Problem 11 - Chemistry: Principles and Reactions

Question

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)^{3}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}}{\left(P_{\mathrm{C}, \mathrm{H}_{4}}\right)\left(P_{\mathrm{O}_{2}}\right)^{5}}\) (b) \(K=\frac{P_{\mathrm{C}_{0} \mathrm{H}_{12}}}{\left(P_{\mathrm{C}, \mathrm{H}_{6}}\right)^{2}}\) (c) \(K=\frac{\left[\mathrm{PO}_{4}^{3-}\right]\left[\mathrm{H}^{+}\right]^{3}}{\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]}\) (d) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)}{\left[\mathrm{CO}_{3}^{2-}\right]\left[\mathrm{H}^{+}\right]^{2}}\)

Step-by-Step Solution

Answer

Question: Based on the given equilibrium expressions, provide the balanced chemical equations for each. a) \(K = \frac{(P_{\mathrm{CO}_{2}})^{3}(P_{\mathrm{H}_{2}\mathrm{O})^{4}}{(P_{\mathrm{C},\mathrm{H}_{4}})(P_{\mathrm{O}_{2}})^{5}}\) b) \(K = \frac{P_{\mathrm{C}_{0}\mathrm{H}_{12}}}{(P_{\mathrm{C},\mathrm{H}_{6}})^2}\) c) \(K = \frac{[\mathrm{PO}_{4}^{3-}][\mathrm{H}^{+}]^3}{[\mathrm{H}_{3}\mathrm{PO}_{4}]}\) d) \(K = \frac{(P_{\mathrm{CO}_{2}})(P_{\mathrm{H}_{2}\mathrm{O})}{[\mathrm{CO}_{3}^{2-}][\mathrm{H}^{+}]^2}\) Answer: a) \(\mathrm{CH_{4}}(g) + 5\,\mathrm{O}_{2}(g) \longleftrightarrow 3\,\mathrm{CO}_{2}(g) + 4\,\mathrm{H}_{2}\mathrm{O}(g)\) b) \(2\,\mathrm{CH}_{6}(g) \longleftrightarrow \mathrm{C}_{2}\mathrm{H}_{12}(g)\) c) \(\mathrm{H}_{3}\mathrm{PO}_{4}(aq) \longleftrightarrow \mathrm{PO}_{4}^{3-}(aq) + 3\,\mathrm{H}^{+}(aq)\) d) \(\mathrm{CO}_{3}^{2-}(aq) + 2\,\mathrm{H}^{+}(aq) \longleftrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g)\)

Step 1: (a) Identifying reactants and products

For expression (a), \(K = \frac{(P_{\mathrm{CO}_{2}})^{3}(P_{\mathrm{H}_{2}\mathrm{O})^{4}}{(P_{\mathrm{C},\mathrm{H}_{4}})(P_{\mathrm{O}_{2}})^{5}}\), we need to find the balanced chemical equation that would result in this expression for the equilibrium constant \(K\).

Step 2: (a) Writing the chemical equation

Based on the equilibrium constant expression, we can write the balanced chemical equation as: \(\mathrm{CH_{4}}(g) + 5\,\mathrm{O}_{2}(g) \longleftrightarrow 3\,\mathrm{CO}_{2}(g) + 4\,\mathrm{H}_{2}\mathrm{O}(g)\)

Step 3: (b) Identifying reactants and products

For expression (b), \(K = \frac{P_{\mathrm{C}_{0}\mathrm{H}_{12}}}{(P_{\mathrm{C},\mathrm{H}_{6}})^2}\), we need to find the balanced chemical equation that would result in this expression for the equilibrium constant (K).

Step 4: (b) Writing the chemical equation

Based on the equilibrium constant expression, we can write the balanced chemical equation as: \(2\,\mathrm{CH}_{6}(g) \longleftrightarrow \mathrm{C}_{2}\mathrm{H}_{12}(g)\)

Step 5: (c) Identifying reactants and products

For expression (c), \(K = \frac{[\mathrm{PO}_{4}^{3-}][\mathrm{H}^{+}]^3}{[\mathrm{H}_{3}\mathrm{PO}_{4}]}\), we need to find the balanced chemical equation that would result in this expression for the equilibrium constant (K).

