Problem 4 - General Chemistry: Principles and Modern Applications

Question

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Step-by-Step Solution

Answer

The equilibrium constant expressions for the given reactions are: \n(a) \(K_{\mathrm{p}} = \frac{(P_{CH4}) \cdot (P_{H2S})^2}{(P_{CS2}) \cdot (P_{H2})^4}\) \n(b) \(K_{\mathrm{p}}=(P_{O2})^{1/2}\) \n(c) \(K_{\mathrm{p}}=(P_{CO2}) \cdot (P_{H2O})\)

Step 1: Writing the equilibrium constant for reaction (a).

The reaction is: \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2}\mathrm{S}(\mathrm{g})\). Apply the formula for \(K_{\mathrm{p}}\): \(K_{\mathrm{p}} = \frac{(P_{CH4})^1 \cdot (P_{H2S})^2}{(P_{CS2})^1 \cdot (P_{H2})^4}\)

Step 2: Writing the equilibrium constant for reaction (b).

The reaction is: \(\mathrm{Ag}_{2}\mathrm{O}(\mathrm{s}) \rightleftharpoons 2\mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\). Notice that solids are omitted. Thus, \(K_{\mathrm{p}}\) expression is: \(K_{\mathrm{p}}=(P_{O2})^{1/2}\)

Step 3: Writing the equilibrium constant for reaction (c).

The reaction is: \(2\mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\mathrm{Na}_{2}\mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\). Notice how we can omit the substances that are solid. Thus, \(K_{\mathrm{p}}\) expression is: \(K_{\mathrm{p}}=(P_{CO2})^1 \cdot (P_{H2O})^1\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is an essential concept in chemistry that describes a state in which the concentrations of reactants and products remain constant over time. This happens when the forward and reverse reactions occur at the same rate.

At equilibrium, no net change is visible, although individual molecules continuously react. The use of equilibrium constants helps in quantifying the ratio of concentrations of products to reactants.

• It allows chemists to predict the direction of reactions
• Enables understanding of reaction conditions
• Helps calculate the concentrations of different species at equilibrium

Chemical equilibrium is dynamic, not static. Though it looks like nothing is happening, reactions happen both ways. It's useful to remember that different reactions reach equilibrium at different concentrations and conditions.

Pressure-based Equilibrium Constants

In chemical reactions involving gases, equilibrium constants can be expressed in terms of pressure, referred to as pressure-based equilibrium constants, or \( K_{p} \).

These constants are derived from the concentrations of gaseous reactants and products, converted using the ideal gas law. Understanding \( K_{p} \) is crucial for reactions in which gases play a significant role.

• \( K_{p} \) is calculated using partial pressures
• Expresses the ratio, similar to concentration-based \( K_{c} \)
• \( K_{p} \) is useful in predicting the behavior of gas-phase reactions

To solve for \( K_{p} \), use the equilibrium expression by inserting each species' partial pressure raised to the power of its stoichiometric coefficient. This approach provides insight into how pressure changes impact reaction equilibria.

Phase Exclusion in Equilibrium Expressions

In equilibrium expressions, only gases and aqueous solutions are generally included, while solids and liquids are excluded. Phase exclusion, specifically the exclusion of solids and liquids, is due to their constant concentrations.

Solids and liquids don't affect the position of equilibrium, as their activities (or effective concentrations) are considered to be constant.

• Solids have constant concentration under given conditions
• Liquids are also omitted if they are pure solvents
• Gas and aqueous phases primarily determine the equilibrium

The simplification from phase exclusion results in easier calculations. Focusing on gaseous and aqueous species narrows the aspects affecting equilibrium. This principle was crucial in the steps of the original solutions, allowing \( K_{p} \) to focus on the relevant measurable quantities.