Question

At a given temperature, the \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) is \(2.4 \times 10^{-3} .\) At the same temperature, the \(K_{\mathrm{C}}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\) \(\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})\) is (1) \(2.4 \times 10^{-3}\) (2) \(2.4 \times 10^{-3}\) (3) \(4.2 \times 10^{2}\) (4) \(4.8 \times 10^{-2}\)

Short answer

Option (3): \( 4.2 \times 10^{2} \)

Step-by-step solution

Understanding the given reaction

The first reaction given is: \[ \mathrm{PCl_{5}}(\mathrm{g}) \rightleftharpoons \mathrm{PCl_{3}}(\mathrm{g}) + \mathrm{Cl_{2}}(\mathrm{g}) \] with equilibrium constant \( K_c = 2.4 \times 10^{-3} \).

Write the reverse reaction

The reverse reaction is: \[ \mathrm{PCl_{3}}(\mathrm{g}) + \mathrm{Cl_{2}}(\mathrm{g}) \rightleftharpoons \mathrm{PCl_{5}}(\mathrm{g}) \]

Relation of equilibrium constants

For a reverse reaction, the equilibrium constant \( K'_c \) is the reciprocal of the equilibrium constant of the forward reaction. Therefore, \[ K'_c = \frac{1}{K_c} \]

Calculate the equilibrium constant for the reverse reaction

Using the given \( K_c = 2.4 \times 10^{-3} \), calculate \( K'_c \). \[ K'_c = \frac{1}{2.4 \times 10^{-3}} \] Simplifying this, we get: \[ K'_c = \frac{1}{2.4 \times 10^{-3}} = \frac{1}{2.4} \times 10^{3} = 0.4167 \times 10^{3} = 4.167 \times 10^{2} \]

Identify the correct option

The value of \( K'_c \) calculated is \( 4.167 \times 10^{2} \). Round off if necessary. The approximate answer is \( 4.2 \times 10^{2} \). Therefore, the correct option is (3).

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical process are equal, resulting in no overall change in the concentrations of the reactants and products over time. In other words, the system is in a state of balance.
In the given exercise, at equilibrium, the concentrations of \(\text{PCl}_5\), \(\text{PCl}_3\), and \(\text{Cl}_2\) remain constant even though both forward and reverse reactions are ongoing. This equilibrium can be represented by an equilibrium constant, \(K_c\), which quantifies the ratio of the concentrations of products to the concentrations of reactants at equilibrium.
For the reaction: \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\), the equilibrium constant is given as \(K_c = 2.4 \times 10^{-3}\). The value of \(K_c\) indicates how far the reaction proceeds before reaching equilibrium. A small \(K_c\) value, as in this case, suggests that at equilibrium, the concentration of reactants (\text{PCl}_5) is higher than the concentration of products (\text{PCl}_3 and \text{Cl}_2).

Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that affect these rates. It involves understanding the speed of the forward and reverse reactions that eventually lead to chemical equilibrium.
In reversible reactions like \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\), both the forward reaction (formation of \text{PCl}_3 and \text{Cl}_2) and the reverse reaction (formation of \text{PCl}_5) occur simultaneously. The rate at which the forward reaction occurs is dependent on the concentration of \text{PCl}_5, while the rate of the reverse reaction depends on the concentrations of \text{PCl}_3 and \text{Cl}_2.
When these rates become equal, the system reaches equilibrium. At this point, the speed at which \text{PCl}_5 turns into \text{PCl}_3 and \text{Cl}_2 is exactly balanced by the speed at which \text{PCl}_3 and \text{Cl}_2 recombine to form \text{PCl}_5. This balance is what characterizes the equilibrium state.

Reversible Reactions

Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. These reactions do not go to completion and can reach a state of equilibrium.
In the example provided, \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\), the reaction is reversible. This means that \text{PCl}_5 can decompose into \text{PCl}_3 and \text{Cl}_2, and at the same time, \text{PCl}_3 and \text{Cl}_2 can recombine to form \text{PCl}_5.
The equilibrium constant \(K_c\) for a reversible reaction is calculated from the concentrations of the reactants and products at equilibrium. For the reverse reaction (\text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5), the equilibrium constant is the reciprocal of the forward reaction’s \(K_c\). Therefore, if the equilibrium constant \(K_c\) for the forward reaction is \(2.4 \times 10^{-3}\), the equilibrium constant for the reverse reaction is \( \frac{1}{2.4 \times 10^{-3}} = 4.2 \times 10^{2}\). This reversal of equilibrium constants is a key concept in understanding reversible reactions.