Step 6: (c) Writing the chemical equation

Based on the equilibrium constant expression, we can write the balanced chemical equation as: \(\mathrm{H}_{3}\mathrm{PO}_{4}(aq) \longleftrightarrow \mathrm{PO}_{4}^{3-}(aq) + 3\,\mathrm{H}^{+}(aq)\)

Step 7: (d) Identifying reactants and products

For expression (d), \(K = \frac{(P_{\mathrm{CO}_{2}})(P_{\mathrm{H}_{2}\mathrm{O})}{[\mathrm{CO}_{3}^{2-}][\mathrm{H}^{+}]^2}\), we need to find the balanced chemical equation that would result in this expression for the equilibrium constant (K).

Step 8: (d) Writing the chemical equation

Based on the equilibrium constant expression, we can write the balanced chemical equation as: \(\mathrm{CO}_{3}^{2-}(aq) + 2\,\mathrm{H}^{+}(aq) \longleftrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g)\)

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a measure that helps us understand the direction and extent of a chemical reaction at equilibrium. At this point, the rate of the reaction in the forward direction equals the rate of the reaction in the reverse direction. The value of \( K \) can either be expressed in terms of concentrations (using brackets \( [ ] \)) or partial pressures (using \( P \)). For example, if we have a reaction involving gases, we often use partial pressures in the expression for \( K \).

The equilibrium constant expression is derived from the balanced chemical equation of the reaction. For instance, in the equation \( \mathrm{CH}_4(g) + 5 \mathrm{O}_2(g) \leftrightarrow 3 \mathrm{CO}_2(g) + 4 \mathrm{H}_2\mathrm{O}(g) \), the expression becomes \( K = \frac{(P_{\mathrm{CO}_2})^3(P_{\mathrm{H}_2\mathrm{O}})^4}{(P_{\mathrm{CH}_4})(P_{\mathrm{O}_2})^5} \). The products are multiplied and divided by the reactants, with each concentration or pressure raised to the power of their respective coefficients from the balanced equation. This setup helps in predicting the composition of the system at equilibrium by comparing the value of \( K \):

• If \( K \gg 1 \), the reaction favors products at equilibrium.
• If \( K \ll 1 \), the reaction favors reactants.
• If \( K \approx 1 \), significant amounts of both reactants and products are present.

Thus, understanding and calculating \( K \) is crucial for controlling chemical reactions in industries and laboratories.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They involve reactants, which are the starting substances, and products, which are formed as a result of the reaction. Writing accurate chemical equations is essential because it helps us visually understand the reactants involved and predict the amount of products that will be made.

A chemical equation consists of chemical formulas for all the reactants and products, along with their states and sometimes the conditions under which the reaction occurs. For instance, in the reaction \( 2 \mathrm{CH}_6(g) \leftrightarrow \mathrm{C}_2\mathrm{H}_{12}(g) \), we see two molecules of \( \mathrm{CH}_6 \) gas are needed to produce one molecule of \( \mathrm{C}_2\mathrm{H}_{12} \) gas. The arrow (\( \leftrightarrow \)) indicates the reversibility, meaning products can react to form reactants over time.

By understanding chemical equations, we can:

• Determine the stoichiometry, which describes the proportional relationships between reactants and products.
• Gain insights into the energetics, kinetics, and mechanism of the reaction.
• Utilize this information to optimize chemical processes and make predictions about how varying conditions might affect the reaction outcome.

Balanced Chemical Reactions

Balancing chemical reactions is a fundamental skill in chemistry. It ensures the conservation of mass in a reaction—meaning the number of atoms for each element in the reactants side equals the number of atoms in the products side. It adheres to the law of conservation of mass and is crucial for accurate stoichiometry calculations.

To balance a chemical equation, coefficients are placed in front of the chemical formulas to equalize the number of atoms of each element on both sides of the equation. Consider the reaction of methane combustion: \( \mathrm{CH}_4(g) + 5 \mathrm{O}_2(g) \leftrightarrow 3 \mathrm{CO}_2(g) + 4 \mathrm{H}_2\mathrm{O}(g) \). Here, we ensure that:

• 1 C atom from \( \mathrm{CH}_4 \) equals 3 C atoms from 3 \( \mathrm{CO}_2 \).
• 4 H atoms from \( \mathrm{CH}_4 \) equals 8 H atoms from 4 \( \mathrm{H}_2\mathrm{O} \).
• 10 O atoms from 5 \( \mathrm{O}_2 \) equals 10 O atoms from 3 \( \mathrm{CO}_2 \) and 4 \( \mathrm{H}_2\mathrm{O} \).

Balancing ensures that chemical equations provide the right basis for calculating quantities and understanding mechanisms in reactions. It is an essential step for any chemical analysis or industrial chemical process, helping to prevent errors and optimize resource use.