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

Short answer

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

Step-by-step solution

Understanding the Problem

We need to calculate the time required for 75% decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(450^{\circ} \mathrm{C}\), knowing that the reaction is first-order and we have the half-life at this temperature.

Calculate the Rate Constant at 450°C

For a first-order reaction, the relationship between the half-life \(t_{1/2}\) and the rate constant \(k\) is given by:\[ t_{1/2} = \frac{0.693}{k} \]Given that \(t_{1/2} = 10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\), we rearrange to find \(k\):\[ k = \frac{0.693}{10.17} = 0.0682 \mathrm{~min}^{-1} \]

Use First-Order Reaction Equation

The time \(t\) required for a certain fraction \(x\) of the substance to decompose for a first-order reaction is given by:\[ t = \frac{1}{k} \ln\left( \frac{1}{1-x} \right) \]For 75% decomposition, \(x = 0.75\).Plug in the values:\[ t = \frac{1}{0.0682} \ln\left( \frac{1}{0.25} \right) \]

Solve for Time to 75% Completion

Calculate \( t \):\[ t = \frac{1}{0.0682} \cdot \ln(4) = \frac{1}{0.0682} \cdot 1.3863 \approx 20.4 \mathrm{~min} \]

Determine the Correct Answer Choice

The time calculated for 75% decomposition at \(450^{\circ} \mathrm{C}\) is \(20.4 \mathrm{~min}\), which corresponds to the answer choice (a).

Key concepts

Half-Life Calculation

The concept of half-life is crucial in understanding reaction kinetics, especially for first-order reactions like the decomposition of hydrogen peroxide (\(\mathrm{H}_{2} \mathrm{O}_{2}\)). For a first-order reaction, the half-life (\(t_{1/2}\)) is defined as the time it takes for half of the reactant to decompose. The formula for calculating the half-life of a first-order reaction is:\[t_{1/2} = \frac{0.693}{k}\]where \(k\) is the reaction rate constant.
For instance, if the half-life of hydrogen peroxide at \(450^{\circ} \mathrm{C}\) is given as \(10.17 \mathrm{~min}\), it directly informs us on how quickly the reaction occurs at that temperature.Understanding how to calculate and interpret half-life is essential for predicting the behavior of the reaction over time. This helps in determining the time needed for a specific percentage of the reactant to decompose, as we did for 75% decomposition in the problem.

Decomposition of Hydrogen Peroxide

The decomposition of hydrogen peroxide is a classic example of a first-order chemical reaction. This is significant because it tells us that the rate of reaction depends solely on the concentration of one reactant - here, the hydrogen peroxide itself.
In practice, the decomposition of \(\mathbf{H}_{\mathbf{2}} \mathbf{O}_{\mathbf{2}}\) involves the breakdown into water and oxygen gas:\[2\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2\mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\]For such reactions, understanding the kinetics and half-life allows us to predict how long it will take for the reactant concentration to decrease to a certain point under various conditions.
In the exercise discussed, knowing the half-life at different temperatures helps us adjust our expectations for the completion of the reaction, especially when the temperature deviates, as seen from the given \(360 \mathrm{~min}\) at \(380^{\circ} \mathrm{C}\) and \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\).
This rapid change highlights the sensitivity of the reaction rate to temperature differences, which is an important aspect of chemical kinetics.

Arrhenius Equation

The Arrhenius equation is fundamental in studying how temperature affects the speed of a chemical reaction:\[k = A e^{-\frac{E_{a}}{RT}}\]where:

• \(k\) is the rate constant
• \(A\) is the pre-exponential factor, a constant
• \(E_{a}\) is the energy of activation
• \(R\) is the gas constant
• \(T\) is the temperature in Kelvin

This equation illustrates that as the temperature rises, the exponential term decreases, thus increasing the rate constant \(k\).
This means reactions speed up because more molecules have the necessary energy to react.
In our exercise, knowing the activation energy (\(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\)) and the changes in half-life with temperature helps us see how the reaction accelerates at higher temperatures, such as \(450^{\circ} \mathrm{C}\). This highlights the usefulness of the Arrhenius equation for predicting how external conditions influence chemical reaction rates, vital for experimental and industrial applications.