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 \(1.7 \mathrm{~atm}\). 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 compound is \(\mathrm{XeF}_2\) (option a).

Step 1: Determine Initial Moles of Gases

Use the ideal gas law to find the initial moles of each gas. Since the volume (V) is 1.0 L and the temperature is constant during calculations, you can use the formula \(n = \frac{P \times V}{R \times T}\) but because the initial pressures are known at the same volume and temperature, you use them directly for proportional calculations.

Step 2: Calculate Moles of F2 Remaining

After the reaction, the pressure of the remaining \(\mathrm{F}_2\) is 4.6 atm. Therefore, the moles of \(\mathrm{F}_2\) left is proportional to its pressure: 4.6 atm compared to the initial 8 atm. This shows how much \(\mathrm{F}_2\) reacted.

Step 3: Determine Moles of F2 Reacted

Subtract the final pressure of \(\mathrm{F}_2\) after cooling to 25°C from the initial pressure to find the pressure change: \(8.0\ \mathrm{atm} - 4.6\ \mathrm{atm} = 3.4\ \mathrm{atm}\). This change corresponds to the amount of \(\mathrm{F}_2\) that reacted.

Step 4: Calculate Initial Moles of Xe

The initial pressure of \(\mathrm{Xe}\) is 1.7 atm. When cooled, all \(\mathrm{Xe}\) converted to solid form, meaning all initial moles of \(\mathrm{Xe}\) participated in the reaction.

Step 5: Mole Ratio for XeF Compound

The ratio of consumed \(\mathrm{F}_2\) to \(\mathrm{Xe}\) moles is 3.4 atm to 1.7 atm (1:0.5). Since the suggested compounds are multiples of the simplest formula, divide both sides by 1.7 atm to get the mole ratio: \(1\ \mathrm{Xe}\) to \(2\ \mathrm{F}_2\), hence the formula.

Step 6: Select the Proposed Formula

Given the mole ratio matches with option (a) \(\mathrm{XeF}_2\). This aligns with our calculated ratio of \(1:2\), which leads to the empirical formula \(\mathrm{XeF}_2\).

Key Concepts

Xenon-Fluorine Compounds

Xenon-fluorine compounds are interesting because of their unique chemical properties. Xenon, a noble gas, has a complete outer electron shell which usually makes it inactive or inert. However, under certain conditions, xenon can form compounds with fluorine, one of the most reactive elements.

• **XeF₂**: This is a diatomic molecule with a linear geometry and is one of the simplest xenon-fluorine compounds.
• **XeF₄**: This compound has a square planar shape. It illustrates how xenon can expand its octet by using its d orbitals.
• **XeF₆**: The most fluorine-rich compound, with a distorted octahedral shape.

Xenon's ability to form these compounds is quite special as it only happens under rigorous conditions like exposure to high temperatures or pressures. This ability also defies the general behavior expected from noble gases. The reactivity depicted in the xenon-fluorine compounds is a fine demonstration of xenon's versatile chemistry.

Chemical Reaction Stoichiometry

In chemistry, stoichiometry involves the calculation of reactants and products in chemical reactions. It helps us understand the amounts of substances involved in the reaction based on the balanced chemical equation.

When applying stoichiometry:

• **Identify the balanced chemical equation**: This shows the stoichiometric coefficients, which indicate the ratio in which the reactants combine and products form.
• **Convert pressure or mass to moles**: Using tools like the ideal gas law helps convert measurements like pressure into moles, essential in stoichiometric calculations when dealing with gases.
• **Reactant calculations**: Based on the initial amounts of reactants and products, you can determine how much of each reactant is consumed and how much of each product is made.
• **Interpreting results**: By comparing the remaining and reacted amounts of reactants, the empirical formula of the product can be determined.

In the exercise, stoichiometry was used to deduce that xenon and fluorine combine in a 1:2 ratio, leading to the formation of \( \mathrm{XeF}_{2} \). Understanding stoichiometry allows us not just to predict amounts but also to clarify formulas for newly formed compounds.

Gas Laws in Chemistry

Gas laws describe how gases behave under different conditions of pressure, volume, and temperature. These laws are pivotal in stoichiometric calculations, especially when dealing with gaseous reactions.

The **ideal gas law** is fundamental in these calculations. Represented as:
\[ PV = nRT \]

• **P**: Pressure of the gas (in atm).
• **V**: Volume of the gas (in liters).
• **n**: Moles of the gas.
• **R**: Universal gas constant (0.0821 L atm K^-1 mol^-1).
• **T**: Temperature of the gas (in Kelvin).

This equation allows us to calculate one property of the gas if we know the others. It's particularly useful when converting between the measurable quantity of pressure and the number of moles needed for stoichiometry.
In the exercise, the ideal gas law helped to relate pressure changes to the moles of xenon and fluorine involved in the reaction. By understanding how to manipulate and apply this law, one can convert practical measurements into meaningful stoichiometric data. This data reflects the physical changes occurring during the chemical reaction.