Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

For a first-order reaction, \(t_{0.9}\) is \(138.6\) seconds. Its specific reaction rate constant (in \(\left.\sec ^{-1}\right)\) is (a) \(10^{-2}\) (b) \(10^{-4}\) (c) \(10^{-5}\) (d) \(10^{-6}\)

Step-by-Step Solution

Answer

The specific reaction rate constant is approximately \(10^{-4} \text{ sec}^{-1}\).

Step 1: Understand the Question

We are given a first-order reaction and are asked to find the specific reaction rate constant (k) based on the given value of \(t_{0.9}\), which is the time taken for the concentration of a reactant to decrease to 90% of its initial value.

Step 2: Use First-Order Reaction Formula

For a first-order reaction, the relationship between the rate constant \(k\) and the time required for the concentration to decrease to a certain fraction of its original value is given by the formula: \[ t_x = \frac{2.303}{k} \log \left( \frac{1}{1-x} \right) \] Here, \(x = 0.1\) since we're considering the time to reach 90% of the initial concentration.

Step 3: Substitute Given Values

We have \(t_{0.9} = 138.6\) seconds and \(x = 0.1\). Substitute these values into the formula:\[ 138.6 = \frac{2.303}{k} \log \left( \frac{1}{0.9} \right) \] Calculate the logarithm: \( \log \left( \frac{1}{0.9} \right) \approx 0.045757\).

Step 4: Solve for Rate Constant k

Rearrange the equation to solve for \(k\):\[ k = \frac{2.303 \times 0.045757}{138.6} \] Calculate:\[ k \approx \frac{0.10527}{138.6} \approx 7.6 \times 10^{-4} \text{ sec}^{-1} \]

Step 5: Choose the Correct Option

Compare the calculated \(k = 7.6 \times 10^{-4} \text{ sec}^{-1}\) with the given options:(a) \(10^{-2}\), (b) \(10^{-4}\), (c) \(10^{-5}\), and (d) \(10^{-6}\). The closest value to our calculation is \(10^{-4}\).

Key Concepts

Rate Constant Calculation

To solve problems involving first-order reactions, it is important to calculate the rate constant, denoted as \( k \). The rate constant is a crucial parameter indicating how fast a reaction proceeds.
The formula to find \( k \) for a first-order reaction where the concentration decreases to a fraction \( x \) of its original value is given by:

• \( t_x = \frac{2.303}{k} \log \left( \frac{1}{1-x} \right) \)

To solve for \( k \), you will first need to know two things:

• The time \( t_x \) it takes to reach fraction \( x \)
• The fraction \( x \) itself

By rearranging the formula, \( k \) can be calculated as follows:

• \( k = \frac{2.303}{t_x} \log \left( \frac{1}{1-x} \right) \)

This formula essentially derives \( k \) by linking it to the percentage of reactant concentration remaining. In this specific problem, for example, \( t_{0.9} = 138.6 \) seconds is crucial for determining \( k \). Remember, knowing how to manipulate this basic formula is a doorway to solving various kinetics problems.

Reaction Kinetics

Understanding reaction kinetics is key to mastering how chemical reactions happen and at what speed. Reaction kinetics not only tells us about rate constants but also the mechanisms and sequences of steps involved.
A first-order reaction is straightforward: it relies on the concentration of a single reactant. The rate of reaction decreases exponentially over time as the reactant is consumed.

• One particular characteristic of first-order reactions is their dependency on the concentration of one reactant only.
• They have a constant half-life, meaning it takes the same amount of time for the concentration to fall from any initial amount to half of that, and then from half to one quarter, etc.

Being able to determine the rate constant helps us predict how the reaction will behave over time. Mastery of these kinetics rules is imperative for problem-solving in chemistry, and understanding their application can illuminate many experimentally observed phenomena.

Logarithmic Relationships in Chemistry

Logarithmic functions are often used in chemistry to express complex relationships efficiently. When dealing with first-order reactions, the logarithmic relationship provides a tool to link time and concentration changes.
The formula \( t_x = \frac{2.303}{k} \log \left( \frac{1}{1-x} \right) \) involves a logarithm to represent how remaining reactant concentrations affect reaction rates, a critical aspect to elucidate with logarithm properties:

• The logarithm, \( \log \left( \frac{1}{1-x} \right) \), expresses how the remaining fraction of a reactant over its initial amount is related to the progression of the reaction.
• It enables simplification because reactions often exhibit exponential decrease, and logarithms provide a linear method to express and calculate these trends accurately.

Always remember, grasp of logarithmic functions in chemistry extends beyond first-order reactions. They frequently arise in calculating pH in acid-base chemistry, examining light absorption in spectrophotometry, and more. Having a strong handle on logarithms equips you to tackle these challenges across different chemistry domains.