Problem 84 - Chemistry and Chemical Reactivity

Question

Chlorine trifluoride, \(\mathrm{CIF}_{3}\), is a valuable reagent because it can be used to convert metal oxides to metal fluorides: \(6 \mathrm{NiO}(\mathrm{s})+4 \mathrm{ClF}_{3}(\mathrm{g}) \rightarrow 6 \mathrm{NiF}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})\) (a) What mass of NiO will react with CIF \(_{3}\) gas if the gas has a pressure of \(250 \mathrm{mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) in a \(2.5-\mathrm{L}\) flask? (b) If the CIF a described in part (a) is completely consumed, what are the partial pressures of \(\mathrm{Cl}_{2}\) and of \(\mathrm{O}_{2}\) in the 2.5 -I. flask at \(20^{\circ} \mathrm{C}\) (in \(\mathrm{mm}\) Hg)? What is the total pressure in the flask?

Step-by-Step Solution

Answer

(a) \(3.81\) g of \(\mathrm{NiO}\) will react. (b) Partial pressures: \(\mathrm{Cl}_2 = 124\) mm Hg and \(\mathrm{O}_2 = 191\) mm Hg. Total pressure: \(315\) mm Hg.

Step 1: Convert Given Pressure to SI Units

Given the initial pressure of \( \mathrm{ClF}_3 \) is 250 mm Hg, convert this pressure to atmospheres (atm) using the conversion factor 1 atm = 760 mm Hg. Thus, \( P = \frac{250}{760} \approx 0.329 \text{ atm} \).

Step 2: Apply the Ideal Gas Law

Using the ideal gas law \( PV = nRT \), solve for the number of moles \( n \). Here, \( R = 0.0821 \text{ L atm/K mol} \), the temperature \( T = 20^{\circ} \text{C} = 293 \text{ K} \), and the volume \( V = 2.5 \text{ L} \). Calculate \( n = \frac{PV}{RT} = \frac{0.329 \times 2.5}{0.0821 \times 293} \approx 0.034 \text{ moles of } \mathrm{ClF}_3 \).

Step 3: Calculate Mass of NiO Required

From the balanced equation, 4 moles of \( \mathrm{ClF}_3 \) react with 6 moles of \( \mathrm{NiO} \). Therefore, 0.034 moles of \( \mathrm{ClF}_3 \) will react with \( \left(\frac{6}{4}\right) \times 0.034 = 0.051 \text{ moles of } \mathrm{NiO} \). The molar mass of \( \mathrm{NiO} \) is \( 74.69 \text{ g/mol} \), hence, mass = \( 0.051 \times 74.69 \approx 3.81 \text{ grams} \).

Step 4: Determine Moles and Partial Pressures of Products

After the reaction, from the balanced equation:- 2 moles of \( \mathrm{Cl}_2 \) and 3 moles of \( \mathrm{O}_2 \) are produced from 4 moles of \( \mathrm{ClF}_3 \).- Calculate moles of produced gases from 0.034 moles of \( \mathrm{ClF}_3 \): - \( \mathrm{Cl}_2 \) moles = \( \left(\frac{2}{4}\right) \times 0.034 = 0.017 \text{ moles} \) - \( \mathrm{O}_2 \) moles = \( \left(\frac{3}{4}\right) \times 0.034 = 0.026 \text{ moles} \).Use the ideal gas law to find each partial pressure:- \( P_{\mathrm{Cl}_2} = \frac{0.017 \times 0.0821 \times 293}{2.5} \approx 0.163 \text{ atm} \)- \( P_{\mathrm{O}_2} = \frac{0.026 \times 0.0821 \times 293}{2.5} \approx 0.251 \text{ atm} \).Convert to mm Hg:- \( P_{\mathrm{Cl}_2} = 0.163 \times 760 \approx 124 \text{ mm Hg} \)- \( P_{\mathrm{O}_2} = 0.251 \times 760 \approx 191 \text{ mm Hg} \).

Step 5: Calculate Total Pressure

The total pressure in the flask is the sum of partial pressures of \( \mathrm{Cl}_2 \) and \( \mathrm{O}_2 \):\[ P_{\text{total}} = P_{\mathrm{Cl}_2} + P_{\mathrm{O}_2} = 124 + 191 = 315 \text{ mm Hg} \].

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and moles of a gas. It is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature in Kelvin.

The ideal gas constant \( R \) is generally accepted as \( 0.0821 \, \text{L atm/K mol} \). To use the ideal gas law effectively, it's crucial to ensure that the units for pressure, volume, and temperature are consistent. For example, if pressure is measured in atmospheres (atm), the volume should be in liters (L), and temperature in Kelvin (K).

When using the ideal gas law in solving chemical problems, you often need to convert temperatures to Kelvin and pressures to atmospheres, as shown in this stoichiometry problem where it is used to find the number of moles of \( \mathrm{ClF}_3 \) given its pressure and volume.

Partial Pressure

Partial pressure is a concept used to describe the pressure exerted by a single gas in a mixture of gases. It is particularly useful when dealing with gaseous reactions and products, such as chlorine gas (\( \mathrm{Cl}_2 \)) and oxygen gas (\( \mathrm{O}_2 \)), which are the products in the reaction involving \( \mathrm{ClF}_3 \).

The partial pressure of a gas is proportional to its mole fraction in the mixture. In calculations, once you determine the moles of the gases produced, you can use the ideal gas law to find their partial pressures. For instance, in the example given, the partial pressures of \( \mathrm{Cl}_2 \) and \( \mathrm{O}_2 \) are calculated after finding their respective moles in the reaction right after using the ideal gas law.

To convert the atmospheric partial pressures to mm Hg, multiply by 760 because 1 atm equals 760 mm Hg. This conversion is vital when the initial conditions are given in mm Hg, as was necessary in this stoichiometry problem.

Chemical Reactions

Chemical reactions are processes in which reactants are transformed into products. In the given example, \( 6 \mathrm{NiO}(\mathrm{s}) + 4 \mathrm{ClF}_{3}(\mathrm{g}) \rightarrow 6 \mathrm{NiF}_{2}(\mathrm{s}) + 2 \mathrm{Cl}_{2}(\mathrm{g}) + 3 \mathrm{O}_{2}(\mathrm{g}) \), we see a complex balanced chemical reaction.

Balancing chemical reactions is critical to ensuring the law of conservation of mass is followed - that is, the atoms entering and leaving a reaction must be the same.

In the reaction, the stoichiometric coefficients provide a ratio of how the moles of reactants relate to the moles of products. This ratio is crucial for stoichiometry calculations, such as determining how many moles of \( \mathrm{NiO} \) react with a given amount of \( \mathrm{ClF}_3 \). Understanding these relationships can help solve problems such as finding the mass of reactants or volumes of gaseous products.

Mole Concept

The mole concept serves as a bridge between the atomic world and the macroscopic world. It allows chemists to count entities like atoms and molecules by weighing out a "handy" amount of material. One mole of a substance contains \( 6.022 \times 10^{23} \) entities, known as Avogadro's number.

In chemical reactions, moles are used to express the number of reactants and products, which can then be converted to grams or other measurable quantities using molar mass.

In this stoichiometry problem, the mole concept helps determine how many grams of nickel oxide (\( \mathrm{NiO} \)) are needed to react with a specified number of moles of \( \mathrm{ClF}_3 \). By using the molar mass of \( \mathrm{NiO} \) and the stoichiometric ratio from the balanced equation, you can convert moles to grams, allowing practical laboratory work.