Question

Balance each of the following examples of heterogeneous equilibria and write each reaction quotient, \(Q_{c}:\) (a) \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) (b) \(\mathrm{KNO}_{3}(s) \rightleftharpoons \mathrm{KNO}_{2}(s)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{S}_{8}(s)+\mathrm{F}_{2}(g) \rightleftharpoons \mathrm{SF}_{6}(g)\)

Short answer

(a) \(Q_c = [\text{H}_2\text{SO}_4] / [\text{SO}_3]\) (b) \(Q_c = [\text{O}_2]\) (c) \(Q_c = [\text{SF}_6]^4 / [\text{F}_2]^4\)

Step-by-step solution

Write the balanced chemical equation for (a)

The given reaction is \(\text{H}_2\text{O}(l) + \text{SO}_3(g) \rightleftharpoons \text{H}_2\text{SO}_4(aq)\). This equation is already balanced as it stands: \(\text{1 H}_2\text{O}(l) + \text{1 SO}_3(g) \rightleftharpoons \text{1 H}_2\text{SO}_4(aq)\).

Write the reaction quotient \(Q_c\) for (a)

For the reaction \(\text{H}_2\text{O}(l) + \text{SO}_3(g) \rightleftharpoons \text{H}_2\text{SO}_4(aq)\), the reaction quotient \(Q_c\) only includes gases and aqueous species. Thus, \(Q_c = [\text{H}_2\text{SO}_4] / [\text{SO}_3]\). Note that pure liquids (like \(\text{H}_2\text{O}(l)\)) are omitted.

Write the balanced chemical equation for (b)

The given reaction is \(\text{KNO}_3(s) \rightleftharpoons \text{KNO}_2(s) + \text{O}_2(g)\). This equation is already balanced as it stands: \(\text{1 KNO}_3(s) \rightleftharpoons \text{1 KNO}_2(s) + \text{1 O}_2(g)\).

Write the reaction quotient \(Q_c\) for (b)

For the reaction \(\text{KNO}_3(s) \rightleftharpoons \text{KNO}_2(s) + \text{O}_2(g)\), the reaction quotient \(Q_c\) only includes gaseous species. Thus, \(Q_c = [\text{O}_2]\). Solids (like \(\text{KNO}_3(s)\) and \(\text{KNO}_2(s)\)) are omitted.

Write the balanced chemical equation for (c)

The given reaction is \(\text{S}_8(s) + \text{F}_2(g) \rightleftharpoons \text{SF}_6(g)\). Firstly, balance the number of fluorine atoms. There are 8 fluorine atoms on the left and 6 on the right, so multiply \(\text{SF}_6\) by 1 and \(\text{F}_2\) by 4/3 to get: \(\text{S}_8(s) + 4\text{F}_2(g) \rightleftharpoons 4\text{SF}_6(g)\)

Write the reaction quotient \(Q_c\) for (c)

For the balanced reaction \(\text{S}_8(s) + 4\text{F}_2(g) \rightleftharpoons 4\text{SF}_6(g)\), the reaction quotient \(Q_c\) only includes gases. Thus, \(Q_c = [\text{SF}_6]^4 / [\text{F}_2]^4\). Solids (like \(\text{S}_8(s)\)) are omitted.

Key concepts

Reaction Quotient

The reaction quotient, denoted as \(Q_c\), is a crucial concept in chemical equilibrium. It helps us determine the direction in which a reaction will proceed to reach equilibrium. Essentially, it’s a ratio that compares the concentrations of the products and the reactants at any point during a reaction.
To calculate \(Q_c\), we use the formula: \[ Q_c = \frac{[\text{products}]}{[\text{reactants}]} \]
Importantly, only gases and aqueous species are included in this calculation. Pure solids and liquids are omitted. This aligns with our example reactions, where only the gaseous and aqueous components were considered for \(Q_c\).
For instance, in the reaction \( \text{H}_2\text{O}(l) + \text{SO}_3(g) \rightleftharpoons \text{H}_2\text{SO}_4(aq) \), \(Q_c\) is calculated as \( [\text{H}_2\text{SO}_4] / [\text{SO}_3] \).Notice the exclusion of \( \text{H}_2\text{O}(l) \), a pure liquid, from the \(Q_c\) expression.

Balancing Chemical Equations

Balancing chemical equations is fundamental in chemistry to ensure the law of conservation of mass is respected. This principle states that matter cannot be created or destroyed in a chemical reaction.
To balance an equation, follow these steps:

• Write the unbalanced equation.
• Identify the number of atoms of each element on both sides of the equation.
• Add coefficients to balance each element one at a time.
• Verify the equation to make sure all elements are balanced.

For example, the reaction \( \text{S}_8(s) + \text{F}_2(g) \rightleftharpoons \text{SF}_6(g) \) initially appears unbalanced. We see that there are 8 sulfur atoms on both sides, but the fluorine atoms are not balanced. By adjusting the coefficients, we get: \( \text{S}_8(s) + 4 \text{F}_2(g) \rightleftharpoons 4 \text{SF}_6(g) \). Now both sulfur and fluorine atoms are balanced.

Phases in Chemical Reactions

Phases of substances play an important role in heterogeneous equilibria, which involve substances in different phases (solid, liquid, gas, or aqueous). Understanding these phases helps us correctly apply equilibrium concepts.
For instance, consider the heterogeneous reaction \( \text{KNO}_3(s) \rightleftharpoons \text{KNO}_2(s) + \text{O}_2(g) \). Here, we have a solid reactant producing a solid and a gas.
In equilibrium expressions such as \( Q_c \), only gases and aqueous species are included. That's why in our example, \(Q_c\) is represented as \( [\text{O}_2] \), ignoring the solid phases (\( \text{KNO}_3 \) and \( \text{KNO}_2 \)).
This concept helps simplify calculations and focuses on the dynamic components of the reaction.

Equilibrium Constants

The equilibrium constant, denoted as \(K_c\), is a value that expresses the ratio of the concentrations of products to reactants at equilibrium. Unlike \(Q_c\), which can be calculated at any point during the reaction, \(K_c\) specifically applies when the reaction is at equilibrium.
The formula for \(K_c\) is the same as for \(Q_c\):
\[ K_c = \frac{[\text{products}]_{eq}}{[\text{reactants}]_{eq}} \]
Since \(K_c\) is constant for a given reaction at a specific temperature, it allows us to predict whether a system is at equilibrium or if it will shift to form more products or reactants.
For example, if for the reaction \( \text{H}_2\text{SO}_4(aq) \rightleftharpoons \text{H}_2\text{O}(l) + \text{SO}_3(g) \), the \(Q_c\) is not equal to \( K_c \), the system will shift either to the products or reactants to achieve equilibrium.