Question

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(3.7 \times 10^{-5} \mathrm{~s}^{-1}\) at \(298 \mathrm{~K}\). (a) What is the half-life (in hours) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(298 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(3.5 \mathrm{~h}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

Short answer

The half-life is approximately 5.19 hours, the concentration of Dinitrogen pentoxide after 3.5 hours will be approximately 0.0494 mol L^-1, and it will take around 126.91 minutes for the concentration to decrease from 0.0567 mol L^-1 to 0.0135 mol L^-1.

Step-by-step solution

Calculating the half-life using the rate constant

The half-life for first-order reaction is calculated using the formula: \(t_{1/2} = \frac{\ln 2}{k}\), where \(k\) is the rate constant. Substitute the given rate constant \(k = 3.7 \times 10^{-5} \text{s}^{-1}\) into the formula to find the half-life in seconds. Then convert the half-life from seconds to hours by dividing by 3600.

Determining concentration after a period of time

For first-order kinetics, the concentration at any time \(t\) can be found using \(\left[ N_2O_5 \right]_t = \left[N_2O_5\right]_0 e^{-kt}\), where \(\left[N_2O_5\right]_0\) is the initial concentration and \(\left[N_2O_5\right]_t\) is the concentration after time \(t\). Substitute \(\left[N_2O_5\right]_0 = 0.0567 \text{mol L}^{-1}\), \(k = 3.7 \times 10^{-5} \text{s}^{-1}\), and \(t = 3.5 \text{h} = 12600 \text{s}\) into this formula to find the concentration after 3.5 hours.

Calculating the time for a decrease in concentration

Using the same first-order equation, with \(k = 3.7 \times 10^{-5} \text{s}^{-1}\), \(\left[N_2O_5\right]_0 = 0.0567 \text{mol L}^{-1}\), and final concentration \(\left[N_2O_5\right] = 0.0135 \text{mol L}^{-1}\), rearrange to solve for time \(t\). After finding the value of \(t\) in seconds, convert it into minutes by dividing by 60.

Key concepts

Chemical Kinetics

Understanding chemical kinetics is crucial for analyzing how reactions occur and predicting the rates at which they proceed. It involves the study of reaction rates and the factors that affect them, including temperature, pressure, concentration, and the presence of catalysts.

For instance, in our textbook example, we're looking at the decomposition of dinitrogen pentoxide \( \mathrm{N}_{2} \mathrm{O}_{5} \), which follows first-order reaction kinetics. This means the rate at which \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposes is directly proportional to its concentration at any given moment. Thus, as \( \mathrm{N}_{2} \mathrm{O}_{5} \) concentration decreases over time, so does the rate of decomposition.

In first-order kinetics, the rate constant \( k \) doesn't change with concentration or time, which allows us to use it to predict reaction progression over time. It's expressed in reciprocal seconds \( s^{-1} \) which indicates how quickly a reaction proceeds toward its completion.

Half-life of a Reaction

The half-life of a reaction is a concept in chemical kinetics that signifies the time required for half the substance to react or decompose. It's represented by \( t_{1/2} \).

For first-order reactions, the half-life is constant and independent of the initial concentration. It can be calculated using the formula \( t_{1/2} = \frac{\ln 2}{k} \), where \( k \) is the rate constant. By using the formula, we can determine how long it takes for half of the \( \mathrm{N}_{2} \mathrm{O}_{5} \) to decompose, which is a pivotal step in understanding the overall reaction kinetics. Moreover, knowing the half-life provides insight into the stability and lifetime of substances, which is particularly important in fields such as pharmacology and environmental science.

In our exercise, the half-life calculation aids us not just in grasping the rate of decomposition but also in predicting concentrations at any given time in the future.

Concentration-Time Relationship

The concentration-time relationship in chemical kinetics allows us to predict the concentration of a reactant or a product at any given time. In a first-order reaction, this relationship is modeled by the equation \( \left[N_2O_5\right]_t = \left[N_2O_5\right]_0 e^{-kt} \), where \( \left[N_2O_5\right]_0 \) is the initial concentration, \( \left[N_2O_5\right]_t \) is the concentration at time \( t \) and \( k \) is the rate constant.

This equation is derived from integrating the rate law for a first-order process and quantifies the exponential decay of a substance over time. By plugging in the values for \( k \) and \( t \) into the equation, we can predict the concentration of \( \mathrm{N}_{2} \mathrm{O}_{5} \) in the future, or determine how long it has been reacting based on its current concentration. These predictions are vital for practical applications, like determining the shelf life of a chemical or formulating a drug dosage regimen based on its decay rate.