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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

Expert verified
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 law for the reaction below is given by the expression \(\mathrm{k}[\mathrm{A}]\) [B] \(\mathrm{A}+\mathrm{B} \rightarrow\) Product If the concentration of \(\mathrm{B}\) is increased from \(0.1\) to \(0.3 \mathrm{~mole}\), keeping the value of \(A\) at \(0.1\) mole, the rate constant will be: [Main Online April 10, 2016] (a) \(3 k\) (b) \(9 k\) (c) \(k / 3\) (d) \(k\)

The rate constant of a zero order reaction is \(2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). If the concentration of the reactant after 25 seconds is \(0.5 \mathrm{M}\). What is the initial concentration? [Main Online April 23, 2013](a) \(0.5 \mathrm{M}\) (b) \(1.25 \mathrm{M}\) (c) \(12.5 \mathrm{M}\) (d) \(1.0 \mathrm{M}\)

A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(C^{14}\) concentration is \(C_{1}\) in nearby areas and \(C_{2}\) in areas far away. If the age of the fossil is determined to be \(T_{1}\) and \(T_{2}\) at the respective places then (a) The age of the fossil will increase at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (b) The age of the fossil will decrease at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (c) The age of fossil will be determined to be same (d) \(\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}\)

According to the Arrhenius equation, (a) a high activation energy usually implies a fast reaction. (b) rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy. (c) higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant. (d) the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.

The rate of a reaction quadruples when the temperature changes from 300 to \(310 \mathrm{~K}\). The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature; \(\left.\ln 2=0.693 ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(107.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(26.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(214.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
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