Problem 100 - Chemistry: An Atoms-Focused Approach

Question

Sulfur dioxide reacts with \(\mathrm{NO}_{2},\) forming \(\mathrm{SO}_{3}\) and \(\mathrm{NO}\) : $$ \mathrm{SO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ If \(K_{c}=2.50\) for the reaction, what are the equilibrium concentrations of the products if the reaction mixture was initially \(0.50 \mathrm{M} \mathrm{SO}_{2}, 0.50 \mathrm{M} \mathrm{NO}_{2}, 0.0050 \mathrm{M} \mathrm{SO}_{3},\) and \(0.0050 M \mathrm{NO} ?\)

Step-by-Step Solution

Answer

Solution: The equilibrium concentrations of the products are 0.0125 M for both SO₃ and NO.

Step 1: Write the balanced chemical equation

First, we need to write down the balanced chemical equation given in the problem: $$ SO_2 (g) + NO_2 (g) \rightleftharpoons SO_3 (g) + NO (g) $$

Step 2: Write the expression for Kc

Now, we should write the expression for Kc using the balanced chemical equation: $$ K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]} $$

Step 3: Set up the ICE table

Set up an ICE table (Initial-Change-Equilibrium) to keep track of the changes in concentrations: ``` SO₂ + NO₂ ⇌ SO₃ + NO Initial (M): 0.50 0.50 0.0050 0.0050 Change (M): -x -x +x +x Equilibrium: 0.50-x 0.50-x 0.0050+x 0.0050+x ```

Step 4: Write the Kc expression using the ICE table

Now, substitute the equilibrium concentrations from the ICE table into the Kc expression: $$ K_c = \frac{ (0.0050 + x) (0.0050 + x) }{ (0.50 - x) (0.50 - x) } $$ Given, \(K_c = 2.50\). Plugging this value into the equation: $$ 2.50 = \frac{ (0.0050 + x) (0.0050 + x) }{ (0.50 - x) (0.50 - x) } $$

Step 5: Solve for x

To solve for x, first, we can simplify the equation as follows: $$ (0.50 - x) ^2 = \frac{ (0.0050 + x) (0.0050 + x) }{2.50} $$ Now, solve for x using any available method (quadratic formula, approximation, or numerical methods). The value we get is: $$ x ≈ 0.0075 $$

Step 6: Find the equilibrium concentrations of SO₃ and NO

Finally, use the value of x to find the equilibrium concentrations of SO₃ and NO: $$ [SO_3]_{eq} = 0.0050 + x = 0.0050 + 0.0075 = 0.0125 \,\text{M} $$ $$ [NO]_{eq} = 0.0050 + x = 0.0050 + 0.0075 = 0.0125 \,\text{M} $$ So, the equilibrium concentrations of the products are \(0.0125 \,\text{M} SO_3\) and \(0.0125 \,\text{M} NO\).

Key Concepts

ICE Table

An ICE table is a useful tool in chemistry that helps keep track of concentrations through the steps of a chemical reaction. The acronym "ICE" stands for Initial, Change, and Equilibrium, representing the stages the concentrations undergo as the reaction progresses.

In our example, we start by listing the Initial concentrations of all reactants and products given in the problem. For instance, both sulfur dioxide \((SO_2)\) and nitrogen dioxide \((NO_2)\) have initial concentrations of 0.50 M, while the products \((SO_3)\) and \((NO)\) start with 0.0050 M each.

The "Change" row indicates how the concentrations change as the reaction moves towards equilibrium. If the reactants decrease by \(x\), the products will increase by \(x\), following the stoichiometry of the balanced reaction equation. These changes are algebraic expressions that we use to represent shifts in concentration.

The "Equilibrium" row combines these calculations to show the concentrations once the system reaches equilibrium. For example, the equilibrium concentration of \([SO_3]\) is calculated as \(0.0050 + x\), accounting for the initial amount plus the increase from the reaction progress. Using these values, we can then solve for unknowns and determine equilibrium concentrations of the products.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a value that expresses the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction. This dimensionless number provides insight into the extent of the reaction.

For a given reaction, like \(SO_2 (g) + NO_2 (g) \rightleftharpoons SO_3 (g) + NO (g)\), the equilibrium expression is formulated by taking the concentration of the products, \([SO_3]\) and \([NO]\), and dividing it by the concentration of the reactants, \([SO_2]\) and \([NO_2]\). Each concentration is raised to the power of its stoichiometric coefficient from the balanced equation:

\[ K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]} \]

In this instance, we use the \(K_c\) value of 2.50 given in the problem to substitute into the equilibrium expression. This helps us determine how far the reaction will proceed before reaching equilibrium and provides a basis to calculate the changes in concentration we've outlined in the ICE table.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves using the coefficients from a balanced chemical equation to determine the ratios in which reactants are consumed and products are formed. This concept is crucial for setting up the ICE table and understanding how changes in one component affect the others in a reaction system.

In our provided reaction, the stoichiometry is simple with coefficients of 1 for all reactants and products:

\[ SO_2 (g) + NO_2 (g) \rightleftharpoons SO_3 (g) + NO (g) \]

This 1:1:1:1 ratio implies that as one mole of \(SO_2\) reacts with one mole of \(NO_2\), one mole each of \(SO_3\) and \(NO\) is produced. This proportionality guides us in setting up the correct expressions in the "Change" row of the ICE table.

Stoichiometry ensures that if there's a decrease by \(x\) in reactants, there's a corresponding increase by \(x\) in products, confirming mass balance within the reaction. Understanding these ratios helps predict the dynamic shifts in concentration until the system achieves equilibrium. By applying stoichiometry, we utilize these relationships quantitatively in calculations involving equilibrium concentrations.