Question

WEB Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Short answer

Question: Write the equilibrium constant expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) Answer: (a) \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\) (b) \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\) (c) \(K_{c}=[\mathrm{CO}_{2}]\) (d) \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\)

Step-by-step solution

Write an equilibrium constant expression for reaction (a)

For reaction (a), the given equation is: \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) To write the expression, we'll use the law of mass action. However, since \(\mathrm{H}_{2} \mathrm{O}(l)\) is a pure liquid, it will not be included in the equilibrium constant expression. So, the equilibrium constant expression is: \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\)

Write an equilibrium constant expression for reaction (b)

For reaction (b), the given equation is: \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) Now, apply the law of mass action to write the expression: \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\)

Write an equilibrium constant expression for reaction (c)

For reaction (c), the given equation is: \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) The two solids, \(\mathrm{BaCO}_{3}\) and \(\mathrm{BaO}\), will not be included in the equilibrium constant expression, so we have: \(K_{c}=[\mathrm{CO}_{2}]\)

Write an equilibrium constant expression for reaction (d)

For reaction (d), the given equation is: \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) As \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) is a solid, it will not be included in the equilibrium constant expression. Thus, we have: \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\) Now all equilibrium constant expressions for the given reactions have been found.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a beautiful balancing act within chemical reactions. When a reaction reaches the point where the rates of the forward and reverse reactions are equal, it is said to be in equilibrium. At this stage, the concentrations of reactants and products remain constant over time. This doesn't mean they are equal but rather that they stabilize at a specific ratio. This state of balance allows us to predict the concentration of various substances at equilibrium using the equilibrium constant.

Key characteristics of chemical equilibrium include:

• The system must be closed, meaning no substances are added or removed.
• It is a dynamic process where substances continue to react, but their concentrations remain unchanged.
• The equilibrium can be shifted by changes in concentration, temperature, or pressure (Le Chatelier's Principle).

Law of Mass Action

The Law of Mass Action is a fundamental principle that helps us understand and quantify chemical equilibrium. This law states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced chemical equation.

This law allows us to write the equilibrium constant expression for a reaction, which quantitatively expresses the ratio of the concentrations of products to reactants at equilibrium. For a general reaction: \[ aA + bB \rightleftharpoons cC + dD \]The equilibrium constant (K) expression is:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]This expression is crucial because:

• It helps determine the direction of the reaction; if \( Q < K \), the reaction proceeds forward, and if \( Q > K \), it proceeds backward.
• Only aqueous and gaseous reactants/products are included in the expression.
• Solids and liquids do not appear in the equilibrium constant expression because their concentrations do not change.

Gaseous Reactions

Gaseous reactions are a special category in chemical equilibria because they involve substances in the gas phase. These reactions often depend heavily on pressure and volume changes according to the ideal gas law. The equilibrium expression for gaseous reactions is similar to other reactions, but it may also include partial pressures.

For example, consider the reaction:\[ 2A(g) + B(g) \rightleftharpoons C(g) + 3D(g) \]The equilibrium constant can also be expressed in terms of partial pressures:\[ K_p = \frac{P_C \, P_D^3}{(P_A^2)(P_B)} \]Here, \( P \) denotes the partial pressure of the gases. Understanding these expressions is important because:

• The equilibrium constant in terms of partial pressures \( (K_p) \) can be related to concentration expressions \( (K_c) \) using the equation \( K_p = K_c(RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas.
• Gaseous equilibria are sensitive to pressure changes; a decrease in volume increases pressure and shifts equilibrium toward fewer gas moles, according to Le Chatelier's Principle.

Solid Reactions

Solid reactions, especially in the context of equilibrium, take a unique approach compared to gases and liquids. Solids do not have a concentration in the same sense as gases or solutions. For this reason, they are not included in the equilibrium constant expression.

When analyzing solid-state equilibria, the focus is often on the gaseous products or reactants. For instance:\[ aX(s) + bY(g) \rightleftharpoons cZ(s) + dW(g) \]The equilibrium expression will be:\[ K = \frac{[W]^d}{[Y]^b} \]Meaning:

• Only the gaseous components \( Y \) and \( W \) appear in the expression, while the solids \( X \) and \( Z \) are omitted.
• The reaction quotient \( K \) provides insights only into the change in gaseous products over time.
• Solubility product constants \( (K_{sp}) \) are often used for slightly soluble solids, focusing on the dissolved ions in the equilibrium expression.