Problem 18 - Chemistry The Molecular Science

Question

Write the expression for \(K_{\mathrm{c}}\) for each reaction. (a) \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(2 \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) \([\mathrm{PCl}_3][\mathrm{Cl}_2]\); (b) \(\frac{[\mathrm{CoCl}_4^{2-}]}{[\mathrm{Co(H}_2\mathrm{O})_6^{2+}][\mathrm{Cl}^-]^4}\); (c) \(\frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]}\); (d) \(\frac{[\mathrm{OF}_2][\mathrm{HF}]^2}{[\mathrm{F}_2]^2[\mathrm{H}_2\mathrm{O}]}\)."}

Step 1: Understand Equilibrium Constant Expression

The equilibrium constant expression, denoted as \(K_c\), is calculated using the concentrations of products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their coefficients. Be aware of the physical states: pure solids and liquids do not appear in the expression.

Step 2: Express K_c for Reaction (a)

The reaction is \( \mathrm{PCl}_{5}(\mathrm{s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \). Since \(\mathrm{PCl}_{5}(\mathrm{s})\) is a solid, it is not included. The expression for \(K_c\) is:\[ K_c = [\mathrm{PCl}_3][\mathrm{Cl}_2] \]

Step 3: Express K_c for Reaction (b)

The reaction is \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq}) + 4\mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{CoCl}_{4}^{2-}(\mathrm{aq}) + 6 \mathrm{H}_2 \mathrm{O}(\ell)\). The water is liquid, thus omitted. The expression for \(K_c\) is:\[ K_c = \frac{[\mathrm{CoCl}_4^{2-}]}{[\mathrm{Co(H}_2\mathrm{O})_6^{2+}][\mathrm{Cl}^-]^4} \]

Step 4: Express K_c for Reaction (c)

The reaction is \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq}) + \mathrm{H}^{+}(\mathrm{aq})\). All species are in aqueous solution, so they all appear in the expression:\[ K_c = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \]

Step 5: Express K_c for Reaction (d)

The reaction is \(2 \mathrm{~F}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_2(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\). All species are gases, so the expression is:\[ K_c = \frac{[\mathrm{OF}_2][\mathrm{HF}]^2}{[\mathrm{F}_2]^2[\mathrm{H}_2\mathrm{O}]} \]

Key Concepts

Chemical Equilibrium

In chemistry, equilibrium refers to a state in which the rates of the forward and reverse reactions are equal. At this point, the concentrations of the reactants and products remain constant over time, even though both reactions continue to occur. This balance results from the dynamic nature of chemical reactions, where reactions are ongoing but occur at the same rate in both directions.

• The concept of equilibrium is fundamental in understanding how reactions progress and reach a state of balance.
• It is important to note that equilibrium does not mean that the quantities of reactants and products are equal; rather, it means their ratios remain constant.

One important aspect of chemical equilibrium is the equilibrium constant, often represented as \(K_c\). This constant is derived from the concentrations of products and reactants at equilibrium.
In the expressions for \(K_c\):

• It includes only the species that are present in a gaseous or aqueous state.
• Solids and pure liquids do not appear in the equilibrium expression because their concentrations are constant and do not affect the equilibrium position.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. The expression for \(Q\) is similar to that of the equilibrium constant \(K_c\), but \(Q\) can be calculated at any point during a reaction and not just at equilibrium.

• If \(Q = K_c\), the system is at equilibrium.
• If \(Q < K_c\), the forward reaction is favored, meaning reactants are converted into products.
• If \(Q > K_c\), the reverse reaction is favored, meaning products are converted back into reactants.

Calculating the reaction quotient involves using the same formula as the equilibrium constant but substituting the current concentrations of the reactants and products.
This helps predict how the reaction will shift to achieve equilibrium, providing valuable insight into the reaction's progress.

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a system at equilibrium responds to disturbances such as changes in concentration, temperature, or pressure. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system shifts its position to counteract the change and re-establish equilibrium.

• Adding more reactants will typically cause the equilibrium to shift toward the product side to reduce the disturbance.
• Increasing the temperature in an exothermic reaction will shift the equilibrium toward the reactants to absorb the excess heat.
• Conversely, decreasing the temperature in an exothermic reaction shifts the equilibrium toward the products.
• Increased pressure will cause the equilibrium to favor the side of the reaction with fewer moles of gas, if gases are involved.

Understanding Le Chatelier's Principle is crucial for controlling chemical reactions in industrial processes, where maximizing yields and minimizing costs are critical. It provides a predictive tool for how conditions can be manipulated to drive the equilibrium position in a desired direction.