Question

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and for \(K_{P}\), if applicable, for the following reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

Short answer

For each reaction, write \(K_c\) excluding solids and liquids; use \(K_P\) for gases only.

Step-by-step solution

Understanding Equilibrium Constant Expressions

The equilibrium constant expression for a reaction depends on the states of matter of the reactants and products involved. For gases, we use partial pressures to write the equilibrium expressions, which are denoted as \(K_P\). For substances in aqueous or gaseous phases, we use concentrations, denoted as \(K_c\). Pure solids and liquids are not included in the equilibrium expression.

Reaction (a) \(K_c\) Expression

For the reaction \(2 \mathrm{NO}_2(g) + 7 \mathrm{H}_2(g) \rightleftarrows 2 \mathrm{NH}_3(g) + 4 \mathrm{H}_2O(l)\), the equilibrium constant expression for \(K_c\) considers only gases. Since \(\mathrm{H}_2O\) is a liquid, it is not included:\[K_c = \frac{{[\mathrm{NH}_3]^2}}{{[\mathrm{NO}_2]^2 [\mathrm{H}_2]^7}}\]

Reaction (a) \(K_P\) Expression

The \(K_P\) expression for the same reaction involves the partial pressures of gaseous reactants and products:\[K_P = \frac{{(P_{\mathrm{NH}_3})^2}}{{(P_{\mathrm{NO}_2})^2 (P_{\mathrm{H}_2})^7}}\]

Reaction (b) \(K_c\) Expression

For \(2 \mathrm{ZnS}(s) + 3 \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{ZnO}(s) + 2 \mathrm{SO}_2(g)\), since solids are involved, only gaseous species \(\mathrm{O}_2\) and \(\mathrm{SO}_2\) are considered in the \(K_c\) expression:\[K_c = \frac{{[\mathrm{SO}_2]^2}}{{[\mathrm{O}_2]^3}}\]

Reaction (b) \(K_P\) Expression

For the gaseous components in the reaction, \(K_P\) can be written using partial pressures:\[K_P = \frac{{(P_{\mathrm{SO}_2})^2}}{{(P_{\mathrm{O}_2})^3}}\]

Reaction (c) \(K_c\) Expression

For \(\mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftarrows 2 \mathrm{CO}(g)\), the solid is not included, so the \(K_c\) expression involves only the gaseous reactants and products:\[K_c = \frac{{[\mathrm{CO}]^2}}{{[\mathrm{CO}_2]}}\]

Reaction (c) \(K_P\) Expression

The \(K_P\) expression focuses on partial pressures for the gases:\[K_P = \frac{{(P_{\mathrm{CO}})^2}}{{(P_{\mathrm{CO}_2})}}\]

Reaction (d) \(K_c\) Expression

For the aqueous reaction \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(aq) \rightleftarrows \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-(aq) + \mathrm{H}^+(aq)\), the \(K_c\) expression is based on concentrations of all species as they are all in aqueous state:\[K_c = \frac{{[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-][\mathrm{H}^+]}}{{[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}]}}\]

No \(K_P\) for Reaction (d)

For the reaction in aqueous solution, there is no \(K_P\) expression because \(K_P\) is only applied to gas phase reactions.

Key concepts

Kc (equilibrium constant with concentrations)

The equilibrium constant expression with concentrations, represented by \( K_c \), is crucial for understanding how chemical reactions behave when they reach equilibrium. Equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. In these scenarios, the concentration of reactants and products remains constant over time.The \( K_c \) expression is derived from the balanced chemical equation and it only includes the concentrations of substances in the gaseous or aqueous state. Pure solids and liquids are omitted from this expression. For instance, for the reaction: \[ 2 \mathrm{NO}_2(g) + 7 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) + 4 \mathrm{H}_2O(l) \] the \( K_c \) expression will be:\[K_c = \frac{{[\mathrm{NH}_3]^2}}{{[\mathrm{NO}_2]^2 [\mathrm{H}_2]^7}} \]Here, \([\mathrm{NH}_3]\), \([\mathrm{NO}_2]\), and \([\mathrm{H}_2]\) represent the concentrations of the gases only, while water is omitted because it is a liquid.

Kp (equilibrium constant with pressures)

Sometimes, a chemical reaction involving gases is easier to work with when expressed using partial pressures rather than concentrations. This is where \( K_p \) comes into play. The equilibrium constant expression \( K_p \) uses the partial pressures of each gaseous species.For the same reaction as above:\[ 2 \mathrm{NO}_2(g) + 7 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) + 4 \mathrm{H}_2O(l) \] the \( K_p \) expression becomes:\[ K_p = \frac{{(P_{\mathrm{NH}_3})^2}}{{(P_{\mathrm{NO}_2})^2 (P_{\mathrm{H}_2})^7}} \]Again, note how only the gaseous substances are included, emphasizing that the use of \( K_p \) is particularly helpful in situations involving gas phase equilibria, as pressure is often a more natural measurement than concentration for gases.

chemical reactions in equilibrium

Chemical reactions in equilibrium occur when the forward and reverse reactions proceed at the same rate, resulting in constant concentrations of reactants and products over time. This dynamic balance means that the reaction has not "stopped" but rather reached a state where both processes are equal.In writing equilibrium expressions:

• Identify whether the substances are gases, liquids, solids, or aqueous.
• Include only gaseous and aqueous reactants/products in \( K_c \) and \( K_p \) unless only one type applies.
• The coefficients in the balanced equation become exponents in the equilibrium expressions.

Observing how equilibrium is achieved and maintained allows scientists to manipulate reaction conditions to favor the formation of either products or reactants, a principle known as Le Châtelier's Principle.

role of states of matter in equilibrium expressions

The states of matter significantly influence which components are included in the equilibrium expressions \( K_c \) and \( K_p \). Understanding this role is crucial for accurately establishing these expressions.

• **Gases** contribute using their concentrations in \( K_c \) and their partial pressures in \( K_p \).
• **Aqueous solutions** use concentrations to form \( K_c \) expressions.
• **Pure solids and liquids** do not appear in these expressions because their concentrations do not change with reaction progression. In equilibrium expressions, their activity is considered constant and thus not included.

Consider the reaction:\[ \mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) \] The \( K_c \) and \( K_p \) use only the gaseous reactants and products:\[ K_c = \frac{{[\mathrm{CO}]^2}}{{[\mathrm{CO}_2]}} \quad \text{and} \quad K_p = \frac{{(P_{\mathrm{CO}})^2}}{{(P_{\mathrm{CO}_2})}} \]Understanding these subtle nuances is essential for setting up correct and meaningful equilibrium constant expressions that accurately reflect the chemical system you are studying.