Question

Nitrogen oxide is formed at night by dissociation of dinitrogen pentoxide. For the first order reaction $$ \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow \mathrm{NO}_{2}+\mathrm{NO}_{3} $$ the rate constant is \(3.14 \times 10^{-2} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\), and \(1.42 \times 10^{-3} \mathrm{~s}^{-1}\) at \(55^{\circ} \mathrm{C}\). Calculate the half life of this molecule at the two temperatures. For a concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) of \(3.6 \mathrm{ppbv}\), calculate the length of time it could take for the concentration to be reduced to \(1.0 \mathrm{ppbv}\) at a constant temperature of \(25^{\circ} \mathrm{C}\). Calculate the Arrhenius parameters, \(A\) and \(E_{2}\). for the reaction.

Short answer

Half-lives: 22.1 s at 25°C, 487.3 s at 55°C. Time to reduce concentration: 37.04 s. Arrhenius parameters: \( A = 1.02 \times 10^{13} \ \text{s}^{-1} \), \( E_a = 98234 \ \text{J/mol} \).

Step-by-step solution

Calculate Half-Life at 25°C

For a first-order reaction, the half-life is given by the formula \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant. At 25°C, \( k = 3.14 \times 10^{-2} \ \text{s}^{-1} \). Substitute the value into the formula: \[ t_{1/2} = \frac{0.693}{3.14 \times 10^{-2}} = 22.1 \ \text{s}. \]

Calculate Half-Life at 55°C

Use the same formula for half-life as in Step 1. At 55°C, the rate constant \( k = 1.42 \times 10^{-3} \ \text{s}^{-1} \). Substitute: \[ t_{1/2} = \frac{0.693}{1.42 \times 10^{-3}} = 487.3 \ \text{s}. \]

Calculate Time to Reduce Concentration from 3.6 ppbv to 1.0 ppbv

For a first-order reaction, the equation \( ln \left( \frac{[A]_0}{[A]} \right) = kt \) applies, where \([A]_0 = 3.6\,\text{ppbv}\) and \([A] = 1.0\,\text{ppbv}\). At 25°C, \( k = 3.14 \times 10^{-2} \ \text{s}^{-1} \). \[ ln \left( \frac{3.6}{1.0} \right) = (3.14 \times 10^{-2}) t \] Calculating gives: \[ t = \frac{ln(3.6) - ln(1.0)}{3.14 \times 10^{-2}} \approx 37.04 \ \text{s}. \]

Calculate Arrhenius Parameters

The Arrhenius equation is \( k = A e^{-E_a/RT} \). From the temperature and rate constant data at 25°C and 55°C:\[ \ln(k_2 / k_1) = (-E_a/R) \left(1/T_2 - 1/T_1\right) \] Where:- \( k_1 = 3.14 \times 10^{-2} \ \text{s}^{-1} \) at \( T_1 = 298\,\text{K} \),- \( k_2 = 1.42 \times 10^{-3} \ \text{s}^{-1} \) at \( T_2 = 328\,\text{K} \),- \( R = 8.314 \ \text{J/mol}\cdot\text{K}. \)Plug values into the formula:\[ \ln \left(\frac{1.42 \times 10^{-3}}{3.14 \times 10^{-2}}\right) = \left(-\frac{E_a}{8.314}\right) \left( \frac{1}{328} - \frac{1}{298} \right) \] Solving the equation provides \( E_a \approx 98234 \, \text{J/mol}. \)To find \( A \), use the Arrhenius equation with either pair of \(T, k\):\[ A = \frac{k}{e^{-E_a/RT}} \] Substitute for 25°C:\[ A = \frac{3.14 \times 10^{-2}}{e^{-98234/(8.314 \times 298)}} \] This gives \( A \approx 1.02 \times 10^{13} \ \text{s}^{-1}. \)

Key concepts

Arrhenius Equation

The Arrhenius Equation is a pivotal formula in chemical kinetics. It links the rate constant, often denoted as \(k\), of a reaction with the temperature at which the reaction occurs. Mathematically, it is expressed as \(k = A e^{-E_a/RT}\). Here, \(A\) is the pre-exponential factor or frequency factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
Understanding the Arrhenius Equation involves recognizing how temperature and energy barriers affect chemical reactions. The pre-exponential factor \(A\) is a constant specific to each reaction, reflecting how often molecules collide with the correct orientation. Meanwhile, activation energy \(E_a\) is the minimum energy the reactants need to transform into products.
The equation indicates that higher temperatures generally increase the rate constant \(k\). This means reactions tend to occur faster as temperature rises because more molecules have sufficient energy to overcome the activation energy barrier. By comparing rate constants at different temperatures, you can compute \(E_a\) using the natural logarithm form: \(\ln(k_2/k_1) = (-E_a/R)(1/T_2 - 1/T_1)\). This form helps determine how sensitive a reaction rate is to temperature changes.

First-order Reactions

First-order reactions are characterized by their dependency on the concentration of one reactant to determine the reaction rate. The rate law for a first-order reaction is expressed as \(rate = k[A]\), where \(k\) is the rate constant and \([A]\) represents the concentration of the reactant. This implies that the reaction rate is directly proportional to the concentration of the single reactant.
In a first-order reaction, a constant half-life is observed no matter the initial concentration. The half-life \(t_{1/2}\) can be determined using the formula \(t_{1/2} = \frac{0.693}{k}\), where \(k\) is the specific rate constant. This characteristic makes first-order reactions unique, as the time required for half of the reactant to react is constant and independent of its original amount.
An example of a first-order reaction found in the exercise is the decomposition of dinitrogen pentoxide \(N_2O_5\) into nitrogen dioxide \(NO_2\) and nitrate \(NO_3\). The process follows the first-order kinetics, where, at a given temperature, the rate of decomposition is dependent only on the concentration of \(N_2O_5\).

Reaction Rate Constants

Reaction Rate Constants, often abbreviated as \(k\), are pivotal values in chemical kinetics that indicate the speed of a chemical reaction. These constants are derived from experimental data and are influenced by various factors like temperature, pressure, and the presence of catalysts.
In the context of chemical reactions, the reaction rate constant serves as a proportionality factor in rate laws. For a first-order reaction such as \(N_2O_5\to NO_2+NO_3\), it helps calculate how quickly \(N_2O_5\) decomposes over time. The value of \(k\) increases with temperature, reflecting the reaction's temperature dependency as per the Arrhenius equation.
Understanding \(k\) is crucial when computing the half-life of reactions and predicting the time required to reach a certain conversion of the reactant. Changes in temperature can alter \(k\), which, in turn, impacts the timing of the half-life and the overall speed of the reaction. The mathematics of reaction rate constants not only provides insights into reaction dynamics but also helps in fine-tuning conditions for industrial processes and experimental setups.