Problem 129 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(380^{\circ} \mathrm{C}\), half-life period for the first-order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{1} .\) Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half-life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). (a) \(20.4 \mathrm{~min}\) (b) \(408 \mathrm{~min}\) (c) \(10.2 \mathrm{~min}\) (d) none of these

Step-by-Step Solution

Answer

The time required for 75% decomposition at 450°C is approximately 20.4 minutes (option a).

Step 1: Understand the given data

We have a first-order reaction where the half-life period at two different temperatures is given. At \( 380^{\circ}C \), the half-life \( t_{1/2} = 360 \) min and at \( 450^{\circ}C \), the half-life \( t_{1/2} = 10.17 \) min. The activation energy \( E_a = 200 \text{ kJ/mol} \), and we need to calculate the time for 75% decomposition at \( 450^{\circ}C \).

Step 2: Understand first-order reaction kinetics

For a first-order reaction, the relationship between the half-life \( t_{1/2} \) and the rate constant \( k \) is \( t_{1/2} = \frac{0.693}{k} \). We will use this formula to find the rate constants at each temperature.

Step 3: Calculate rate constants at both temperatures

First, find \( k_1 \) at \( 380^{\circ}C \): \[ k_1 = \frac{0.693}{360 \text{ min}} \approx 0.001925 \text{ min}^{-1} \] Then find \( k_2 \) at \( 450^{\circ}C \): \[ k_2 = \frac{0.693}{10.17 \text{ min}} \approx 0.06815 \text{ min}^{-1} \]

Step 4: Apply Arrhenius equation to relate rate constants

The Arrhenius equation relates rate constants and temperature: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] with \( R = 8.314 \text{ J/(mol K)} \), \( T_1 = 380 + 273 = 653 \text{ K} \), and \( T_2 = 450 + 273 = 723 \text{ K} \).Calculate:\[ \ln \left( \frac{k_2}{0.06815} \right) \]

Step 5: Calculate time for 75% decomposition

For a first-order reaction, the relationship between time \( t \) and the fraction decomposed \( (1-x) = 0.25 \) at 75% decomposition is given by:\[ t = \frac{1}{k_2} \ln \left( \frac{1}{1-x} \right) \]Using \( k_2 = 0.06815 \text{ min}^{-1} \):\[ t = \frac{1}{0.06815} \ln \left( \frac{1}{0.25} \right) \]\[ t = \frac{1}{0.06815} \cdot 1.386 \approx 20.4 \text{ min} \]

Step 6: Choose the correct answer

The calculated time for 75% decomposition at \( 450^{\circ}C \) is approximately 20.4 minutes. Therefore, the correct answer is option (a) 20.4 minutes.

Key Concepts

Arrhenius Equation

The Arrhenius equation is a critical formula in chemistry that describes how temperature affects the rate constant of a reaction. It can be represented as: \[ k = A \exp \left( -\frac{E_a}{RT} \right) \] where

• \( k \) is the rate constant,
• \( A \) is the frequency factor, which relates to the frequency of collisions and their proper orientation,
• \( E_a \) is the activation energy, which is the minimum energy required to initiate the reaction,
• \( R \) is the universal gas constant \( 8.314 \text{ J/(mol K)} \),
• \( T \) is the temperature in Kelvin.

This equation highlights two primary relationships:

• The higher the temperature, the higher the rate constant, making reactions faster.
• A higher activation energy results in a lower rate constant at a given temperature, making the reaction slower.

In practice, using the Arrhenius equation allows us to predict how changing temperature will influence reaction rates by calculating the new rate constant. This is especially useful in chemical kinetics when studying reactions at varying temperatures.

Rate Constant Calculation

For a first-order reaction, the rate constant \( k \) is a crucial parameter that defines the speed of the reaction. It relates to how quickly a reactant decomposes over time. The formula for calculating \( k \) in first-order reactions when the half-life is known is:\[ k = \frac{0.693}{t_{1/2}} \] where \( t_{1/2} \) is the half-life of the reaction, the time taken for half of the initial amount of a reactant to decompose.To calculate the rate constant at different temperatures:- Determine the half-life period for the reaction at each temperature.- Substitute the half-life into the formula above to find \( k \).This is beautifully demonstrated in the example problem where we calculate the rate constant at two distinct temperatures \( 380^{\circ}C \) and \( 450^{\circ}C \), resulting in different \( k \) values, showing the effect of temperature on reaction rates.

Half-Life of Decomposition

The half-life of a reaction is a very important concept in first-order kinetics because it provides insight into how long a substance will take to reach a certain level of decomposition. In the context of first-order reactions, the half-life remains constant regardless of the initial concentration of the reactant.In our exercise, the half-life of hydrogen peroxide's decomposition was given for two temperatures, \( 380^{\circ}C \) and \( 450^{\circ}C \). This allowed us to calculate the rate constants at these temperatures using the relation: \[ t_{1/2} = \frac{0.693}{k} \] Understanding half-life helps in predicting how long it will take for a certain fraction of a reactant to remain. When calculating the time required for 75% decomposition, knowing the rate constant derived from the half-life enables precise computation of the time needed for such decomposition, as shown in the problem solution.

Activation Energy Calculation

Activation energy \( E_a \) is important because it represents the energy barrier that must be overcome for a reaction to proceed. It can be expressed using the Arrhenius equation rearranged in logarithmic form:\[ \ln(k) = \ln(A) - \frac{E_a}{R} \left( \frac{1}{T} \right) \] The value of \( E_a \) is provided in the exercise at \( 200 \text{ kJ/mol} \), which plays a crucial role in understanding the temperature dependence of the reaction rate.To calculate the activation energy, one can use data from experiments at different temperatures, substituting into the Arrhenius equation to find \( E_a \). This provides a means of predicting how the reaction rate will change if the temperature changes. The exercise illustrates how different temperatures (like \( 380^{\circ}C \) and \( 450^{\circ}C \)) lead to different outcomes based on the calculated rate constants, with the given \( E_a \) linking these outcomes to temperature changes.