Problem 10 - Chemistry: Principles and Reactions

Question

Given the following descriptions of reversible reactions, write a balanced net ionic equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Liquid acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) is in equilibrium with its vapor. (b) Hydrogen gas reduces nitrogen dioxide gas to form ammonia and steam. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

Step-by-Step Solution

Answer

Question: Write the equilibrium constant expressions for the following reactions: (a) Liquid acetone in equilibrium with its vapor (b) Hydrogen gas reduces nitrogen dioxide gas to form ammonia and steam (c) Hydrogen sulfide gas bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions. Answer: (a) $$K_p = \mathrm{P_{C_{3}H_{6}O (g)}}$$ (b) $$K_p = \frac{\mathrm{P^2_{NH_{3}}\,\cdot\,P_{H_{2}O}}}{\mathrm{P_{H_{2}}\,\cdot\,P^2_{NO_{2}}}}$$ (c) $$K_c = \frac{\mathrm{[H^+]^2}}{\mathrm{[H_{2}S (g)]\,\cdot\,[Pb^{2+}]}}$$

Step 1: Write the balanced net ionic equation.

The given reaction is the vaporization of liquid acetone to form acetone vapor. The balanced net ionic equation can be written as: $$\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O (l)} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O (g)}$$

Step 2: Write the equilibrium constant expression.

For this reaction, the equilibrium constant \((K)\) is expressed in terms of partial pressures, \(K_p\): $$K_p = \frac{\mathrm{P_{C_{3}H_{6}O (g)}}}{\mathrm{P_{C_{3}H_{6}O (l)}}}$$ However, the pressure of the liquid is not included in the calculation, so we can simplify this expression to: $$K_p = \mathrm{P_{C_{3}H_{6}O (g)}}$$ (b) Hydrogen gas reduces nitrogen dioxide gas to form ammonia and steam

Step 1: Write the balanced net ionic equation.

First, we write the unbalanced chemical equation for the reaction: $$\mathrm{H}_{2 (\mathrm{g})} + \mathrm{NO}_{2 (\mathrm{g})} \rightleftharpoons \mathrm{NH}_{3 (\mathrm{g})} + \mathrm{H}_{2} \mathrm{O (g)}$$ To balance the equation, we place a coefficient of 2 in front of \(\mathrm{NO}_2\) and \(\mathrm{NH}_3\). The balanced net ionic equation becomes: $$\mathrm{H}_{2 (\mathrm{g})} + 2\,\mathrm{NO}_{2 (\mathrm{g})} \rightleftharpoons 2\,\mathrm{NH}_{3 (\mathrm{g})} + \mathrm{H}_{2} \mathrm{O (g)}$$

Step 2: Write the equilibrium constant expression.

The equilibrium constant \((K)\) can be expressed in terms of partial pressures, \(K_p\): $$K_p = \frac{\mathrm{P^2_{NH_{3}}\,\cdot\,P_{H_{2}O}}}{\mathrm{P_{H_{2}}\,\cdot\,P^2_{NO_{2}}}}$$ (c) Hydrogen sulfide gas bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions

Step 1: Write the balanced net ionic equation.

First, we write the unbalanced chemical equation for the reaction: $$\mathrm{H}_{2 \mathrm{S (g)}} + \mathrm{Pb^{2+} (aq)} \rightleftharpoons \mathrm{PbS (s)} + \mathrm{H}^{+} \mathrm{(aq)}$$ The equation has to be balanced in terms of charges too. Thus, we need to place a coefficient of 2 in front of \(\mathrm{H}^+\): $$\mathrm{H}_{2 \mathrm{S (g)}} + \mathrm{Pb^{2+} (aq)} \rightleftharpoons \mathrm{PbS (s)} + 2\,\mathrm{H}^{+} \mathrm{(aq)}$$

Step 2: Write the equilibrium constant expression.

