Problem 147 - A Complete Resource Book in Chemistry for JEE Main

Question

Two substance 'A' and 'B' are present such that \(\left[\mathrm{A}_{0}\right]=\) \(4\left[\mathrm{~B}_{0}\right]\), and half-life of ' \(^{4} \mathrm{~A}\) ' is 5 minutes and that of ' \(\mathrm{B}\) ' is 15 minute. If they start decaying at the same time following first order, how much time later will the concentration of both of them would be same. (a) \(10 \mathrm{~min}\) (b) \(12 \mathrm{~min}\) (c) \(5 \mathrm{~min}\) (d) \(15 \mathrm{~min}\)

Step-by-Step Solution

Answer

The concentrations will be the same after approximately 6 minutes.

Step 1: Understand the Problem

We are given two substances, A and B, with initial concentrations related by \( \left[\mathrm{A}_{0}\right] = 4\left[\mathrm{~B}_{0}\right] \). The half-life of A is 5 minutes, and for B, it's 15 minutes. Both decay following first-order kinetics. We need to determine when their concentrations will equal each other.

Step 2: Write the First-order Decay Equation

The concentration of a substance undergoing first-order decay is given by the equation:\[ [X] = [X_0] e^{-kt} \]where \([X_0]\) is the initial concentration, \(k\) is the decay constant, and \(t\) is time. We'll use this for both A and B.

Step 3: Calculate Decay Constants

The decay constant \(k\) can be found using the half-life formula \( t_{1/2} = \frac{0.693}{k} \).For A: \( k_A = \frac{0.693}{5} \) min⁻¹.For B: \( k_B = \frac{0.693}{15} \) min⁻¹.

Step 4: Set Up the Equation for Equal Concentration

To find when the concentrations are equal, set up the equation based on \([A_0] e^{-k_A t} = [B_0] e^{-k_B t}\). Given \([A_0] = 4[B_0]\), substitute and simplify:\(4[B_0] e^{-k_A t} = [B_0] e^{-k_B t}\). Cancel \([B_0]\) from both sides to get:\(4 e^{-k_A t} = e^{-k_B t}\).

Step 5: Solve for Time (t)

Taking natural logs on both sides:\( \ln(4) - k_A t = -k_B t \).Substitute values for \(k_A\) and \(k_B\):\( 1.386 - 0.693/5 \cdot t = -0.693/15 \cdot t \).Solve for \(t\):\( 1.386 = (0.693/5 + 0.693/15) \cdot t \).Simplify to find \(t\):\( 1.386 = 0.231 \cdot t \).\( t \approx 6 \) minutes.

Key Concepts

Half-life

The concept of half-life is crucial for understanding first-order kinetics in chemical reactions. Half-life refers to the amount of time it takes for the concentration of a substance to reduce to half of its initial value. This concept is pivotal, especially when dealing with radioactive decay or any first-order reaction.

Here's how to think about it:

• For any first-order reaction, the half-life remains constant irrespective of the initial concentration. This is because the decay process only depends on the proportion of the substance present, not its full concentration.
• The mathematical expression relating half-life to the decay constant is: \[ t_{1/2} = \frac{0.693}{k} \] where \(t_{1/2}\) is the half-life and \(k\) is the decay constant.
• Keeping the half-life constant helps predict how long a substance will remain active or present in a system, which is essential for applications in chemistry, pharmacology, and environmental science.

Understanding half-life helps in predicting and comparing the rates of decay processes of different substances. In our example, substance A has a half-life of 5 minutes, while B has a half-life of 15 minutes, illustrating that A decays much faster.

Decay Constant

The decay constant is a fundamental parameter in first-order kinetics and provides insight into the rate at which a substance undergoes decay.

Here's a deeper look at this concept:

• The decay constant \( (k) \) indicates the probability per unit time that a single substance molecule will decay. A higher decay constant means faster decay.
• Mathematically, to find the decay constant, we use the relationship with half-life: \[ k = \frac{0.693}{t_{1/2}} \] This shows how directly the decay constant is linked to the half-life.
• In a practical scenario, knowing the decay constant allows you to calculate the remaining concentration over time using the formula: \( [X] = [X_0]e^{-kt} \).

This parameter plays a crucial role in determining how quickly a process unfolds. For example, with the provided half-life values, decay constants for A and B are used to calculate the time at which their concentrations become equal.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value, making it a hallmark of first-order kinetics.

Some key points about exponential decay include:

• It implies that the rate of decay decreases over time because the process continually reduces the substance present.
• This behavior is characterized by the equation: \( [X] = [X_0]e^{-kt} \), where the exponents ensure that the function continually decreases as time increases.
• Exponential decay is inherently predictable, making it invaluable in fields such as nuclear physics and pharmacokinetics where understanding the rate of decay is essential.

In practice, for substances A and B in our problem, the use of exponential decay equations is crucial to determine the time at which their concentrations match. This application is indicative of why exponential decay models are so prevalent in scientific analyses.