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For the decomposition of \(\mathrm{PCl}_{5}\) (g) in a closed vessel which is the correct relation between totalpressure (P), equilibrium constant \(\left(\mathrm{K}_{\mathrm{p}}\right)\) and degree of dissociation \((\alpha)\) a. \(\alpha=\sqrt{\left(K_{p}+P\right)}\) b. \(\alpha=\sqrt{\left[1 /\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right)\right]}\) c. \(\alpha=\sqrt{\left[K_{p} /\left(\mathrm{K}_{\mathrm{p}}+\mathrm{P}\right)\right]}\) d. \(\left.\alpha=\sqrt{[}\left(\mathrm{K}_{\mathrm{n}}+\mathrm{P}\right) / \mathrm{K}_{\mathrm{n}}\right]\)

Short Answer

Expert verified
The correct relation is option c: \( \alpha = \sqrt{ \frac{K_p }{K_p + P} } \).

Step by step solution

01

Understand the Reaction

The decomposition of \( \mathrm{PCl}_5 \) can be represented as \( \mathrm{PCl}_5 (g) \rightleftharpoons \mathrm{PCl}_3 (g) + \mathrm{Cl}_2 (g) \). Initially, we have a concentration \( c \) of \( \mathrm{PCl}_5 \); it decomposes to form \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \).
02

Define Initial and Equilibrium Conditions

Initially, if the concentration of \( \mathrm{PCl}_5 \) is \( c \), then at equilibrium the concentration of \( \mathrm{PCl}_5 \) is \( c(1 - \alpha) \), and the concentrations of \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) are \( c\alpha \) each, where \( \alpha \) is the degree of dissociation.
03

Write Expression for Total Pressure

The total pressure \( P \) at equilibrium is the sum of the partial pressures: \[ P = (c(1-\alpha) + c\alpha + c\alpha) RT = cRT(1 + \alpha) \] assuming the reaction is at constant temperature and using the ideal gas law.
04

Write the Expression for Equilibrium Constant (Kp)

The equilibrium constant \( K_p \) for the reaction is given by the expression: \[ K_p = \frac{p_{\mathrm{PCl}_3} \cdot p_{\mathrm{Cl}_2}}{p_{\mathrm{PCl}_5}} = \frac{(c\alpha)^2}{c(1-\alpha)} \cdot (RT) \] simplifying, \[ K_p = \frac{c^2 \alpha^2}{c (1-\alpha)} \cdot \frac{1}{P} \] which further simplifies to \[ K_p = \frac{c\alpha^2}{(1-\alpha) P} \]
05

Relate the Expressions and Solve for \( \alpha \)

Rearrange the expression for \( K_p \): \[ \alpha^2 = \frac{K_p(1-\alpha) P}{c} \] and considering the total pressure relation in Step 3, \( c \) can be replaced: \[ \alpha^2 = \frac{K_p(1-\alpha)}{(1+\alpha)} \] Simplifying further: \[ \alpha = \sqrt{\frac{K_p }{K_p+ P}} \]
06