The equilibrium constant \((K)\) can be expressed in terms of concentrations, \(K_c\): $$K_c = \frac{\mathrm{[PbS (s)]\,\cdot\,[H^+]^2}}{\mathrm{[H_{2}S (g)]\,\cdot\,[Pb^{2+}]}}$$ However, the concentration of the solid, \(\mathrm{[PbS(s)]}\), is omitted from the expression, so we can simplify it to: $$K_c = \frac{\mathrm{[H^+]^2}}{\mathrm{[H_{2}S (g)]\,\cdot\,[Pb^{2+}]}}$$

Key Concepts

Net Ionic Equation

Understanding the net ionic equation is crucial when studying chemical reactions in solution. A net ionic equation represents only the species that actually participate in the reaction, eliminating the spectator ions that remain unchanged. It simplifies the overall chemical equation to focus on the core changes occurring during the reaction.

The process of deriving the net ionic equation involves writing the balanced molecular equation, dissociating all strong electrolytes into their respective ions, and then cancelling out the common ions on both sides to reveal the net change.

For example, in the vaporization of liquid acetone, the balanced net ionic equation depicts the phase change from liquid to gas:\[\begin{equation}\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O (l)} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O (g)}\end{equation}\]This equation highlights the pure transformation without any ion dissociation, emphasizing the importance of context in determining the nature of the net ionic equation.

Vaporization of Liquid Acetone

The physical transformation from liquid to gas is known as vaporization, and it is a reversible process often encountered in chemistry. When liquid acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O_{(l)}}\), becomes acetone vapor, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O_{(g)}}\), this process is occurring.It is important for students to recognize that vaporization does not involve a chemical reaction but rather a physical change. The molecules in the liquid phase gain energy, typically from heat, which allows them to overcome intermolecular forces and escape into the gas phase.

When we write the equilibrium expression for a vaporization process, we exclude the concentration of the liquid phase because it is effectively constant and thus does not affect the position of equilibrium:

Reduction Reaction

Reduction reactions are chemical processes where a substance gains electrons, typically accompanied by a decrease in oxidation state. In the context of a reaction like the reduction of nitrogen dioxide (\(\mathrm{NO}_{2}\)) by hydrogen gas to form ammonia and water, the net ionic equation is:\[\begin{equation}\mathrm{H}_{2 (\mathrm{g})} + 2\mathrm{NO}_{2 (\mathrm{g})} \rightleftharpoons 2\mathrm{NH}_{3 (\mathrm{g})} + \mathrm{H}_{2}\mathrm{O (g)}\end{equation}\]This demonstrates a reduction because the nitrogen in \(\mathrm{NO}_{2}\) receiving electrons from \(\mathrm{H}_{2}\) to become \(\mathrm{NH}_{3}\).

Reduction is one half of a redox reaction, where the other half is oxidation—the loss of electrons. These two processes always occur together, leading to the mnemonic 'OIL RIG' which stands for 'Oxidation Is Loss, Reduction Is Gain' of electrons. Understanding these types of reactions is fundamental for grasping concepts in electrochemistry and metabolism in biological systems.

Precipitation Reaction

A precipitation reaction is an aqueous chemical reaction that results in the formation of an insoluble product, known as the precipitate. When a solution containing ions forms a precipitate, the solid substance is typically an insoluble ionic compound.

In the example of hydrogen sulfide gas mixed with an aqueous solution of lead(II) ions, the reaction produces a solid lead sulfide precipitate: \[\begin{equation}\mathrm{H}_{2 \mathrm{S (g)}} + \mathrm{Pb^{2+} (aq)} \rightleftharpoons \mathrm{PbS (s)} + 2\mathrm{H}^{+} \mathrm{(aq)}\end{equation}\]This equation illustrates that the lead (II) ions from the aqueous phase react with the sulfide ions from the hydrogen sulfide gas to form solid lead sulfide, which falls out of the solution as a precipitate. Precipitation reactions are widely used in qualitative chemical analysis, water treatment, and as a method for preparing solid compounds in the laboratory.