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

Short answer

The time required for 75% decomposition at 450°C is 20.4 min.

Step-by-step solution

Understanding the First Order Kinetics

For first order reactions, the relation between half-life \( t_{1/2} \) and rate constant \( k \) is given by the equation: \( t_{1/2} = \frac{0.693}{k} \). We will use this relationship to find \( k \) at the different temperatures given.

Calculating Rate Constant at 450°C

At \(450^{\circ} \mathrm{C}\), the half-life is 10.17 min. Using the formula: \[ k_{450} = \frac{0.693}{10.17} \approx 0.068 \mathrm{~min}^{-1} \]

Using Integrated Rate Law for First Order Reactions

The integrated rate law for a first order reaction is \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \). For 75% decomposition, \([A] = 0.25[A]_0\), and \( \frac{[A]_0}{[A]} = 4 \). Thus: \[ \ln(4) = k_{450} \cdot t \] \[ 1.386 = 0.068 \cdot t \]

Solving for Time at 450°C

Solving the equation from Step 3 gives: \[ t = \frac{1.386}{0.068} \approx 20.4 \mathrm{~min} \]

Key concepts

Half-life Calculation

To understand the concept of half-life, think of it as the time required for half of a given amount of a substance to decompose or change into another state. For reactions following first order kinetics, which is common in chemical reactions, the relationship between half-life ( \( t_{1/2} \)), and the rate constant ( \( k \)) is crucial. This relationship is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
- Here, \( t_{1/2} \) represents the half-life of the substance.- The constant 0.693 derives from the natural logarithm of 2, reflecting the exponential decay in first order reactions.- \( k \) is the reaction rate constant that varies with temperature.
For example, in the decomposition of \( \text{H}_2 \text{O}_2 \), finding \( k \) helps predict how quickly the reaction progresses at a particular temperature. Knowing half-life helps in calculating the necessary time for reactions to reach specific completion stages, like 75% decomposition in this exercise.

Decomposition Reaction

Decomposition reactions involve breaking down a compound into simpler substances. Specifically, the decomposition of hydrogen peroxide \( (\text{H}_2 \text{O}_2) \) breaks down into water (\( \text{H}_2\text{O} \)) and oxygen (\( \text{O}_2 \)). This type of reaction often follows first order kinetics, where the rate of decomposition is directly proportional to the concentration of the reactant.
In first order decomposition:

• The higher the concentration of the substance, the faster the rate of its decomposition.
• The rate constant, \( k \), is essential as it determines the speed of the reaction at a given temperature.

Moreover, changes in temperature can significantly impact both the rate constant and the overall rate of decomposition. Therefore, accurately calculating the rate constant at different temperatures, as shown in the exercise, allows for precise prediction of how fast a decomposition reaction will occur. This understanding is vital for practical applications, such as safely handling and storing reactive substances like \( \text{H}_2 \text{O}_2 \).

Activation Energy

Activation energy is the minimum energy required to initiate a chemical reaction. For first order reactions, such as the decomposition of \( \text{H}_2 \text{O}_2 \), knowing the activation energy helps in understanding how easy it is to trigger the reaction.
The relation between activation energy and rate constant is typically explored using the Arrhenius Equation:
\[ k = A \times e^{-\frac{E_a}{RT}} \]

• \( E_a \) represents the activation energy in \( \text{kJ mol}^{-1} \).
• \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
• \( A \) is the pre-exponential factor, which indicates the frequency of collisions with correct orientation.

By calculating the activation energy, as given in the problem (\( 200 \text{ kJ mol}^{-1} \)), we gain insight into the sensitivity of the reaction rate to temperature changes. High activation energy suggests the reaction is more temperature-dependent, hence slower at lower temperatures. Conversely, lower activation energy often signifies a reaction can proceed swiftly even with minor temperature variations, highlighting why the decomposition might be more vigorous at elevated temperatures like 450°C. In real-world scenarios, understanding activation energy is fundamental to controlling reaction conditions and ensuring safety in chemical processes.