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

Question

Two substance ' \(\mathrm{A}^{\prime}\) and ' \(\mathrm{B}^{\prime}\) are present such that \(\left[\mathrm{A}_{0}\right]=\) \(4\left[\mathrm{~B}_{0}\right]\), and half-life of ' \(\mathrm{A}^{\prime}\) 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

(b) 12 min

Step 1: Understand the Problem

We need to find the time after which the concentrations of substances A and B become equal when both decay according to first-order kinetics. The initial concentration of A is four times that of B. We have their half-lives: \(t_{1/2,A} = 5 \text{ min}\) and \(t_{1/2,B} = 15 \text{ min}\).

Step 2: Write the First-Order Kinetics Formula

For a first-order reaction, the concentration \([X]\) at time \(t\) is given by \([X] = [X]_0 e^{-kt}\), where \([X]_0\) is the initial concentration and \(k\) is the rate constant. We will apply this for both A and B.

Step 3: Calculate the Rate Constants

The rate constant \(k\) is related to the half-life by \(k = \frac{0.693}{t_{1/2}}\). Let's calculate the rate constants for A and B: \(k_A = \frac{0.693}{5}\) and \(k_B = \frac{0.693}{15}\).

Step 4: Set Up the Equality Equation

We are looking for the time \(t\) when the concentrations of A and B are equal: \([A]_0 e^{-k_A t} = [B]_0 e^{-k_B t}\). Using \([A]_0 = 4[B]_0\), we have: \(4[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t}\).

Step 5: Simplify and Solve for t

Divide both sides by \([B]_0\) and simplify:\(4 e^{-k_A t} = e^{-k_B t}\).Take natural logarithms on both sides:\(\ln 4 = (k_B - k_A) t\).Substitute \(k_A = \frac{0.693}{5}\) and \(k_B = \frac{0.693}{15}\):\(t = \frac{\ln 4}{\frac{0.693}{15} - \frac{0.693}{5}}\).

Step 6: Calculate the Time

Calculate the numerical value:\(t = \frac{\ln 4}{\frac{0.693}{15} - \frac{0.693}{5}}\).Simplifying gives us \(t = \frac{\ln 4}{0.0462 - 0.1386}\),\(= \frac{\ln 4}{0.0462}\),\(\approx 12 \text{ min.}\)

Step 7: Conclusion

The concentrations of substances A and B become equal after approximately 12 minutes.

Key Concepts

Half-Life Calculation

Understanding the concept of half-life is essential to solving many problems in first-order kinetics. The half-life of a substance is the time required for its concentration to reduce to half of its initial value. For first-order reactions, the half-life is constant, irrespective of the initial concentration. This characteristic makes it easier to predict the behavior of substances over time.

To find the half-life (\(t_{1/2}\)) for any first-order reaction, we use the formula:\[ t_{1/2} = \frac{0.693}{k} \]where \(k\) is the rate constant of the reaction. This formula is derived from the first-order kinetics equation and illustrates how the rate constant inversely affects the half-life. A larger rate constant results in a shorter half-life, indicating a faster reaction. It is crucial to remember this relationship when analyzing the decay of substances.

In the given exercise, substance A has a half-life of 5 minutes, while substance B has a half-life of 15 minutes. Applying this formula helps calculate each substance's rate constant, which is a critical step in determining when their concentrations equalize.

Rate Constant Determination

The rate constant (\(k\)) is a pivotal factor in understanding first-order reactions. It provides insight into the speed of the reaction. For first-order kinetics, the rate of reaction depends solely on the concentration of one reactant. The formula for calculating the rate constant from the half-life is:\[ k = \frac{0.693}{t_{1/2}} \]This relationship shows that the rate constant can be easily determined once the half-life is known.

In the problem, we calculate:

• For substance A, the rate constant \(k_A = \frac{0.693}{5}\).
• For substance B, the rate constant \(k_B = \frac{0.693}{15}\).

These calculations reveal that substance A decays faster, as indicated by its larger rate constant. Understanding this helps predict not only how fast a reaction proceeds but also assists in computing crucial time-dependent changes such as when two substances reach equal concentrations.

Concentration Equality in Reactions

Achieving concentration equality in a reaction requires setting both substances' concentrations to be equivalent at a particular time. In first-order reactions, the concentration at any time \(t\) is given by:\[ [X] = [X]_0 e^{-kt} \]Here, \([X]_0\) is the initial concentration, and \(e^{-kt}\) describes how the concentration decreases over time.

For equal concentrations in our problem, we set:\[[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \]With the initial concentration relation \([A]_0 = 4[B]_0\), we can simplify this to:\[4[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \]By dividing through by \([B]_0\) and taking the natural logarithm, we find:\[ \ln 4 = (k_B - k_A) t \]This key equation allows us to solve for the time \(t\) when the concentrations equate. Substituting the previously calculated rate constants and solving yields approximately 12 minutes as the time when substances A and B have equal concentrations.

This solution illustrates how minor mathematical manipulation of the rate equations can lead to significant predictions on reaction progress and equal concentration times.