Question

Based on these descriptions, write a balanced equation and the corresponding $K_c$ expression for each reversible reaction. (a) Carbonyl fluoride, ${\mathrm{COF}}_2(\mathrm{g}),$ decomposes into gaseous carbon dioxide and gaseous carbon tetrafluoride. (b) Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion. (c) Peroxodisulfate ion, ${\mathrm{S}}_2{\mathrm{O}}_8^{2-}$, oxidizes iron(II) ion to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.

Short answer

The balanced chemical equations and their corresponding $K_c$ expressions are: (a) $2{\mathrm{COF}}_2(\mathrm{g})\leftrightarrow {\mathrm{CO}}_2(\mathrm{g})+{\mathrm{CF}}_4(\mathrm{g})$, $K_c=\frac{[{\mathrm{CO}}_2][{\mathrm{CF}}_4]}{[{\mathrm{COF}}_2{]}^2}$; (b) $\mathrm{Cu}(\mathrm{s})+2{\mathrm{Ag}}^{+}(\mathrm{aq})\leftrightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})$, $K_c=\frac{[{\mathrm{Cu}}^{2+}]}{[{\mathrm{Ag}}^{+}{]}^2}$; (c) $2{\mathrm{S}}_2{\mathrm{O}}_8^{2-}+2{\mathrm{Fe}}^{2+}\leftrightarrow 4{\mathrm{SO}}_4^{2-}+2{\mathrm{Fe}}^{3+}$, $K_c=\frac{[{\mathrm{SO}}_4^{2-}{]}^4[{\mathrm{Fe}}^{3+}{]}^2}{[{\mathrm{S}}_2{\mathrm{O}}_8^{2-}{]}^2[{\mathrm{Fe}}^{2+}{]}^2}$

Step-by-step solution

Balance Equation and Compute $K_c$ for (a)

Firstly, the balanced equation for the decomposition of Carbonyl fluoride into Carbon dioxide and Carbon tetrafluoride is: $$2{\mathrm{COF}}_2(\mathrm{g})\leftrightarrow {\mathrm{CO}}_2(\mathrm{g})+{\mathrm{CF}}_4(\mathrm{g})$$. For the $K_c$ expression, it is given by the ratio of the concentrations of the products over the reactants: $$K_c=\frac{[{\mathrm{CO}}_2][{\mathrm{CF}}_4]}{[{\mathrm{COF}}_2{]}^2}$$

Balance Equation and Compute $K_c$ for (b)

The balanced equation for the displacement of silver(I) ion by copper metal is: $$\mathrm{Cu}(\mathrm{s})+2{\mathrm{Ag}}^{+}(\mathrm{aq})\leftrightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})$$. The $K_c$ expression becomes: $$K_c=\frac{[{\mathrm{Cu}}^{2+}]}{[{\mathrm{Ag}}^{+}{]}^2}$$. Notice that Cu(s) and Ag(s) are not included in the $K_c$ expression, because pure solids and liquids are not included.

Balance Equation and Compute $K_c$ for (c)

The balanced equation for the oxidation of iron(II) ion by peroxodisulfate ion is: $$2{\mathrm{S}}_2{\mathrm{O}}_8^{2-}+2{\mathrm{Fe}}^{2+}\leftrightarrow 4{\mathrm{SO}}_4^{2-}+2{\mathrm{Fe}}^{3+}$$. The $K_c$ expression in this case is: $$K_c=\frac{[{\mathrm{SO}}_4^{2-}{]}^4[{\mathrm{Fe}}^{3+}{]}^2}{[{\mathrm{S}}_2{\mathrm{O}}_8^{2-}{]}^2[{\mathrm{Fe}}^{2+}{]}^2}$$

Key concepts

Reversible Reactions

In chemistry, reversible reactions are processes where the reactants can transform into products and vice versa. This means the reaction can proceed in both forward and backward directions. When writing reversible reactions, a double-headed arrow, $\leftrightarrow$, is used to indicate that equilibrium can occur.

Reversible reactions play an essential role in chemical equilibrium. In these reactions, both the reactants and products are present in the system. As the reaction progresses, conditions such as concentration, temperature, and pressure can influence the direction of the reaction. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in a constant concentration of products and reactants.

Understanding reversible reactions helps predict product formation and reactant consumption over time. It is essential in fields ranging from industrial chemical synthesis to biological systems, like enzyme reactions in metabolism. Thus, mastering the concept of reversible reactions is fundamental in studying chemical processes.

Balanced Chemical Equations

Balanced chemical equations reflect the principle of mass conservation. In a balanced equation, the number of atoms for each element is the same on both sides of the equation. This ensures that matter is neither created nor destroyed during the chemical reaction. Balancing equations is crucial as it provides the correct proportions of reactants and products involved in the reaction.

For example, consider the reaction of the decomposition of Carbonyl Fluoride:
$$2{\mathrm{COF}}_2(\mathrm{g})\leftrightarrow {\mathrm{CO}}_2(\mathrm{g})+{\mathrm{CF}}_4(\mathrm{g})$$

This equation is balanced because there are equal numbers of carbon, oxygen, and fluorine atoms on both sides. The coefficients (numbers in front of molecules) bring the equation into balance and indicate the molar proportions required for the reaction. Balancing chemical equations requires practice and a good understanding of how to adjust coefficients without altering the compounds themselves.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as $K_c$, is a fundamental concept in chemical equilibrium. It quantifies the relationship between the concentrations of products and reactants at equilibrium for a given reaction. Essentially, $K_c$ provides insight into the position of equilibrium, indicating whether the reactants or products are favored.

The expression for $K_c$ is derived from the balanced chemical equation. For a general reaction $aA+bB\leftrightarrow cC+dD$, the equilibrium constant is expressed as: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ where $[C]$, $[D]$, $[A]$, and $[B]$ symbolize the molar concentrations of the substances involved.

• A large $K_c$ value suggests that products are favored at equilibrium.
• A small $K_c$ indicates that reactants are favored.

It is important to remember that $K_c$ values are constant for a specific reaction at a given temperature, but changing the temperature can shift the equilibrium, thereby altering $K_c$. Recognizing these dynamics enables chemists to control and predict chemical reactions accurately in various applications.