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

The equilibrium constant is (a) \(0.32 \times 10^{-2}\).

Step 1: Write Down Given Equilibria

We have two equilibria: For the first reaction, \(\text{Ag}^{+} + 2\text{NH}_3 \rightleftharpoons \text{Ag(NH}_3\text{)}_2^{+}\), the equilibrium constant \(K_1 = 1.8 \times 10^{7}\).For the second reaction, \(\text{Ag}^{+} + \text{Cl}^- \rightleftharpoons \text{AgCl}\), the equilibrium constant is \(K_2 = 5.6 \times 10^{9}\).

Step 2: Calculate the Overall Equilibrium

The overall reaction is: \[ \text{AgCl} + 2 \text{NH}_3 \rightleftharpoons \text{Ag(NH}_3\text{)}_2^{+} + \text{Cl}^- \]This overall reaction can be achieved by reversing the second given reaction and then adding it to the first reaction.

Step 3: Reverse the Second Reaction and Find its New Constant

Reversing the second reaction \( \text{Ag}^{+} + \text{Cl}^- \rightleftharpoons \text{AgCl} \) means the new reaction will be: \( \text{AgCl} \rightleftharpoons \text{Ag}^{+} + \text{Cl}^- \).When a reaction is reversed, the new equilibrium constant is the reciprocal: \(K_{2'} = \frac{1}{5.6 \times 10^9}\).

Step 4: Combine Equilibrium Constants

The overall equilibrium constant \(K_{overall}\) for the reaction is the product of the equilibrium constants \(K_1\) and \(K_{2'}\):\[K_{overall} = K_1 \times K_{2'} = 1.8 \times 10^7 \times \frac{1}{5.6 \times 10^9}\].

Step 5: Calculate the Value

Now calculate:\[K_{overall} = \frac{1.8 \times 10^7}{5.6 \times 10^9} = 3.21 \times 10^{-3}\].

Step 6: Compare and Give the Answer

The calculated \(K_{overall} = 3.21 \times 10^{-3}\) matches option (a) after considering significant figures, which is \(0.32 \times 10^{-2}\).

Key Concepts

Equilibrium Constants

The equilibrium constant, symbolized as \( K \), determines the extent to which a chemical reaction proceeds. For reversible reactions, the equilibrium constant quantifies the ratio of products to reactants when the system is in equilibrium.

- It is temperature-dependent and can vary if conditions like temperature change.
- The value of \( K \) helps predict the direction of the reaction.
- A large \( K \) (\( K \gg 1\)) suggests products are favored, while a small \( K \) (\( K \ll 1\)) indicates reactants are favored.

In our exercise, the equilibrium constants \( K_1 \) and \( K_2 \) represent the balance of two different reversible reactions involving silver ions (\( \text{Ag}^+ \)) with ammonia and chloride ions, respectively. By combining these equilibria, the overall constant (\( K_{overall} \)) is determined by the product of these constants, adjusted according to the manipulation of these reactions (such as reversing one of them). Using the formula \( K_{overall} = K_1 \times K_{2'} \), where \( K_{2'} \) is the reciprocal of the reversed second reaction's constant, we get a complete picture of the overall reaction dynamics.

Reversible Reactions

Reversible reactions, as their name suggests, can proceed in both forward and backward directions.

- In a dynamic equilibrium, the rates at which the forward and backward reactions occur are equal.
- Concentrations of reactants and products remain constant over time, though not necessarily equal.

The most important feature of a reversible reaction is the concept of equilibrium consciousness: recognizing that reactants can form products, and products can reform into reactants. In chemical notation, we illustrate these reactions with a double arrow (\(\rightleftharpoons\)).

In the given exercise, this concept is exemplified through the reaction of silver ions with chloride ions and ammonia to form complex ions and AgCl. By manipulating and combining the given reversible reactions, we understand how equilibrium helps us reach an accurate chemical profile of the reaction process. Understanding reversible reaction dynamics aids in calculating equilibrium constants, as evidenced in our problem-solving approach.

Ionic Equations

Ionic equations display the various ions and molecules directly involved in a chemical reaction. These are crucial in understanding reactions that occur in solutions, especially where electrolytes are involved, dissociating into respective ions.

- Overall ionic equations show all reactants and products in the ionic form.
- Net ionic equations present only those ions and molecules that directly participate in the reaction, omitting spectator ions.

In the example from the exercise, the original equations are essentially ionic equations showing how \(\text{Ag}^+\) interacts with \(\text{NH}_3\) and \(\text{Cl}^-\) to form various complexes and precipitates. Understanding ionic equations is critical for writing correct expressions of equilibrium, allowing us to calculate equilibrium constants accurately, as seen in the manipulation of \( K_1\) and \( K_2\). By distinguishing between complete and net ionic equations, students can better grasp the core reactions and their impact on overall chemical equilibrium.