Question

The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant \(\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) at \(298 \mathrm{~K}\).

Short answer

The equilibrium constant for the required equilibrium, \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) at \(298 \mathrm{~K}\), is approximately \(2.857 \times 10^{-3}\).

Step-by-step solution

Combine the Equilibria

Let the given equilibria be Reaction 1 and Reaction 2: Reaction 1: \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s), K_{c1}=5.6 \times 10^{9}\) Reaction 2: \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q), K_{c2}=1.6 \times 10^{7}\) Required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) To combine Reaction 1 and Reaction 2 to get the required equilibrium, we need to reverse Reaction 1 and multiply Reaction 2 by 1, since the stoichiometry of the reactions is maintained throughout the process. The new combined reaction will be"> Reverse Reaction 1: \(\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q)\) Multiply Reaction 2 by 1: \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q)\) Combine these to get the required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) This new equilibrium is what we were asked to find, so now let's manipulate the equilibrium constants.

Apply Rules for Manipulating Equilibrium Constants

When we reverse a reaction, we must take the reciprocal of its equilibrium constant. Additionally, when we multiply a reaction by a factor, we raise its equilibrium constant to the power of the same factor. Since we reversed Reaction 1 and multiplied Reaction 2 by 1 (which doesn't affect the equilibrium constant), we get the new equilibrium constants: Reverse Reaction 1: \(K_{c1'} = \frac{1}{K_{c1}} = \frac{1}{5.6 \times 10^{9}}\) Now, multiply the two equilibrium constants together: Combined equilibrium constant: \(K_{c} = K_{c1'} \times K_{c2} = \frac{1}{5.6 \times 10^{9}} \times 1.6 \times 10^{7}\)

Calculate the Combined Equilibrium Constant

Finally, let's multiply the two constants together and find the combined equilibrium constant: \(K_c = \frac{1.6 \times 10^{7}}{5.6 \times 10^{9}}\) \(K_c = 2.857 \times 10^{-3}\) So, the equilibrium constant for the required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) at \(298 \mathrm{~K}\) is approximately \(2.857 \times 10^{-3}\).

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), provides insight into the balance of materials present at equilibrium in a chemical reaction. It is a vital value determined only by the temperature and nature of the reaction.

**What Does \( K_c \) Mean?** * A large \( K_c \) means the reaction heavily favors the products. * A small \( K_c \) means the reaction leans towards the reactants. * A \( K_c \) approximately equal to 1 indicates a fairly balanced reaction at equilibrium.

To find \( K_c \) for a net reaction, you need to consider the \( K_c \) of the individual reactions involved. For the reaction from the exercise, we use the relationship: \[ K_{c, \text{new}} = \frac{1}{K_{c1}} \times K_{c2} \] This was because Reaction 1 was reversed, requiring the reciprocal of its original \( K_c \). As neither reaction was multiplied by a coefficient, their \( K_c \) values remain otherwise unchanged. This understanding is crucial in predicting the direction and extent of reversible reactions, particularly under varying temperatures.

Reversible Reactions

Reversible reactions are processes that occur in both forward and backward directions simultaneously. This means the reactants convert to products, and the products revert back to reactants. Let's delve further into what this means.

**Characteristics of Reversible Reactions*** **Dynamic State**: At dynamic equilibrium, these reactions do not show an overall change because the forward and reverse reaction rates equalize.* **Represented by Double Arrow**: The double arrow (\(\rightleftharpoons\)) signifies that both transformations occur.* **Pathway to Equilibrium**: Initially, the forward reaction occurs more prominently. As the concentration of products increases, the reverse reaction begins to match this pace, culminating in equilibrium. The reaction discussed, involving silver ions and ammonia formation, achieves such an equilibrium where the concentration ratios stay constant. Understanding when and how these equilibria are established helps predict the system behavior under different conditions, an essential aspect when analyzing systems like the ones in the exercise.

Thermodynamics

Thermodynamics, the study of energy transformations, plays a pivotal role in understanding chemical equilibria. Let's examine some fundamental aspects that link thermodynamics to equilibrium constants and reactions.

**Link Between Thermodynamics and Equilibrium*** **Energy Exchange**: Reversible reactions involve energy changes, typically enthalpy (heat content) or Gibbs free energy.* **Gibbs Free Energy (\( \Delta G \))**: At equilibrium, \( \Delta G = 0 \) which indicates no net change in free energy and the system is least excited.* **Temperature Dependency**: The value of \( K_c \) is inherently dependent on temperature, as it links to \( \Delta G \) through the relation \[ \Delta G = -RT \ln K_c \] where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. This equation connects thermodynamics with the constants used to predict equilibrium positions. Thermodynamics allows you to appreciate why reactions prefer certain directions, particularly how changes in temperature can tip the equilibrium scale. Understanding these factors is essential for grasping why in the example given, equilibrium constants are calculated at a specified temperature to reflect stability.