Question

Dinitrogen pentoxide decomposes in chloroform at a rate of \(2.48 \times 10^{-4} \mathrm{mol} /(\mathrm{L} \cdot \mathrm{min})\) at a particular temperature according to the equation \(2 \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) The reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) . Given an initial concentration 0.400 \(\mathrm{mol} / \mathrm{L}\) , what is the rate constant for the reaction? What is the approximate \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) after the reaction proceeds for 1.30 \(\mathrm{h}\) ?

Short answer

The rate constant for the reaction is \(6.2 \times 10^{-4} \) min⁻¹, and the approximate concentration of N₂O₅ after the reaction proceeds for 1.30 h is 0.397 mol/L.

Step-by-step solution

Write down the first-order rate law

For a first-order reaction, the rate law equation is given by: Rate = k[N₂O₅] where Rate is the rate of decomposition, k is the rate constant, and [N₂O₅] is the concentration of N₂O₅.

Find the rate constant (k)

Given that the Rate of decomposition = \(2.48 \times 10^{-4} \) mol/(L·min) and the initial concentration of N₂O₅= 0.400 mol/L, we can plug in these values into the rate law equation to find the rate constant k: \(2.48 \times 10^{-4} \) = k(0.400) k = \( \frac{2.48 \times 10^{-4}}{0.400}\) k = \(6.2 \times 10^{-4} \) min⁻¹ (rate constant)

Use the integrated rate law equation to calculate the concentration of N₂O₅ after 1.30 h

For a first-order reaction, the integrated rate law equation is: \(\ln (\frac{[\mathrm{N}_{2} \mathrm{O}_{5}]_t}{[\mathrm{N}_{2} \mathrm{O}_{5}]_0}) = -kt\) Where [N₂O₅]ₜ is the concentration of N₂O₅ at time t, [N₂O₅]₀ is the initial concentration of N₂O₅, k is the rate constant, and t is the time in minutes. Given that the reaction proceeds for 1.30 h, let's convert that to minutes: t = 1.30 h × 60 min/h = 78 min Then, substitute the values of k, [N₂O₅]₀, and t into the equation: \(\ln (\frac{[\mathrm{N}_{2} \mathrm{O}_{5}]_t}{0.400}) = -(6.2 \times 10^{-4})(78)\) Now, calculate [N₂O₅]ₜ: \(\ln (\frac{[\mathrm{N}_{2} \mathrm{O}_{5}]_t}{0.400}) = -0.04836\) \( [\mathrm{N}_{2} \mathrm{O}_{5}]_t = 0.400 \times e^{-0.04836}\) \[ [\mathrm{N}_{2} \mathrm{O}_{5}]_t = 0.397 \ \mathrm{mol}/\mathrm{L} \] (approximate concentration after 1.30 h) So, the rate constant for the reaction is \(6.2 \times 10^{-4} \) min⁻¹, and the approximate concentration of N₂O₅ after the reaction proceeds for 1.30 h is 0.397 mol/L.

Key concepts

Understanding Chemical Kinetics

Chemical kinetics is the branch of physical chemistry that studies the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which the reactions proceed. A central concept in this field is the reaction rate, a measure of how quickly reactants are consumed or products are formed over time. The rate can be influenced by several factors, including reactant concentration, temperature, the presence of catalysts, and the physical state of the reactants.

In the given exercise, we examined the decomposition of dinitrogen pentoxide in chloroform, which is a first-order reaction. This implies that the rate at which the reaction occurs is directly proportional to the concentration of the single reactant, dinitrogen pentoxide. By understanding the reaction order and the rate law, students can predict how changing conditions, like the concentration of reactants, affects the speed of the reaction.

Rate Constant Calculation

The rate constant, represented by the symbol 'k', is an essential component of the rate law in chemical kinetics. It is a unique value for each chemical reaction at a given temperature, reflecting the intrinsic speed of the reaction. The rate constant links the reaction rate to the concentration(s) of reactant(s) in the rate law equation.

As illustrated in the problem, to find the rate constant for a first-order reaction, we take the observed rate of decomposition and divide it by the concentration of the reactant. It's important to emphasize here that the units of the rate constant will vary depending on the order of the reaction. In the case of a first-order reaction, the units are typically reciprocal time, such as s⁻¹ or min⁻¹. For students, grasping the concept of the rate constant and its calculation is vital for understanding other aspects of chemical kinetics like half-life and reaction mechanisms.

Integrated Rate Law for First-Order Reactions

The integrated rate law expresses the relationship between reactant concentrations and time for different reaction orders. For first-order reactions, the integrated rate law has a distinctive form that allows us to calculate the concentration of a reactant at any time given its initial concentration and the rate constant.

Using the natural logarithm (ln), the equation \( \ln (\frac{[\mathrm{N}_{2} \mathrm{O}_{5}]_t}{[\mathrm{N}_{2} \mathrm{O}_{5}]_0}) = -kt \) relates the concentrations at initial time and time t for a first-order reaction. Calculating the concentration of the reactant after a certain time involves solving for \( [\mathrm{N}_{2} \mathrm{O}_{5}]_t \) after rearranging the formula and exponentiating both sides to remove the natural log. This shows a decay in reactant concentration over time, which is exponential for a first-order reaction. Understanding and being able to apply the integrated rate law is crucial for predicting the course of a reaction and designing processes accordingly.