Problem 62 - General Chemistry: Principles and Modern Applications

Question

Equation \((23.18),\) which represents the chromatedichromate equilibrium, is actually the sum of two equilibrium expressions. The first is an acid-base reaction, \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\). The second reaction involves elimination of a water molecule between two \(\mathrm{HCrO}_{4}^{-}\) ions (a dehydration reaction), \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} .\) If the ionization constant, \(K_{\mathrm{a}},\) for \(\mathrm{HCrO}_{4}^{-}\) is \(3.2 \times 10^{-7},\) what is the value of \(K\) for the dehydration reaction?

Step-by-Step Solution

Answer

The equilibrium constant, \( K \), for the dehydration reaction is approximately \( \frac{1}{3.2 \times 10^{-7}} \)

Step 1: Identify relevant equations

This problem involves two reactions-\n\nReaction 1: \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\) with a known ionization constant, \( K_a = 3.2 \times 10^{-7}\)\n\nReaction 2: \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+\mathrm{H}_{2}\mathrm{O}\), for which we need to find the equilibrium constant, \( K \).\n\nThe sum of these reactions is the complete chromatedichromate equilibrium reaction, defined in Equation (23.18).

Step 2: Use properties of equilibrium constants

The equilibrium constant for the net reaction obtained by adding multiple reactions is the product of the equilibrium constants for the individual reactions. Suppose \( K' \) is the equilibrium constant for the overall reaction, and \( K \) and \( K_a \) are equilibrium constants for reactions 1 and 2, respectively. Then, by the property of equilibrium constants, we have: \[ K' = K \cdot K_a \]

Step 3: Use the given ionisation constant

Substitute the ionization constant \( K_a \) of the first reaction into the formula: \[ K' = K \cdot 3.2 \times 10^{-7} \]

Step 4: Calculate K

As we know the equilibrium constant for the complete reaction can only be 1, we can substitute this into the formula and solve for \( K \). \[ 1 = K \cdot 3.2 \times 10^{-7} \] Solving for \( K \), we get \( K = \frac{1}{3.2 \times 10^{-7}} \)

Step 5: Final Answer

Calculate the value using a calculator to get the final result.

Key Concepts

Understanding Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where an acid donates a proton (H\(^+\)) to a base. This exchange can lead to the formation of new compounds. In the given exercise, the reaction \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\) is an example of an acid-base equilibrium. Here, \(\mathrm{HCrO}_{4}^{-}\) is formed when \(\mathrm{H}^{+}\) combines with \(\mathrm{CrO}_{4}^{2-}\).
This process highlights the dynamic nature of equilibrium, where molecules continually react in forward and reverse reactions. In equilibrium, the rate of the forward reaction equals that of the reverse, and concentrations of reactants and products remain constant over time.
Understanding these dynamic interactions is essential for fields like biochemistry and environmental science. It explains phenomena like acid rain formation or cellular pH regulation.

• Acid - Proton donor.
• Base - Proton acceptor.
• Equilibrium - State of balance in a reversible reaction.

Unveiling the Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a critical value that represents the ratio of concentrations of products to reactants at equilibrium for a given reaction. It is unique for every reaction condition and serves as a guide to predict the direction of a reaction.
In the context of the exercise, the equilibrium constant \(K_a = 3.2 \times 10^{-7}\) for the acid-base reaction signifies how far the reaction proceeds to form products.
For combined reactions, equilibrium constants multiply. For example, if \(K'\) is the constant for a series of reactions, then it can be expressed as:
\[ K' = K \cdot K_a \]
This multiplication helps determine the equilibrium state of complex systems like the chromate-dichromate equilibrium. The equilibrium constant not only predicts the predominance of reactants or products but also greatly influences industrial and laboratory processes.

• Equilibrium Constant - Measures reaction tendency toward completion.
• Predictive Value - Indicates reaction progress and equilibrium position.
• Multiplicative Property - Combine constants for sequential reactions.

Role of Dehydration Reactions

Dehydration reactions involve the removal of a water molecule during a chemical process. These reactions are crucial in organic chemistry, often leading to the formation of anhydrides or larger molecular structures.
In the provided exercise, the dehydration reaction \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+\mathrm{H}_{2}\mathrm{O}\) shows how two compounds combine, releasing water. This type of reaction alters structures by eliminating small molecules like \(\mathrm{H}_2\mathrm{O}\).
They play an essential role in biological systems, such as forming complex carbohydrates and in chemical industries for synthesizing polymers. Understanding dehydration is vital because it offers insight into larger compound synthesis and stability.

• Dehydration - Removal of water during a reaction.
• Synthesis - Builds complex structures from smaller molecules.
• Applicability - Key in biological and industrial chemistry.