Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constants for the reactions, \(\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons{\mathrm{XeOF}}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) (i) and \(\mathrm{XeO}_{4}(\mathrm{~s})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons{=} \mathrm{XeOF}_{4}(\mathrm{~g})+\) \(\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g}) \ldots .\) (ii) are \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) respectively. The equilibrium constant \((\mathrm{K})\) for the reaction \(\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is (a) \(\mathrm{K}_{1} / \mathrm{K}\) (b) \(\mathrm{K}_{2} / \mathrm{K}_{1}\) (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}\) (d) \(\mathrm{K}_{1} / \mathrm{K}_{2}\)

Step-by-Step Solution

Answer

The answer is (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}\).

Step 1: Understand the given reactions

We have two reactions with equilibrium constants: \(\mathrm{K}_1\) for \(\mathrm{XeF}_6(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons{\mathrm{XeOF}}_4(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) and \(\mathrm{K}_2\) for \(\mathrm{XeO}_{4}(\mathrm{~s})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons{=}{\mathrm{XeOF}}_4(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_2(\mathrm{~g})\). We need to find \(K\) for the reaction \(\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Step 2: Combine the given reactions

To find the equilibrium constant \(K\), we must manipulate the given reactions to arrive at the target reaction. Subtract reaction (ii) from reaction (i) to eliminate \(\mathrm{XeF}_6\) and \(\mathrm{XeOF}_4\), arriving at: \(\mathrm{XeO}_{4}(\mathrm{~g}) + 2\mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{g})\).

Step 3: Calculate the equilibrium constant for the resulting reaction

When subtracting one equilibrium reaction from another, the resulting equilibrium constant \(K\) is the quotient of the individual constants: \(K = \frac{K_1}{K_2}\).

Step 4: Select the correct option

According to the derived expression \(K = \frac{K_1}{K_2}\), we match this with the given options. The correct option is (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}\).

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. These transformations involve the breaking and forming of chemical bonds, which change the arrangement of atoms. Reactants and products in a chemical reaction are represented by a chemical equation. For example, the reaction \[\text{XeF}_6(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{XeOF}_4(\text{g}) + 2 \text{HF}(\text{g})\]shows how xenon hexafluoride reacts with water to form xenon oxytetrafluoride and hydrogen fluoride. In the context of equilibrium, reactions can proceed forward and backward until they reach a state where the rate of the forward reaction equals the rate of the reverse reaction. This state is what we call chemical equilibrium. At this point, the concentrations of reactants and products remain constant over time, not necessarily because they are equal, but because they are balanced in a dynamic state where reactions are still occurring.

• Reactants transform into products.
• In equilibrium, the forward and reverse reaction rates are equal.

Thermodynamics

Thermodynamics in chemistry involves studying the energy changes that occur during chemical reactions. Two key aspects to focus on are enthalpy and entropy. Enthalpy (\(\Delta H\)) represents the heat content of a system. If a reaction releases heat, it is exothermic (\(\Delta H < 0\)). If it absorbs heat, it is endothermic (\(\Delta H > 0\)). Entropy (\(\Delta S\)) measures the disorder or randomness of a system. More disorder generally means higher entropy. Together, enthalpy and entropy dictate the spontaneity of a reaction through Gibbs free energy (\(\Delta G\)). The equation is \(\Delta G = \Delta H - T\Delta S\), where \(T\) is the temperature in Kelvin.

• A negative \(\Delta G\) indicates a spontaneous reaction.
• A positive \(\Delta G\) means the reaction is non-spontaneous.

In equilibrium scenarios, the change in Gibbs free energy (\(\Delta G\)) is zero, highlighting that the system is at a no-net-change state. Understanding these concepts helps explain why certain reactions occur and others do not, based on the changes in heat and disorder involved.

Gaseous Equilibrium

When dealing with chemical reactions involving gases, understanding gaseous equilibrium is crucial. At gaseous equilibrium, the concentrations of gaseous reactants and products have reached a state where they no longer change over time. This balance is described by the equilibrium constant \(K\). For any general reaction: \[aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\]The equilibrium constant is given by the expression:\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]Here, square brackets denote the concentration of the substances, and \(a, b, c, d\) are the stoichiometric coefficients from the balanced equation. In problems that involve adding or subtracting gaseous reactions, such as in the exercise, the overall equilibrium constant can be determined from the individual constants by either multiplying or dividing them, depending on the operations used to combine the reactions.

• Equilibrium constants describe the ratio of concentrations at equilibrium.
• Changes in pressure, temperature, or concentration can shift the equilibrium position.

It is important to remember that equilibrium does not mean the reactants and products are equal in amount; rather, their concentrations remain steady over time, due to the balanced forward and reverse reaction rates.