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Chapter 18: Problem 2

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives for \(\mathrm{A}\) and \(\mathrm{B}\) are 300 \(\mathrm{s}\) and \(180 \mathrm{~s}\), respectively. If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, the time required for the concentration of \(A\) to be four times that of \(\mathrm{B}\) (in s) is: (Use \(\ln 2=0.693\) ) [Main Sep. \(\mathbf{0 5}, \mathbf{2 0 2 0}\) (I)](a) 180 (b) 900 (c) 300 (d) 120

Short Answer

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The time required is 900 seconds.

Step by step solution

01

Understanding First-Order Kinetics

The decomposition of compounds \(A\) and \(B\) are governed by first-order kinetics. For a first-order reaction, the rate of decay can be expressed using the formula \( [A] = [A_0]e^{-kt} \), where \([A]\) is the concentration at time \(t\), \([A_0]\) the initial concentration, and \(k\) is the rate constant.
02

Calculating Rate Constants

The rate constant \(k\) for first-order reactions is related to the half-life \(t_{1/2}\) by the equation \( k = \frac{0.693}{t_{1/2}} \). For \(A\), \( t_{1/2} = 300 \) s, hence \( k_A = \frac{0.693}{300} = 0.00231 \) s\(^{-1}\). For \(B\), \( t_{1/2} = 180 \) s, giving \( k_B = \frac{0.693}{180} = 0.00385 \) s\(^{-1}\).
03

Setting Up the Problem

We start with equal concentrations of \(A\) and \(B\), hence \([A_0] = [B_0]\). We need to find the time \(t\) when \([A] = 4[B]\). This gives us the relationship \( e^{-k_A t} = 4e^{-k_B t} \).
04

Solving for Time \(t\)

Substitute \( k_A \) and \( k_B \) into the relationship: \( e^{-0.00231t} = 4e^{-0.00385t}\). Taking natural logs, we get \(-0.00231t = \ln 4 - 0.00385t\). Rearranging gives \( t(0.00385 - 0.00231) = \ln 4\).\[ t = \frac{2 \ln 2}{0.00154}\].
05

Calculating Time \(t\)

Substitute the values: \( \ln 2 = 0.693 \). Therefore, \[ t = \frac{2 \times 0.693}{0.00154} \approx 900 \] seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Half-Life of a Reaction
Half-life is an important concept when studying the kinetics of chemical reactions, especially for first-order reactions. In simple terms, the half-life of a reaction is the time required for the concentration of a reactant to decrease to half its initial concentration.
This metric is particularly useful because it remains constant regardless of the starting concentration, as long as the reaction follows first-order kinetics. Understanding that the half-life is independent of initial concentration can simplify calculations.
For instance, in the original problem, the half-lives of compounds A and B are given as 300 seconds and 180 seconds, respectively. This indicates how quickly each compound will decompose over time.
  • For compound A: - Half-life ( t_{1/2}) = 300 seconds
  • For compound B: - Half-life ( t_{1/2}) = 180 seconds
These different half-lives help determine the rate at which each compound decomposes and subsequently assist in figuring out how long it will take for one concentration to dominate the other.
Rate Constant Calculation
The rate constant ( k) is a crucial part of understanding first-order kinetics. It's a measure of the speed at which a chemical reaction occurs.
For first-order reactions, there's a straightforward relationship between the half-life and the rate constant: \[ k = \frac{0.693}{t_{1/2}} \].This equation implies that knowing the half-life allows us to determine the rate constant and vice versa.In the problem provided:
  • For compound A: - Rate constant ( k_A) is calculated as follows: \[ k_A = \frac{0.693}{300} = 0.00231 \text{ s}^{-1} \]This value tells us how fast compound A decomposes.
  • For compound B: - Rate constant ( k_B) is calculated as \[ k_B = \frac{0.693}{180} = 0.00385 \text{ s}^{-1} \]This higher rate constant for B indicates it decomposes faster than compound A.
Understanding and calculating the rate constant is crucial for solving time-related problems in chemical kinetics.
Role of Natural Logarithms in Kinetics
In chemical kinetics, natural logarithms ( \ln) play an essential role, particularly for first-order reactions. These logarithms help simplify exponential decay equations into a more manageable format.
In the context of kinetics, when solving problems involving concentrations over time, taking the natural logarithm of both sides of the equation allows for straightforward algebraic manipulation.For example, in the provided solution:
  • The relationship between the concentrations of compounds A and B after a certain time can be expressed in exponential form.
  • This can be difficult to solve directly, so taking the natural logarithm of both sides converts the exponential equation \[e^{-k_At} = 4e^{-k_Bt}\] into a linear form: \[-0.00231t = \ln 4 - 0.00385t\].
This transformation makes it much simpler to isolate the variable (in this case, time) and solve the problem efficiently.Thus, natural logarithms are not just mathematical tools, but also practical necessities in the realm of first-order kinetics.

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Most popular questions from this chapter

The rate constant of a reaction is \(1.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(50^{\circ} \mathrm{C}\) and \(4.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). Evaluate the Arrhenius parameters \(A\) and \(E_{a}\).

\(\mathrm{A}+2 \mathrm{~B} \rightarrow \mathrm{C}\), the rate equation for this reaction is given as Rate \(=\mathrm{k}[\mathrm{A}]\) [B] [Main Online April 11, 2015] If the concentration of \(\mathrm{A}\) is kept the same but that of \(\mathrm{B}\) is doubled what will happen to the rate itself? (a) halved (b) the same (c) doubled (d) quadrupled

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The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at \(25^{\circ} \mathrm{C}\) are \(3.0 \times 10^{-4} \mathrm{~s}^{-1}, 104.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(6.0\) \(\times 10^{14} \mathrm{~s}^{-1}\) respectively. The value of the rate constant as \(T \rightarrow \infty\) is, (a) \(2.0 \times 10^{18} \mathrm{~s}^{-1}\) (b) \(6.0 \times 10^{14} \mathrm{~s}^{-1}\) (c) infinity (d) \(3.6 \times 10^{30} \mathrm{~s}^{-1}\)

A first order reaction has \(k=1.5 \times 10^{-6}\) per second at \(200^{\circ} \mathrm{C}\). If the reaction is allowed to run for 10 hours, what percentage of the initial concentration would have changed in the product? What is the half life of this reaction? [1987 - 5 Marks]
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