Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{PCl}_{5}\) is \(50 \%\) dissociated at \(20{ }^{\circ} \mathrm{C}\) and 1 atm pressure. The value of \(\mathrm{K}_{p}\) is (a) \(0.444\) (b) \(0.555\) (c) \(0.333\) (d) \(0.666\)

Step-by-Step Solution

Answer

The value of \(K_p\) is 0.333, option (c).

Step 1: Understanding the Chemical Equation

The dissociation of \(\mathrm{PCl}_{5}\) can be represented by the equilibrium reaction: \(\mathrm{PCl}_5 (g) \rightleftharpoons \mathrm{PCl}_3 (g) + \mathrm{Cl}_2 (g)\). The initial moles of \(\mathrm{PCl}_{5}\) are 1.

Step 2: Calculating Moles at Equilibrium

Since \(\mathrm{PCl}_{5}\) is 50% dissociated, at equilibrium we have 0.5 moles of \(\mathrm{PCl}_{5}\), 0.5 moles of \(\mathrm{PCl}_{3}\), and 0.5 moles of \(\mathrm{Cl}_{2}\).

Step 3: Applying the Ideal Gas Law

At 1 atm and assuming ideal gas behavior, the total pressure is 1 atm with a total number of moles equal to 1.5 (0.5 of \(\mathrm{PCl}_5\), 0.5 of \(\mathrm{PCl}_3\), and 0.5 of \(\mathrm{Cl}_2\)).

Step 4: Calculating Partial Pressures

The partial pressure of each component at equilibrium can be calculated as follows: - \( P_{\mathrm{PCl}_5} = 0.5/1.5 \times 1 = \frac{1}{3} \) atm - \( P_{\mathrm{PCl}_3} = 0.5/1.5 \times 1 = \frac{1}{3} \) atm - \( P_{\mathrm{Cl}_2} = 0.5/1.5 \times 1 = \frac{1}{3} \) atm.

Step 5: Calculating \(K_p\)

\(K_p\) for the reaction is given by the expression: \[ K_p = \frac{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} = \frac{\left( \frac{1}{3} \right) \times \left( \frac{1}{3} \right)}{\frac{1}{3}} \]Simplifying, we find: \[ K_p = 0.333 \]

Step 6: Selecting the Correct Answer

Option (c) \(0.333\) is the correct value for \(K_p\).

Key Concepts

Dissociation Equilibrium

In a chemical context, dissociation equilibrium occurs when a compound splits into two or more components, reaching a state where the rate of dissociation is equal to the rate of recombination. For the given chemical reaction of \[\mathrm{PCl}_5 (g) \rightleftharpoons \mathrm{PCl}_3 (g) + \mathrm{Cl}_2 (g),\]dissociation equilibrium involves the breaking of the \(\mathrm{PCl}_5\) molecule into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\). At this equilibrium, the concentrations of \(\mathrm{PCl}_5, \mathrm{PCl}_3,\) and \(\mathrm{Cl}_2\) remain constant over time.Key features of dissociation equilibrium include:

• The ratio of the concentrations of products to reactants remains constant at equilibrium conditions.
• For the \(50\%\) dissociation of \(\mathrm{PCl}_5\), this suggests that initially 1 mol of \(\mathrm{PCl}_5\) splits into 0.5 mol of \(\mathrm{PCl}_3\) and 0.5 mol of \(\mathrm{Cl}_2\), with 0.5 mol remaining unreacted.
• The equilibrium state is dynamic; molecules continue to dissociate and recombine, balancing out the rates of both processes.

By understanding dissociation equilibrium, we can predict the behavior of substances in reversible reactions and calculate important properties such as the equilibrium constant, \(K_p\).

Partial Pressure

Partial pressure is an essential concept in understanding chemical equilibria in gaseous systems. Simply put, the partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature.In the \(\mathrm{PCl}_5\) dissociation example, we have a mixture of \(\mathrm{PCl}_5, \mathrm{PCl}_3,\) and \(\mathrm{Cl}_2\), contributing to the total pressure of 1 atm. Partial pressures can be calculated using the mole fractions as shown:

• First, find the mole fraction for each gas, resulting from the equilibrium moles divided by the total moles \( (1.5)\).
• The partial pressure for each gas can then be determined by multiplying the mole fraction by the total pressure (1 atm).
• Following this method, each component \(\mathrm{PCl}_5, \mathrm{PCl}_3,\) and \(\mathrm{Cl}_2\) has a partial pressure equal to \(\frac{1}{3}\) atm.

Partial pressure calculations are crucial to find equilibrium constants and describe the behavior of each gas in a mixture. Understanding them helps predict how changes in pressure affect equilibrium positions.

Equilibrium Constant Kp

The equilibrium constant, \(K_p\), gives us insight into the position of equilibrium for reactions involving gases. It is calculated using the partial pressures of the gases involved in the reaction. For the dissociation of \(\mathrm{PCl}_5\), the calculation of \(K_p\) involves:

• Using the expression: \[ K_p = \frac{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} \]
• Substituting the known partial pressures: \(\frac{1}{3}\) atm for each component.
• Simplifying the expression: \[ K_p = \frac{\left( \frac{1}{3} \right) \times \left( \frac{1}{3} \right)}{\frac{1}{3}} = 0.333 \]

This value of \(K_p\) signifies that under the given conditions, the reaction stands in a particular balance between forward and reverse processes. A smaller or larger \(K_p\) would indicate more significant dissociation or recombination, respectively.In summary, \(K_p\) helps chemists compare the stability and tendencies of reactions under various conditions and is a fundamental part of chemical equilibrium studies.