Question

Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, \(Q_{\mathrm{c}}\) (a) \(\mathrm{Al}(s)+\mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \operatorname{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right](a q)+\mathrm{H}_{2}(g) $$ (b) \(\mathrm{CO}_{2}(s) \rightleftharpoons \mathrm{CO}_{2}(g)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{5}(s) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\)

Short answer

(a) \( Q_c = \frac{[\text{Na[Al(OH)}_4](aq)] \cdot [\mathrm{H}_2(g)]^3}{[\mathrm{NaOH}(aq)]^3} \)(b) \( Q_c = [\mathrm{CO}_2(g)] \)(c) \( Q_c = \frac{[\mathrm{NO}_2(g)]^2 \cdot [\mathrm{O}_2(g)]}{1} \)

Step-by-step solution

- Write the balanced equation for part (a)

For part (a), the reaction is: \[ \mathrm{Al}(s) + 3\mathrm{NaOH}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \text{Na}\text{[Al(OH)}_4\text{]}(aq) + \mathrm{3H}_2(g) \]Identify the coefficients that balance the equation by ensuring that there are an equal number of each type of atom on both sides.

- Write the reaction quotient for part (a)

The reaction quotient for an equilibrium involving gases and aqueous species is given by: \[ Q_c = \frac{[\text{Na[Al(OH)}_4](aq)] \cdot [\mathrm{H}_2(g)]^3}{[\mathrm{NaOH}(aq)]^3} \]Note that pure solids and pure liquids are not included in the reaction quotient.

- Write the balanced equation for part (b)

For part (b), the reaction is simple: \[ \mathrm{CO}_2(s) \rightleftharpoons \mathrm{CO}_2(g) \]There is no need to balance as both sides are naturally balanced.

- Write the reaction quotient for part (b)

The reaction quotient for part (b) where a solid is in equilibrium with its gas phase is: \[ Q_c = [\mathrm{CO}_2(g)] \]Pure solids are not included in the reaction quotient.

- Write the balanced equation for part (c)

For part (c), the balanced equation is: \[ \mathrm{N}_2\mathrm{O}_5(s) \rightleftharpoons 2\mathrm{NO}_2(g) + \mathrm{O}_2(g) \]Ensure that the number of atoms for each element is the same on both sides.

- Write the reaction quotient for part (c)

The reaction quotient for part (c) involving gaseous products is: \[ Q_c = \frac{[\mathrm{NO}_2(g)]^2 \cdot [\mathrm{O}_2(g)]}{1} \]Pure solids are not included in the reaction quotient.

Key concepts

reaction quotient

In chemistry, the reaction quotient (\(Q_{\text{c}}\)) is a measure of the relative amounts of products and reactants present during a reaction at a given point in time. It is calculated using the concentrations (or partial pressures) of the reactants and products, just like the equilibrium constant (\(K_{\text{c}}\)), but applies to non-equilibrium conditions. To calculate \(Q_{\text{c}}\), you use the coefficients of the balanced chemical equation to form a ratio of the products over the reactants, each raised to the power of their respective coefficients. If you have the generalized reaction: \[ a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \], then \(Q_{\text{c}}\) is expressed as: \[ Q_{\text{c}} = \frac{[\text{C}]^c \cdot [\text{D}]^d}{[\text{A}]^a \cdot [\text{B}]^b} \] This helps to determine the direction in which the reaction will proceed to reach equilibrium.

equilibrium constants

The equilibrium constant (\(K_{\text{c}}\)) is a fixed value that relates the concentrations of the reactants and products in a reaction mixture at equilibrium. It is defined by the same expression as the reaction quotient \(Q_{\text{c}}\), but only when the reaction is at equilibrium. The magnitude of \(K_{\text{c}}\) indicates the extent to which a reaction will proceed to form products from reactants. A large \(K_{\text{c}}\) value means that at equilibrium, products are favored, while a small \(K_{\text{c}}\) implies that reactants are favored. For example, in the balanced reaction \[ \text{N}_2\text{O}_5(s) \rightleftharpoons 2\text{NO}_2(g) + \text{O}_2(g) \], the equilibrium constant \(K_{\text{c}}\) is \[ K_{\text{c}} = \frac{[\text{NO}_2(g)]^2 \cdot [\text{O}_2(g)]}{1} \] Note that \(K_{\text{c}}\) does not include the concentrations of pure solids or liquids.

balancing chemical equations

Balancing chemical equations is crucial for accurately describing chemical reactions. The goal is to have the same number of each type of atom on both sides of the equation. This adheres to the law of conservation of mass. To balance an equation, follow these steps:

• Write down the unbalanced equation.
• Count the atoms of each element in the reactants and products.
• Add coefficients to balance the number of atoms of each element.

For example, to balance the reaction \[ \text{Al}(s) + \text{NaOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{Na[Al(OH)}_4](aq) + \text{H}_2(g) \], we adjust coefficients to get \[ \text{Al}(s) + 3\text{NaOH}(aq) + 3\text{H}_2\text{O}(l) \rightleftharpoons \text{Na[Al(OH)}_4](aq) + 3\text{H}_2(g) \].

pure solids and liquids exclusion

In equilibrium expressions, pure solids and pure liquids are excluded because their concentrations remain constant. This is due to their density and molar volume staying consistent, which implies their activity is 1. For instance, in the heterogeneous equilibrium \[ \text{CO}_2(s) \rightleftharpoons \text{CO}_2(g) \], the equilibrium constant is given by \[ K_{\text{c}} = [\text{CO}_2(g)] \] Here, \(K_{\text{c}}\) does not include \( \text{CO}_2(s) \) because it is a pure solid. Similarly, in the reaction \[ \text{Al}(s) + 3\text{NaOH}(aq) + 3\text{H}_2\text{O}(l) \rightleftharpoons \text{Na[Al(OH)}_4](aq) + 3\text{H}_2(g) \], neither \( \text{Al}(s) \) nor \( \text{H}_2\text{O}(l) \) are included in the reaction quotient.