Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

In a first-order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times 10^{4}\) sec. The rate constant of reaction in \(\mathrm{sec}^{-1}\) is (a) \(2 \times 10^{4}\) (b) \(3.45 \times 10^{-5}\) (c) \(1.386 \times 10^{-4}\) (d) \(2 \times 10^{-4}\)

Step-by-Step Solution

Answer

The rate constant is approximately \(1.386 \times 10^{-4}\) sec\(^{-1}\), which is option (c).

Step 1: Understand the First-order Reaction Kinetics

For a first-order reaction, the rate equation can be written as \( ext{ln}([A]_t) = -kt + ext{ln}([A]_0) \). Here, \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), and \(k\) is the rate constant. Our goal is to solve for \(k\).

Step 2: Recognize the Given Information

We know that \([A]_0 = 800 \, \text{mol/dm}^3\), \([A]_t = 50 \, \text{mol/dm}^3\), and \(t = 2 \times 10^4\) seconds. We'll use these values in the first-order rate equation to calculate \(k\).

Step 3: Substitute Values into the First-order Rate Equation

Substitute the given values into the equation \( ext{ln}([A]_t) = -kt + ext{ln}([A]_0) \):\[ ext{ln}(50) = -k (2 \times 10^4) + ext{ln}(800) \].

Step 4: Rearrange Equation to Solve for the Rate Constant \(k\)

Rearrange the equation to solve for \(k\):\[ k = \frac{ ext{ln}(800) - ext{ln}(50)}{2 \times 10^4} \].

Step 5: Calculate the Natural Logarithms

Calculate the natural logarithms for \(800\) and \(50\):\( ext{ln}(800) \approx 6.6846\) and \( ext{ln}(50) \approx 3.912\).

Step 6: Solve for \(k\)

Substitute the values from step 5 into the equation:\[ k = \frac{6.6846 - 3.912}{2 \times 10^4} \approx \frac{2.7726}{2 \times 10^4} \].Thus, \( k \approx 1.386 \times 10^{-4} \) sec\(^{-1}\).

Step 7: Choose the Correct Option

From our calculations, the rate constant \(k\) is approximately \(1.386 \times 10^{-4}\) sec\(^{-1}\). This matches with option (c).

Key Concepts

Rate Constant Calculation

In first-order reaction kinetics, the rate constant, often represented as \(k\), is a measure of how quickly a reaction proceeds. It's crucial to calculate \(k\) to predict how long a reaction will take or to understand the reaction's dynamics. To determine the rate constant of a reaction, we use a specific rate equation that applies to first-order reactions:

• \( \ln([A]_t) = -kt + \ln([A]_0) \)

This equation involves:

• \([A]_0\): The initial concentration of the reactant
• \([A]_t\): The concentration of the reactant at time \(t\)
• \(k\): The rate constant

By rearranging this equation, you can solve for \(k\) as follows:

• \( k = \frac{\ln([A]_0) - \ln([A]_t)}{t} \)

This formula highlights the inverse relationship between \(k\) and time \(t\): as \(k\) increases, the time needed for the concentration to change decreases. Understanding this concept helps in calculating \(k\) using experimental data of concentrations at different times.

Natural Logarithm in Kinetics

The natural logarithm, represented by \( \ln \), is an essential part of the first-order kinetics equation. Its importance comes from its properties that simplify the complex math of reaction rates. When dealing with exponential decay of reactant concentrations, the natural logarithm helps to linearize the reaction's exponential nature.
In our study of kinetics, the natural logarithm is used as follows:

• Converts exponential concentration data into a form that can be easily manipulated with linear math.
• Allows the calculation of \( k \) when initial and final concentrations are known.

For instance, in our example, we calculate

• \( \ln(800) \approx 6.6846 \)
• \( \ln(50) \approx 3.912 \)

These values insert into our rearranged first-order equation to isolate \( k \). This process of using logs to manipulate exponential equations is a fundamental technique in chemical kinetics, simplifying the visualization and calculation of reaction rates.

Concentration Change in Reactions

In chemical reactions, understanding how and why concentrations change over time is vital. In first-order reactions, the concentration of reactants decreases exponentially. The characteristics of this change are explained by:

• Each half-life is constant; the time it takes for the concentration to halve is always the same.
• The rate of concentration change depends linearly on the concentration itself.

In the scenario given, the concentration drops from \( 800 \mathrm{~mol/dm}^3 \) to \( 50 \mathrm{~mol/dm}^3 \) over \( 2 \times 10^4 \) seconds. This significant decrease highlights the exponential nature of first-order reactions.
To understand this further, we utilize the first-order reaction equation to track concentration changes over time. This equation allows us to predict how concentration will decrease as reactions progress, making it a powerful tool in chemistry. Recognizing these changes is crucial for experiments and industrial applications where specific concentrations of reactants at certain time intervals are critical for desired outcomes.