Question

The reaction for the decomposition of dinitrogen monoxide gas to form an oxygen radical is: \(\mathrm{N}_{2} \mathrm{O}\) \((\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g})\). If the activation energy is 250 \(\mathrm{kJ} / \mathrm{mol}\) and the frequency factor is \(8.0 \times 10^{11} \mathrm{~s}^{-1}\), what is the rate constant for the first order reaction at \(1000 \mathrm{~K} ?\) a. \(7.0 \times 10^{-2} \mathrm{~s}^{-1}\) b. \(3.7 \times 10^{-2} \mathrm{~s}^{-1}\) c. \(0.71 \times 10^{-3} \mathrm{~s}^{-1}\) d. \(9.7 \times 10^{-6} \mathrm{~s}^{-1}\)

Short answer

The rate constant is approximately \(7.0 \times 10^{-2} \text{ s}^{-1}\), matching option (a).

Step-by-step solution

Identify the Arrhenius Equation

To find the rate constant for a chemical reaction, we use the Arrhenius equation:\[ k = A \times e^{-\frac{E_a}{RT}} \]where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant (8.314 \ \( ext{J}\text{{mol}}^{-1}\text{{K}}^{-1} \)), and \( T \) is the temperature in Kelvin.

Convert Activation Energy to Compatible Units

Given that \( E_a = 250 \, \text{kJ/mol} \), convert it to Joules per mole because the gas constant \( R \) is expressed in Joules.\[ E_a = 250 \, \text{kJ/mol} = 250,000 \, \text{J/mol} \]

Substitute Known Values into the Arrhenius Equation

Substitute the known values into the Arrhenius equation:- \( A = 8.0 \times 10^{11} \, \text{s}^{-1} \)- \( E_a = 250,000 \, \text{J/mol} \)- \( R = 8.314 \, \text{J/mol K} \)- \( T = 1000 \, \text{K} \)Plug them into the equation:\[ k = (8.0 \times 10^{11} \, \text{s}^{-1}) \times e^{-\frac{250,000 \, \text{J/mol}}{8.314 \, \text{J/mol} \cdot 1000 \, \text{K}}} \]

Calculate the Exponential Term

Compute the exponent:\[ \frac{250,000}{8.314 \times 1000} \approx 30.06 \]Now calculate the exponential:\[ e^{-30.06} \approx 9.65 \times 10^{-14} \]

Calculate the Rate Constant \( k \)

With the exponential value calculated, determine \( k \):\[ k = 8.0 \times 10^{11} \times 9.65 \times 10^{-14} \approx 7.72 \times 10^{-2} \, \text{s}^{-1} \]

Match with Provided Options

The calculated \( k \) value \( 7.72 \times 10^{-2} \, \text{s}^{-1} \) is closest to option (a) \( 7.0 \times 10^{-2} \, \text{s}^{-1} \). Therefore, the correct answer is option (a).

Key concepts

Activation Energy

Activation energy is the minimum energy needed for a reaction to occur. It acts as an energy barrier. Breaking or forming simple bonds needs input energy and this starts reactions. Measured in kJ/mol, it shows how picky reactions are: higher values mean energy-hungry reactions.
Imagine rolling a ball up a hill. You need enough push to get it over the top; that push represents activation energy. Lower hills require less energy! In chemical terms, higher activation energy means fewer molecules have enough energy to reach the top of that hill.
In the case of the decomposition of dinitrogen monoxide, the activation energy is 250 kJ/mol. This energy is what molecules need to shake up, break apart, and form nitrogen gas and an oxygen radical.

Rate Constant

The rate constant, denoted by \( k \), is a crucial part of the dazzling Arrhenius equation. It quantifies the speed of a reaction under specific conditions. Its value changes with temperature and the presence of a catalyst.
Every reaction has its own unique rate constant. The higher the rate constant, the faster the reaction proceeds. Think of \( k \) as the speed limit of a route, dictating how swiftly a particular reaction flows.
For the first order decomposition of dinitrogen monoxide at 1000 K, after doing some math with the Arrhenius equation, our calculated \( k \) is about \( 7.72 \times 10^{-2} \, \text{s}^{-1} \). This fits option (a) in the provided answers.
By understanding \( k \), you unlock the potential to predict and control reaction speeds, which is particularly useful in both laboratory and industrial settings.

Frequency Factor

The frequency factor, symbolized as \( A \), is a component of the Arrhenius equation and addresses how often particles bump into each other with the right orientation for a reaction.
It's an indicator of the likelihood of reactant collisions leading to products when conditions are right and energy is not limiting.The value of \( A \) is often acquired experimentally, so it remains constant in calculations.
In simple terms, think of \( A \) as the number of times a runner sprints towards the finish line. More sprints i.e., interactions mean a better chance of winning, even without considering hurdles (activation energy).In our example, \( A \) is given as \( 8.0 \times 10^{11} \, \text{s}^{-1} \), indicating a high probability of molecule collisions leading to a successful reaction.

First Order Reaction

A first order reaction is one where the rate depends on the concentration of a single reactant raised to the first power.This means that if you double the concentration of the reactant, the reaction rate also doubles.
These reactions have a constant half-life that is independent of the reactant concentration. This makes mathematical analysis simpler and more predictable.
In the decomposition of dinitrogen monoxide gas, because it follows first order kinetics, the rate constant \( k \) can be directly used to determine how fast the reaction proceeds at a specific temperature.
Armed with this information, mapping out the kinetics of such reactions lets chemists explain reaction mechanisms, design better catalysts, and fine-tune conditions for optimal yields.