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 \(1.386 \times 10^{-4} \text{ sec}^{-1}\), option (c).

Step 1: Understanding First-Order Kinetics

For a first-order reaction, the rate law is given as \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right) \), where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\). Here, \([A]_0 = 800\, \text{mol/dm}^3\) and \([A]_t = 50\, \text{mol/dm}^3\), with \(t = 2 \times 10^4\, \text{sec}\).

Step 2: Applying the First-Order Rate Law

Substitute the given values into the formula: \[ k = \frac{1}{2 \times 10^4} \ln\left(\frac{800}{50}\right) \]. This provides the structural basis for calculating the rate constant \(k\).

Step 3: Simplifying the Logarithm

Calculate the ratio inside the logarithm: \( \frac{800}{50} = 16 \). Then compute \( \ln(16) \). Utilizing a calculator, \( \ln(16) \approx 2.7726 \).

Step 4: Completing the Rate Constant Calculation

Substitute the logarithmic result back into the rate constant formula: \[ k = \frac{1}{2 \times 10^4} \times 2.7726 \]. Calculating, we get \( k \approx 1.386 \times 10^{-4} \text{ sec}^{-1} \).

Step 5: Selecting the Correct Option

Compare the calculated value of \(k\) with the options: \(2 \times 10^{4}\), \(3.45 \times 10^{-5}\), \(1.386 \times 10^{-4}\), and \(2 \times 10^{-4}\). Our result \(1.386 \times 10^{-4} \text{ sec}^{-1}\) matches option (c).

Key Concepts

Rate Constant Calculation

The rate constant is a critical component in understanding how fast a reaction occurs. In first-order kinetics, calculating the rate constant, symbolized by \( k \), helps us determine the speed of a reaction given certain conditions. The formula for first-order reactions is:

• \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right) \)

Here, \([A]_0\) represents the initial concentration of the reactant, and \([A]_t\) is the concentration at time \( t \). Time itself is denoted as \( t \). In the context of our example, the initial concentration \([A]_0\) is 800 mol/dm³, the remaining concentration \([A]_t\) is 50 mol/dm³, and the time \( t \) is \( 2 \times 10^4 \) seconds.

To find the rate constant, we need to calculate the natural logarithm of the initial concentration divided by the remaining concentration, and then divide that by the time. It's important to always match units, here using seconds for time and mol/dm³ for concentration, ensuring consistency and accuracy in calculations. The end result is the rate constant in sec⁻¹, providing a standardized measure we can compare against others in chemical kinetics.

Natural Logarithm in Kinetics

Natural logarithms are an essential mathematical tool used extensively in chemical kinetics, especially when dealing with first-order reactions. In our given formula \( k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right) \), the \( \ln() \) function transforms the concentration ratio into a growth scale that naturally accounts for the exponential nature of reactant decay.

By computing the natural logarithm of the ratio \( \frac{[A]_0}{[A]_t} \), we're simplifying a potentially complex and rapid decrease into a manageable value. This simplification arises from the properties of natural logarithms, which are ideal for handling exponential decay. For instance, in our case, we calculated \( \frac{800}{50} = 16 \) and then found \( \ln(16) \approx 2.7726 \). This result can then be input into our formula for further calculations.

Understanding natural logarithms makes grasping first-order kinetics more intuitive, as many chemical reactions naturally follow exponential decay patterns. The logarithm helps in capturing this behavior precisely, allowing for more accurate predictions and analyses.

Concentration Decrease in Reactions

In first-order reactions, one characteristic feature is the exponential decrease of reactant concentration over time. This pattern results in a logarithmic relationship between concentrations observed over time intervals, illustrated in the rate constant equation.

As seen in the example, the concentration of a reactant decreases from 800 mol/dm³ to 50 mol/dm³ within a specified period. First-order reactions typically show this exponential decline behavior because the rate at which the reaction proceeds is directly proportional to the reactant concentration. As the concentration diminishes, the rate of reaction slows down as well, reflecting this direct relationship.

• Initial concentration \([A]_0\): 800 mol/dm³
• Concentration after time \([A]_t\): 50 mol/dm³
• Time interval \( t \): \( 2 \times 10^4 \) seconds

This understanding allows chemists to predict concentration at any given time, determine how quickly products appear, or how fast starting materials are consumed. By applying first-order kinetics, we gain insight into broader chemical behavior and how alterations in conditions could affect reaction speed and efficiency.