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

(a) 20.4 min

Step 1: Understanding Half-Life and Rate Constant Relationship

For a first-order reaction, the half-life, \( t_{1/2} \), is related to the rate constant \( k \) by the formula \( t_{1/2} = \frac{0.693}{k} \). Using this formula, we can find the rate constant at a given temperature.

Step 2: Calculate Rate Constant at 450°C

Given the half-life at \( 450^{\circ}\text{C} \) is \(10.17 \text{ min}\), use \( k = \frac{0.693}{t_{1/2}} \) to find \( k \). This gives \( k = \frac{0.693}{10.17} = 0.0681 \text{ min}^{-1} \).

Step 3: First-Order Reaction Time for Percentage Decomposition

The time \( t \) required to decompose a certain percentage of a reactant in a first-order reaction is given by \( t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right) \), where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \( t \).

Step 4: Calculation for 75% Decomposition

For \( 75\% \) decomposition, \([A] = 0.25[A]_0\). Thus, \( t = \frac{1}{0.0681} \ln\left(\frac{1}{0.25}\right) = \frac{1}{0.0681} \ln(4) = \frac{1}{0.0681} \times 1.386 = 20.35 \text{ min} \).

Step 5: Final Answer Verification

Compare the calculated time with the given options. The closest option to \(20.35\text{ min}\) is \(20.4\text{ min}\), which is option (a).

Key Concepts

First-Order Reaction

A first-order reaction is a type of chemical reaction where the rate of reaction is directly proportional to the concentration of one reactant. In simpler terms, if you have a reactant A converting into products, the rate at which A is used up only depends on how much A you currently have.
First-order reactions are particularly interesting because they follow a neat logarithmic pattern. When graphed, the natural logarithm of the concentration of the reactant over time gives a straight line. This is very useful for calculating other important parameters like the reaction's half-life or rate constant.
Key Characteristics of First-Order Reactions:

• Rate = rate constant (k) × [A]
• Concentration decreases exponentially over time.
• Has a constant half-life, regardless of the initial concentration.

Half-Life

The half-life of a reaction is the time it takes for half of the reactant to be consumed or transformed into product. It's like cutting what's left over in half each time you check.
For first-order reactions, the half-life is particularly simple to calculate and fascinating to study. It doesn't change as the reaction progresses, meaning it's a constant value for the specific reaction conditions.
For first-order reactions, the half-life (\( t_{1/2} \)) can be determined by the equation:\[t_{1/2} = \frac{0.693}{k},\]where \( k \) is the rate constant. This simplicity is what makes first-order kinetics so straightforward. You plug in the known half-life or rate constant, and you can quickly learn about the time frames involved in your reaction.
In the exercise provided, understanding half-life was key to finding out how fast the decomposition happens, even at different temperatures.

Rate Constant

The rate constant, symbolized as \( k \), is a crucial value in chemical kinetics that indicates how fast a reaction occurs. It's important to remember that the rate constant is specifically affected by temperature.
What's wonderful about the rate constant is that it seamlessly links to other elements such as half-life. For a first-order reaction, once you know the half-life, the rate constant can be easily calculated using the formula:\[k = \frac{0.693}{t_{1/2}}.\]The rate constant is not just a number; it tells you how furious or sluggish a reaction proceeds. Increasing the temperature typically increases the rate constant, indicating a faster reaction.
In our example problem, the rate constant was calculated at \( 450^{\circ} \text{C} \) using the half-life of 10.17 minutes, showing the adaptability of first-order kinetics in determining reaction speeds.

Activation Energy

Activation energy is akin to the energy barrier a reaction must overcome to proceed. Imagine it as the push needed to get a ball rolling uphill before it tumbles down the other side.
This energy is crucial because it influences the rate at which a reaction will occur. The higher the activation energy, the slower the reaction, since more energy is needed to get things started.
The Arrhenius Equation comes into play here, helping to relate activation energy to the temperature and rate constant, allowing predictions on how temperature changes affect reaction rates.\[k = A \exp\left(-\frac{E_a}{RT}\right),\]
where:

• \( k \) is the rate constant,
• \( A \) is the pre-exponential factor,
• \( E_a \) is the activation energy,
• \( R \) is the universal gas constant,
• \( T \) is the temperature in Kelvin.

These concepts provide a comprehensive view of the reaction's energy dynamics. In the provided exercise, knowing the activation energy helps understand how increasing temperature speeds up the decomposition of \( \text{H}_2\text{O}_2 \).