Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{Ag}^{+}+2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+} ; \mathrm{K}_{1}=1.8 \times 10^{7}\) \(\mathrm{Ag}^{+}+\mathrm{Cl} \rightleftharpoons \mathrm{AgCl} ; \mathrm{K}_{2}=5.6 \times 10^{9}\) Then for \(\mathrm{AgCl}+2 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+\mathrm{Cl}^{-}\), Equilibrium constant will be (a) \(0.32 \times 10^{-2}\) (b) \(3.11 \times 10^{2}\) (c) \(10.08 \times 10^{16}\) (d) \(1.00 \times 10^{-17}\)

Step-by-Step Solution

Answer

Option (a) \( 0.32 \times 10^{-2} \).

Step 1: Understand the Reactions

We have two given equilibrium reactions with their constants:1. \( \mathrm{Ag}^{+}+2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+} \) with \( K_1=1.8 \times 10^7 \).2. \( \mathrm{Ag}^{+}+\mathrm{Cl}^- \rightleftharpoons \mathrm{AgCl} \) with \( K_2=5.6 \times 10^9 \).We need to determine \( K \) for the combined reaction: \( \mathrm{AgCl}+2 \mathrm{NH}_{3} \rightleftharpoons \left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+\mathrm{Cl}^{-} \).

Step 2: Write the Net Reaction

The net reaction involves the forward and reverse of the given reactions to lead to \( \mathrm{AgCl}+2 \mathrm{NH}_{3} \rightleftharpoons \left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}+\mathrm{Cl}^{-} \).This is effectively the reverse reaction of \( \mathrm{Ag}^{+}+\mathrm{Cl}^-=\mathrm{AgCl} \) combined with \( \mathrm{Ag}^{+}+2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+} \).

Step 3: Calculate the Equilibrium Constant for the Desired Reaction

For the desired equilibrium, the constant \( K \) can be calculated as \( K = \frac{K_1}{K_2} \) because:- \( K_1 \) corresponds to forming \( \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+} \).- \( K_2 \) corresponds to forming \( \mathrm{AgCl} \), but we need the reverse reaction, which is \( \frac{1}{K_2} \). Therefore, \( K = K_1 \times \frac{1}{K_2} \).

Step 4: Compute the Result

Calculate \( K \):\[ K = \frac{1.8 \times 10^7}{5.6 \times 10^9} \approx 3.21 \times 10^{-3} \].

Step 5: Compare with the Options

The options given are:(a) \( 0.32 \times 10^{-2} \),(b) \( 3.11 \times 10^{2} \),(c) \( 10.08 \times 10^{16} \),(d) \( 1.00 \times 10^{-17} \).Our calculated value \( 3.21 \times 10^{-3} \) matches \( 0.32 \times 10^{-2} \), which is option (a).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical process where the concentrations of reactants and products remain constant over time. This does not mean that reactions stop occurring, but rather that the rates of the forward and reverse reactions are equal. In this balanced state, the system is essentially "at rest."

• In the context of the given exercise, equilibrium is crucial because it pertains to the reversibility of the reactions containing silver ions and ammonia or chloride ions.
• The equilibrium constant ( K ) is key; it quantifies the position of equilibrium, indicating the ratio of product concentrations to reactant concentrations at equilibrium.

Understanding how to manipulate these equations is essential for predicting how changing conditions can affect reaction dynamics.
When you are asked to find the equilibrium constant for a new reaction derived from given reactions, as in this exercise, it requires rearranging and combining the original equilibrium constants to reflect the new net reactions.

Coordination Chemistry

Coordination chemistry deals with metal complexes, where a central metal ion is bonded to surrounding molecules or ions known as ligands.

• In our example, silver ( Ag^{+} ) acts as the central metal ion.
• Ammonia ( NH_{3} ) molecules serve as ligands that bond to the silver ion to form a coordination compound, Ag(NH_{3})_{2}^{+} .

The study of these structures explains how these molecules are bonded.

• Coordination number, which indicates the number of ligand bonds to the central atom, is a crucial aspect.
• Transition metals like silver commonly form such complexes due to their ability to accept electron pairs from ligands, resulting in highly structured compounds often leading to unique properties.

Understanding coordination chemistry helps to explain the behavior and stability of complexes such as Ag(NH_{3})_{2}^{+} when in equilibrium.

Reaction Kinetics

Reaction kinetics involves studying the speed or rate of chemical reactions and the factors controlling them. This provides insight into how quickly equilibrium is reached.

• The rate of a reaction is influenced by factors such as concentration, temperature, presence of catalysts, and the nature of the reactants and products.
• In our specific reaction dynamics, the speed at which silver ions react with NH_{3} to form Ag(NH_{3})_{2}^{+} or with chloride ions to form AgCl can differ significantly.

Understanding kinetics involves plotting concentration–time graphs and calculating rate constants.
Although this exercise focuses more on equilibrium constants than kinetic factors, appreciating how fast reactions approach equilibrium is essential for managing chemical processes effectively.

Silver Complexes

Silver complexes are an important part of inorganic chemistry, especially in coordination compounds and reaction mechanics. In such complexes, silver acts as a central metal ion that can form stable compounds.

• Silver ions readily form complexes with ligands such as ammonia due to their electron-pair acceptance capability.
• The formation and stability of silver complexes like Ag(NH_{3})_{2}^{+} can influence the reactions' equilibrium and overall chemistry.

Silver complexes often exhibit distinctive properties which are explored in material sciences and catalysis.

• These complexes are typically studied for their color changes, solubility, and reactivity, which can be predictive of their behavior in a broader chemical context.
• The reactivity of a silver complex in a reaction like the one in our exercise significantly affects the equilibrium constant calculations.

Learning about these complexes expands understanding of reaction mechanisms and predicts the possible outcomes of chemical synthesis involving silver.