm{s^{-1}}\) and \(k_B = 3.70 \times 10^{-3}\, \mathrm{s^{-1}}\)): \[t = \frac{ln(4)}{3.70 \times 10^{-3} \mathrm{s^{-1}} - 4.50 \times 10^{-4} \mathrm{s^{-1}}}\]

Step 4: Calculate the time for the given condition

Now, calculate the value of \(t\): \[t = \frac{ln(4)}{3.25 \times 10^{-3} \mathrm{s^{-1}}}\] \[t \approx 340.12\, s\] So, approximately \(340.12\, s\) must pass to reach a condition where the concentration of A is 4 times the concentration of B.

Key Concepts

Rate Constant

In first-order kinetics, the rate constant is a crucial figure that determines how quickly a reaction occurs. It is represented by the symbol \( k \). For any first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The rate constant takes a unit of \( ext{time}^{-1} \), such as \( ext{s}^{-1} \) (seconds inverse). This means that the speed of the reaction is measured in relation to time passing.

• The larger the rate constant, \( k \), the faster the reaction proceeds.
• Conversely, a smaller rate constant indicates a slower reaction.

When comparing two reactions, as in the given problem, the rate constant helps us understand how quickly different substances decay over time relative to each other. Here, \( k_A = 4.50 \times 10^{-4} \ \text{s}^{-1} \) and \( k_B = 3.70 \times 10^{-3} \ \text{s}^{-1} \) suggest that substance B decomposes faster than substance A.

Reaction Time Calculation

Calculating the time required for a reaction to reach a specific condition is a fundamental exercise in kinetics. In our scenario, we're tasked with finding the time when the concentration of substance A reaches four times that of substance B. For first-order reactions, we use the formula\[ [A] = A_0 e^{-k_At}\]and similarly for \([B]\).

• We aim to solve the equation \( A_0 e^{-k_At} = 4 \times B_0 e^{-k_Bt} \) to find time, \( t \).
• Given equal initial concentrations, \( A_0 = B_0 \), we simplify the equation by focusing on the exponential terms.

Substituting the known values and manipulating the equation allows us to derive \( t = \frac{\ln(4)}{k_B - k_A} \). Plugging in the given constants and solving yields \( t \approx 340.12 \, ext{s} \). This means it takes a little over 340 seconds for the target condition to be fulfilled.

Exponential Decay

Exponential decay is a core concept in first-order kinetics, describing how the concentration of a substance decreases exponentially over time. The rate at which this decay happens is governed by the rate constant, \( k \). The decrease follows the equation\[ [A] = A_0 e^{-k_At}\]

• The exponential function \( e^{-kt} \) characterizes the decay behavior.
• As time \( t \) increases, the exponent becomes more negative, leading the function \( e^{-kt} \) closer to zero.

This represents a gradual reduction in the concentration of the substance. In the exercise, both A and B undergo exponential decay. Comparing their rate constants shows that B decreases faster than A due to its greater \( k \) value. This is why the decay dynamics influence the condition \([A] = 4[B]\) at a particular time point.

Natural Logarithm

The natural logarithm, \( \ln \), is an essential mathematical function in chemistry, especially when dealing with exponential relationships such as those seen in first-order reactions. \( \ln \) is the logarithm to the base \( e \), an irrational constant approximately equal to 2.718.
When solving for reaction time or other kinetic parameters, the natural logarithm helps linearize the equations from their exponential forms. For the exercise, taking \( \ln \) on both sides of the equation: \[-k_At = \ln(4) - k_Bt\]allowed us to solve for \( t \). The logarithmic properties facilitated simplifying the function to\[t = \frac{\ln(4)}{k_B - k_A}\]

• This step is crucial as it transforms the nonlinear equation into a form that’s solvable using algebraic manipulation.
• Understanding \( \ln \) and its use is central to grasping the mathematics behind kinetic problems in chemistry.