Compare with Given Options

The simplified expression we found is \( \alpha = \sqrt{ \frac{K_p }{K_p + P} } \). This matches option (c), which states \( \alpha = \sqrt{ \frac{K_p }{K_p + P} } \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degree of Dissociation
The degree of dissociation, denoted as \( \alpha \), is an indicator of how much of a substance has decomposed in a chemical reaction. It's a crucial concept when discussing reactions at equilibrium, especially in gas-phase decompositions like that of \( \mathrm{PCl}_5 \). For a reaction starting with a concentration \( c \) of a compound, the degree of dissociation tells us the fraction of the initial compound that has decomposed. This is important for calculating the concentrations of products formed at equilibrium. In our reaction, \( \mathrm{PCl}_5 \) decomposes into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \), and the degree of dissociation determines how much of these products are formed. A higher \( \alpha \) means more products and less of the original reactant. Thus, \( \alpha \) is vital for understanding the reaction's position at equilibrium.
Partial Pressure
Partial pressure is the pressure exerted by an individual gas in a mixture of gases. In the context of the decomposition of \( \mathrm{PCl}_5 \), partial pressure helps us understand how each component in the reaction mixture contributes to the total pressure at equilibrium. When a reaction occurs in a closed container, each gas involved will exert pressure independently of the others. In our specific case, the gases \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) are formed as \( \mathrm{PCl}_5 \) decomposes, each contributing to the total pressure of the system. By the Ideal Gas Law, the pressure is a function of the number of moles, temperature, and volume. So, as \( \mathrm{PCl}_5 \) decomposes and produces \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \), their partial pressures can be expressed in terms of the degree of dissociation, \( \alpha \). This allows us to calculate the total pressure and relate it back to the equilibrium constant.
Ideal Gas Law
The Ideal Gas Law is represented by the equation \( PV = nRT \), where \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles of the gas, \( R \) is the ideal gas constant, and \( T \) is temperature. This law provides the framework for understanding how gases behave under different conditions and is especially useful in reactions where gases are involved. In the decomposition of \( \mathrm{PCl}_5 \), the Ideal Gas Law helps us relate the total pressure in the system to the concentrations of the gases at equilibrium. Since the reaction is conducted in a closed system at constant temperature, the law aids in simplifying the equations used to determine the degree of dissociation and equilibrium constant. By substituting expressions for concentrations obtained using the law into the equilibrium equation, we can derive expressions relating \( \alpha \), \( K_p \), and total pressure \( P \).
Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in constant concentrations of the reactants and products. It's a dynamic state where reactions continue to occur, but there is no net change in the concentrations of substances.In the decomposition of \( \mathrm{PCl}_5 \), equilibrium is reached when the formation of \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) from \( \mathrm{PCl}_5 \) is balanced by their recombination back into \( \mathrm{PCl}_5 \). Equilibrium constants like \( K_p \) allow us to calculate the ratio of product pressures to reactant pressure at this point of balance. Understanding chemical equilibrium is essential for predicting how the system behaves when conditions change, such as variations in pressure or concentration, using Le Chatelier's Principle. It's the equilibrium constant \( K_p \) that allows us to relate the degree of dissociation \( \alpha \) to the total pressure \( P \), giving us insights into the system's behavior at equilibrium.

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Most popular questions from this chapter

Which of the following statement is/are correct for a reversible reaction? a. At a given temperature both \(Q\) and \(K\) vary with the progress of the reaction. b. When \(Q>K\), the reaction proceeds in backward direction before coming to stand still. c. Reaction quotient (Q) is the ratio of the product or arbitrary molar concentrations of the products to those of the reactants. d. \(Q\) may be \(<>=K\).

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane (MCP) according to the equation: \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{MCP}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, predict the direction the reaction will shift, if the initial concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{MCP}\) are \(0.200 \mathrm{M}\) and \(0.100 \mathrm{M}\), respectively. The system a. is already at equilibrium b. will shift right c. will shift left d. is not at equilibrium and will remain in an unequilibrated state.

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane (MCP) according to the equation: \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{MCP}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, predict the direction the reaction will shift, if the initial concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{MCP}\) are \(0.200 \mathrm{M}\) and \(0.100 \mathrm{M}\), respectively. The system a. is already at equilibrium b. will shift right c. will shift left d. is not at equilibrium and will remain in an unequilibrated state.

(A): On addition of \(\mathrm{NH}_{4} \mathrm{Cl}\) to \(\mathrm{NH}_{4} \mathrm{OH}\) solution, \(\mathrm{pH}\) decreases but remains greater than \(7 .\) (R): Addition of \(\mathrm{NH}_{4}^{+}\)ion decreases ionization of \(\mathrm{NH}_{4} \mathrm{OH}\) thus, \(\left[\mathrm{OH}^{-}\right]\)is decreased, hence, \(\mathrm{pH}\) decreases.

(A): On mixing equal volumes of \(1 \mathrm{M} \mathrm{HCl}\) and \(2 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\), an acidic buffer solution is formed. \((\mathbf{R}):\) Resultant mixture contains \(\mathrm{CH}_{3} \mathrm{COOH}\) and CH, COONa which are parts of acidic buffer.

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