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Chemical kinetics is the study of reaction rates and the steps involved in chemical processes. When we discuss a first-order reaction like the decomposition of dinitrogen pentoxide (\(\mathrm{N}_{2} \mathrm{O}_{5}\)), we refer to a process where the rate depends on the concentration of a single reactant. In this context, understanding the kinetics helps us predict how fast a reaction proceeds and how the concentration of reactants changes over time.

For a first-order reaction, the rate equation follows the expression \(A = A_0 \mathrm{e}^{-kt}\), indicating that the concentration of the reactant decreases exponentially over time due to its dependency on a constant rate (\(k\), also known as the rate constant) and time elapsed (\(t\)). This exponential decay is a hallmark of first-order kinetics.

This insight into chemical kinetics allows chemists to predict the behavior of chemical systems and design conditions to control reaction speeds. It's crucial in various fields such as pharmaceuticals and environmental science.

Reaction rate equations describe how the concentration of reactants and products change over time. In the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), the rate equation helps determine how much of the compound has decomposed after a certain period. By using the formula \(A = A_0 \mathrm{e}^{-kt}\), where \(A_0\) is the initial concentration, we estimate the concentration of reactants at any given time.

For the exercise in question, we start with an initial concentration of 0.600 M \(\mathrm{N}_{2} \mathrm{O}_{5}\). After 20 hours, or 72000 seconds, using the given rate constant of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\), we find the remaining concentration. The difference between the initial and remaining amounts of \(\mathrm{N}_{2} \mathrm{O}_{5}\) provides the extent of decomposition.

These calculations not only underline the insight we can gain from reaction rate equations but also demonstrate their utility in planning and executing reactions safely and efficiently.

The ideal gas law connects pressure, volume, temperature, and moles of gas in a system, formulated as \(PV = nRT\). This fundamental equation allows us to calculate the pressure of gases produced or consumed in a chemical reaction at any given temperature and volume.

In this exercise, after calculating the moles of oxygen \((\mathrm{O}_2)\) produced from \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposition, we use the ideal gas law to find the oxygen's partial pressure in a 10.0- liter container. Here, \(P\) is pressure, \(V\) is the container volume, \(n\) is the number of moles of \(\mathrm{O}_2\), \(R\) is the gas constant (0.08206 L atm/mol K), and \(T\) is the temperature in Kelvin (45 + 273.15 = 318.15 K).

By substituting these values, we apply the ideal gas law to find the pressure contributed by the produced oxygen. This approach shows the practicality of the ideal gas law in solving real-life chemical problems, especially when considering gaseous reactions.