Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

A reaction mixture was prepared at \(25^{\circ} \mathrm{C}\) by filling a \(1.0\) litre nickel vessel with \(\mathrm{F}_{2}\) gas at \(8 \mathrm{~atm}\) and \(\mathrm{Xe}\) gas at \(1.7\) atm. The reaction mixture was maintained at \(400^{\circ} \mathrm{C}\) for one hour. Then it was cooled to \(25^{\circ} \mathrm{C}\) and the contents of the nickel container were analysed. All the xenon gas had reacted to form a solid Xe-F compound, but some of the fluorine gas had not reacted. The pressure of the \(\mathrm{F}_{2}\) gas is \(4.6 \mathrm{~atm}\). What formula would you propose for the xenon-fluorine compound? (a) \(\mathrm{XeF}_{2}\) (b) \(\mathrm{XeF}_{4}\) (c) \(\mathrm{XeF}_{6}\) (d) none of these

Step-by-Step Solution

Answer

The proposed compound is \( \mathrm{XeF}_6 \).

Step 1: Determine Initial Moles of Each Gas

Use the ideal gas law \( PV = nRT \) to find the initial moles of F_2 and Xe gases. At 25°C and 1.0 L, the temperature is 298 K. Use \( R = 0.0821 \) L·atm/mol·K. Thus, \( n_{F_2} = \frac{8 \times 1.0}{0.0821 \times 298} \) and \( n_{Xe} = \frac{1.7 \times 1.0}{0.0821 \times 298} \).

Step 2: Calculate Moles of Remaining F2 Gas After Reaction

After reacting at 400°C and cooling back to 25°C, the fluorine gas pressure is 4.6 atm. Calculate the moles of F_2 remaining using \( n'_{F_2} = \frac{4.6 \times 1.0}{0.0821 \times 298} \).

Step 3: Determine Moles of F2 Consumed

Find the difference between initial moles of F_2 and moles remaining to determine moles of F_2 consumed: \( n_{F_2, \,consumed} = n_{F_2} - n'_{F_2} \).

Step 4: Establish Mole Ratio of F2 to Xe

Since all the Xe gas reacts, \( n_{Xe} = \text{moles of } Xe \text{ that reacted} \). Use the moles of F_2 consumed to determine the stoichiometry of the reaction by finding the ratio \( \frac{n_{F_2, \,consumed}}{n_{Xe}} \).

Step 5: Determine the Formula of the Xe-F Compound

From the ratio calculated in Step 4, establish the number of fluorine atoms (subscript) that reacted per xenon atom, and propose the chemical formula. If the ratio is roughly 2, 4, or 6, propose \( \mathrm{XeF}_2 \), \( \mathrm{XeF}_4 \), or \( \mathrm{XeF}_6 \) accordingly. Otherwise, pick option (d) none of these.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, expressed as \( PV = nRT \). It relates the pressure \( P \), volume \( V \), temperature \( T \), and amount in moles \( n \) of an ideal gas through the gas constant \( R \). This equation is helpful in predicting and understanding the behavior of gases under different conditions.
In this scenario, we first used the Ideal Gas Law to calculate the initial moles of \( \mathrm{F}_2 \) and \( \mathrm{Xe} \). At room temperature (298 K) and a fixed volume of 1.0 litre, the pressure readings of each gas enabled us to solve for \( n \), the moles present.

• For \( \mathrm{F}_2 \), with a pressure of 8 atm, the calculation was \( n_{\mathrm{F}_2} = \frac{8 \times 1.0}{0.0821 \times 298} \).
• For \( \mathrm{Xe} \), with a pressure of 1.7 atm, the calculation was \( n_{\mathrm{Xe}} = \frac{1.7 \times 1.0}{0.0821 \times 298} \).

This calculation set up our understanding of the initial conditions before the chemical reaction took place.

Xenon Fluorine Compound

Xenon fluorides are known for their intriguing chemical properties. In this problem, the xenon gas reacts entirely to form a xenon-fluoride compound, a solid product. Notably, fluorine is one of the few elements that can bond with xenon to form stable compounds.
Common xenon-fluorine compounds include \( \text{XeF}_2 \), \( \text{XeF}_4 \), and \( \text{XeF}_6 \). Each of these compounds possesses a distinct structure and stoichiometry:

• \( \text{XeF}_2 \) — linear shape, formed with 2 fluorine atoms.
• \( \text{XeF}_4 \) — square planar shape, formed with 4 fluorine atoms.
• \( \text{XeF}_6 \) — octahedral shape, formed with 6 fluorine atoms.

Our task is to determine which compound is formed based on the stoichiometry of the reaction with the consumed \( \mathrm{F}_2 \). Knowing the stoichiometry helps us propose the correct formula and comprehend the nature of bonding and molecule formation in xenon compounds.

Moles Calculation

Calculating the number of moles is pivotal to understanding chemical reactions. Moles indicate the quantity of substances present and are used in conjunction with the Ideal Gas Law. In the exercise, moles calculation helped determine the amount of \( \mathrm{F}_2 \) consumed and how much \( \mathrm{Xe} \) participated in the reaction.
Initially, we calculated the moles of \( \mathrm{F}_2 \) and \( \mathrm{Xe} \) using their respective pressures at 298 K. After the reaction, we needed to find how much \( \mathrm{F}_2 \) was left. The remaining pressure of \( \mathrm{F}_2 \) was 4.6 atm, thus:
\( n'_{\mathrm{F}_2} = \frac{4.6 \times 1.0}{0.0821 \times 298} \)
By comparing initial and remaining moles of \( \mathrm{F}_2 \), we were able to calculate the moles that were reacted. This informs us about the number of fluorine atoms that bonded with xenon, clearly showing the importance of moles in chemical analyzation.

Chemical Reaction Analysis

In chemical reaction analysis, we figure out how reactants are converted to products, and importantly, in what proportions. Stoichiometry involves establishing molar ratios, which are core to writing balanced chemical equations.
During the reaction, \( \mathrm{Xe} \) and \( \mathrm{F}_2 \) combined to form a xenon-fluoride compound. By analyzing the difference between initial and remaining amounts of \( \mathrm{F}_2 \), we found the moles consumed.
The moles of \( \mathrm{Xe} \) are equal to \( n_{\mathrm{Xe}} \). The ratio \( \frac{n_{\mathrm{F}_2, consumed}}{n_{\mathrm{Xe}}} \) indicates the number of fluorine atoms reacting per xenon atom. If the ratio approximated 2, 4, or 6, the compound formed would be \( \mathrm{XeF}_2 \), \( \mathrm{XeF}_4 \), or \( \mathrm{XeF}_6 \). Understanding and analyzing this ratio is key to identifying the newly formed xenon-fluoride compound.