Problem 10 - General Chemistry: Principles and Modern Applications

Question

At \(65^{\circ} \mathrm{C}\), the half-life for the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) is 2.38min. $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If \(1.00 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is introduced into an evacuated \(15 \mathrm{L}\) flask at \(65^{\circ} \mathrm{C}\) (a) What is the initial partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) ?\) (b) What is the partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) after \(2.38 \mathrm{min} ?\) (c) What is the total gas pressure, in \(\mathrm{mm} \mathrm{Hg}\), after \(2.38 \mathrm{min} ?\)

Step-by-Step Solution

Answer

The initial pressure of \(N_{2}O_{5}(g)\) is 11.92mmHg, the partial pressure after 2.38min is 5.96mmHg and the total gas pressure after 2.38min is 23.84mmHg.

Step 1: Calculation of initial partial pressure of N2O5

To calculate the initial partial pressure, the Ideal Gas Law is used. The formula for Ideal Gas Law is \(PV=nRT\). Here \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is universal gas constant and \(T\) is the temperature. Given, mass of \(N_{2}O_{5}\) = 1.00 g, Volume = 15L, and temperature = \(65^{\circ}C = 65 + 273 = 338K\). Molecular weight of \(N_{2}O_{5} = 2*14(N) + 5*16(O) = 108 g/mol.\) Thus, number of moles \(n = \frac{Mass}{Molar\, mass} = \frac{1.00 g}{108 g/mol} = 0.00926 mol\). To get the pressure in \(\mathrm{mmHg}\), the value for \(R\) we are going to use is 62.4 \(L.mmHg/K.mol\). The equation then becomes \(P = \frac{nRT}{V} = \frac{0.00926 mol * 62.4 L.mmHg/K.mol * 338K}{15L} = 11.92mmHg\).

Step 2: Calculation of partial pressure of N2O5 after 2.38 min

Given that \(N_{2}O_{5}(g)\) decomposes with a half-life of 2.38min, after one half-life, half of the original \(N_{2}O_{5}(g)\) remains. Therefore, the pressure of the remaining \(N_{2}O_{5}(g)\) would also be halved. Hence, partial pressure of \(N_{2}O_{5}\) after 2.38min = \(11.92mmHg / 2 = 5.96mmHg\).

Step 3: Calculation of total gas pressure after 2.38 min

The reaction given implies that for each molecule of \(N_{2}O_{5}\) that decomposes, two molecules of \(NO_{2}(g)\) and half a molecule of \(O_{2}(g)\) are generated. Therefore, for half of the \(N_{2}O_{5}(g)\) that decomposes, an equal pressure of \(O_{2}(g)\) and twice that pressure of \(NO_{2}(g)\) is produced. Therefore, the total pressure after 2.38 min = remaining \(N_{2}O_{5}\) pressure + \(O_{2}(g)\) pressure + \(NO_{2}(g)\) pressure = 5.96mmHg (from Step 2) + 5.96mmHg + 2*5.96mmHg = 23.84mmHg.

Key Concepts

First-Order Reactions

In the field of chemical kinetics, the understanding of first-order reactions is crucial. A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of a single reactant. This is often represented through the equation:

• Rate = k[A]

Here, *k* represents the rate constant, and *[A]* is the concentration of the reactant. Such reactions are characterized by a constant half-life, which means the time required for half of the reactant to be consumed remains constant throughout the reaction.

In the context of the decomposition of \(\mathrm{N_{2}O_{5}}\), the half-life is given as 2.38 minutes. This means that every 2.38 minutes, half of the original quantity of \(\mathrm{N_{2}O_{5}}\) \(for this particular reaction\) will have decomposed into the products \(\mathrm{NO_{2}}\)and\(\mathrm{O_{2}}\).

• This behavior is a hallmark of first-order processes, where the concentration at any time, \(t\), can be calculated using the formula:
• \(\[\left[ ext{A at time t}\right] = \left[ ext{A at time 0}\right] \cdot e^{-kt}\]\)

Understanding this process allows us to predict the amount of a reactant remaining at any given point in time, making it a powerful tool in both laboratory and industrial chemistry.

Partial Pressure Calculation

Partial pressure is an essential concept in chemistry, particularly when dealing with gases in a mixture. It describes the pressure exerted by a single type of gas within a mixture of gases. Each gas in a mixture behaves as if it is alone in the container, applying a pressure independently of other gases present. Using Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of the individual gases.

For the exercise at hand, the initial partial pressure of\(\mathrm{N_{2}O_{5}}\)was calculated using the ideal gas law. This initial pressure was determined to be **11.92 mmHg**. After 2.38 minutes, which corresponds to one half-life, only half of the original amount of\(\mathrm{N_{2}O_{5}}\)remains. As a result, its partial pressure decreases to\(\mathrm{11.92 mmHg / 2 = 5.96 mmHg}\).

• Calculating partial pressures is pivotal when determining the components of gaseous reactions.
• This knowledge allows chemists to understand how pressure relates to reaction kinetics and the behavior of gases.

Ideal Gas Law

The Ideal Gas Law is a cornerstone in understanding the behavior of gases. It is represented by the equation:

• \(PV = nRT\)

In this relationship, *P* stands for pressure, *V* represents volume, *n* is the number of moles of gas, and *T* is the temperature in Kelvin. The constant *R* is the ideal gas constant, which can vary in units depending on the situation.

In the solution of this exercise, the Ideal Gas Law was used to calculate the initial condition of the reactant gas in the system. By rearranging the equation to solve for pressure \(\(P\)\), the above law provides the initial partial pressure in mmHg:

• \(P = \frac{nRT}{V}\)

Understanding and applying the Ideal Gas Law allows us to bridge the gap between macroscopic physical properties and the microscopic molecular properties of gases.

• With this tool, chemists can predict how a gas will respond to different changes in conditions such as temperature, volume, and quantity.

This makes it invaluable for predicting the effects of changing conditions on gas reactions, contributing to efficient and successful chemical processes.