Question

The Arrhenius parameters for the reaction $$ \mathrm{N}_{2} \mathrm{O}-\mathrm{N}_{2}+\mathrm{O} $$ are \(A=7.94 \times 10^{11} \mathrm{~s}^{-1}\) and \(E_{a}=250 \mathrm{kj} \mathrm{mol}^{-1}\). The reaction is first order. Calculate the rate constant and half-life of nitrous oxide assuming a tropospheric mixing ratio of \(310 \mathrm{ppbv} \mathrm{N}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) and comment on the environmental significance of these results.

Short answer

Under these conditions, \( k \approx 0 \mathrm{~s}^{-1} \) and the half-life is infinite; \( \mathrm{N}_2\mathrm{O} \) is environmentally persistent.

Step-by-step solution

Calculate the Rate Constant

The Arrhenius equation is given by \( k = A e^{-\frac{E_a}{RT}} \). Here, we need to use the given parameters:- \( A = 7.94 \times 10^{11} \mathrm{~s}^{-1} \)- \( E_{a} = 250,000 \text{ J mol}^{-1} \) (Note: Converted to Joules)- \( R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \) (universal gas constant)- Temperature \( T = 20^{\circ} \text{C} \) which is \( 293.15 \text{ K} \) when converted to Kelvin.Substitute these values into the equation:\[ k = 7.94 \times 10^{11} e^{-\frac{250,000}{(8.314)(293.15)}} \]Calculate the exponent:\[ \frac{250,000}{(8.314)(293.15)} = \frac{250,000}{2437.7221} \approx 102.58 \]Then calculate \( e^{-102.58} \), which is a very small number, essentially zero for practical purposes. Thus, the rate constant \( k \approx 0 \mathrm{~s}^{-1} \) under these conditions.

Determine the Half-life

The half-life \( t_{1/2} \) for a first order reaction is calculated using:\[ t_{1/2} = \frac{0.693}{k} \]With \( k \approx 0 \mathrm{~s}^{-1} \), the half-life tends to be infinitely large, indicating that the reaction proceeds extremely slowly or practically not at all under standard tropospheric conditions.

Environmental Significance

Under standard tropospheric conditions at \( 20^{\circ} \text{C} \), the extremely low rate constant suggests that the decomposition of \( \mathrm{N}_2\mathrm{O} \) into \( \mathrm{N}_2 + \mathrm{O} \) is negligible. \( \mathrm{N}_2\mathrm{O} \) persists in the atmosphere, which contributes significantly to its role as a greenhouse gas and ozone-depleting substance.

Key concepts

Rate Constant Calculation

The rate constant, symbolized as \( k \), is an important factor that determines how fast a reaction occurs. To calculate the rate constant for a reaction, we often use the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}} \] Where:

• \( A \) is the pre-exponential factor or frequency factor, which suggests the number of collisions that result in a reaction per unit time.
• \( E_a \) stands for the activation energy, in joules per mole, required to start the reaction.
• \( R \) is the universal gas constant, \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \).
• \( T \) is the temperature in Kelvin.

By substituting known values from the problem into the equation, we derive that the reaction's rate constant is almost zero under typical conditions. This demonstrates the reaction's extremely slow rate in the natural environment, leading to larger implications for chemical persistence.

First Order Reaction

In chemistry, a first-order reaction is a type of reaction where the reaction rate is directly proportional to the concentration of a single reactant. This means that if the concentration decreases, the rate of reaction decreases by the same proportion, and vice versa. The mathematical expression for a first-order reaction is:

• The rate \( r = k[A] \)

Where \([A]\) is the concentration of the reactant. In first-order reactions, the half-life, the time required to reduce the concentration of a substance to half of its initial value, does not depend on its initial concentration. This feature makes it unique and independent of how much reactant is present initially. The constant half-life is one reason why these reactions are significant in environmental chemistry, despite the slow decomposition of harmful substances like nitrous oxide.

Environmental Chemistry

Environmental chemistry is the study of how chemicals interact with the environment. It plays a vital role in understanding the impacts of human activities and natural processes on the surroundings. When examining chemical reactions in this field, such as the decomposition of nitrous oxide, factors like reaction rates and persistence in the atmosphere become critical. Nitrous oxide challenges with its slow decomposition rate due to a nearly zero rate constant at typical tropospheric conditions. This persistence means nitrous oxide can travel long distances, contributing significantly to greenhouse gas effects and ozone layer depletion over time. Environmental chemistry seeks to mitigate such effects, often influencing policies and regulations to reduce emissions and enhance atmospheric health.

Nitrous Oxide Decomposition

Nitrous oxide (\( \mathrm{N}_2\mathrm{O} \)) is a potent greenhouse gas, often produced by agricultural activities and waste management. Its decomposition in the atmosphere breaks it down into nitrogen (\( \mathrm{N}_2 \)) and oxygen (\( \mathrm{O} \)), processes which occur naturally but at extremely slow rates under normal conditions. The decomposition of nitrous oxide is crucial because it limits the amount that can have harmful effects on the environment. The negligible rate constant calculated in the exercise indicates that nitrous oxide does not decompose rapidly, allowing it to persist in the atmosphere. This persistence is worrying as it contributes to global warming and ozone layer depletion, calling for careful monitoring and reduction efforts in emissions. Understanding and addressing these issues are key in environmental chemistry and crucial for global ecological health.