Problem 5 - Chemistry in 30 days

Question

At a certain temperature, \(K_{C}\) is 16 for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g})+\mathrm{NO}(\mathrm{g})\). If we take 1 mol of each of the four gases in a 11 container, what would be the equilibrium concentrations of NO and \(\mathrm{NO}_{2} ?\) (a) \(1.6,0.4\) (b) \(0.4,1.6\) (c) \(1.6,0.8\) (d) \(0.8,1.6\)

Step-by-Step Solution

Answer

The correct answer is (a) NO: 1.6 M, NO2: 0.4 M.

Step 1: Write the Equilibrium Expression

For the given reaction \( \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \), the equilibrium constant \( K_C \) can be expressed as \[ K_C = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]} = 16. \]

Step 2: Initial Concentrations

We begin with 1 mol of each gas in a 1 L container. Thus, the initial concentrations are \([\mathrm{SO}_2] = 1\, \text{M}\), \([\mathrm{NO}_2] = 1\, \text{M}\), \([\mathrm{SO}_3] = 1\, \text{M}\), and \([\mathrm{NO}] = 1\, \text{M}\).

Step 3: Set Up Change in Concentrations

Let the change in concentration of \( \mathrm{SO}_2 \) and \( \mathrm{NO}_2 \) at equilibrium be \( x \). Therefore, \([\mathrm{SO}_2] = 1 - x\), \([\mathrm{NO}_2] = 1 - x\), \([\mathrm{SO}_3] = 1 + x\), and \([\mathrm{NO}] = 1 + x\).

Step 4: Substitute in Equilibrium Expression

Substitute these values into the equilibrium expression to get: \[ 16 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)}. \]

Step 5: Solve the Equation

We simplify to \[ 16 = \frac{(1 + x)^2}{(1 - x)^2}. \]Taking the square root of both sides gives: \[ 4 = \frac{1 + x}{1 - x}. \]Cross-multiply to solve for \( x \): \[ 4(1 - x) = 1 + x, \] which simplifies to \[ 4 - 4x = 1 + x. \]Solving for \( x \), we find \[ 5x = 3 \Rightarrow x = 0.6. \]

Step 6: Calculate Equilibrium Concentrations

Substitute back to find equilibrium concentrations: \([\mathrm{NO}_2] = [\mathrm{SO}_2] = 1 - x = 1 - 0.6 = 0.4 \) M and \([\mathrm{NO}] = [\mathrm{SO}_3] = 1 + x = 1 + 0.6 = 1.6 \) M.

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, often denoted as \( K_C \), plays a crucial role in understanding how a reaction proceeds in a closed system. It is a numerical value that provides insights into the ratio of the concentrations of products to reactants at equilibrium.
This ratio is maintained when a reaction has reached a state where the forward and backward reactions occur at equal rates. For a given reaction:

• \( ext{aA} + ext{bB} \rightleftharpoons ext{cC} + ext{dD} \).

The equilibrium constant is expressed as:\[K_C = \frac{[ ext{C}]^c [ ext{D}]^d}{[ ext{A}]^a [ ext{B}]^b}\]
This equation shows that \( K_C \) expresses the balance of concentrations between the reactants and products.
If \( K_C \) is greater than 1, the products are favored at equilibrium. Conversely, a value less than 1 indicates a reaction mixture that has more reactants than products at equilibrium.

Reaction Concentrations

Reaction concentrations refer to the molarity of reactants and products in a chemical reaction, which changes as the reaction progresses.
In solutions,

• Molarity is defined as the number of moles of solute per liter of solution.

For the reaction \( ext{SO}_2( ext{g}) + ext{NO}_2( ext{g}) \rightleftharpoons ext{SO}_3( ext{g}) + ext{NO}( ext{g}) \), initial concentrations were set at 1 M for each gas, being in a 1 L container.
As the reaction progresses towards equilibrium, concentrations change to minimize Gibbs free energy and reach a stable state where the concentrations of reactants and products no longer change. It is crucial to understand that these concentrations are dynamic until equilibrium is reached. The path involves changes that are dependent on the initial amounts and the equilibrium constant.

Equilibrium Expression

The equilibrium expression is a mathematical representation of the balance between reactants and products in a reaction at equilibrium.For our example, the equilibrium expression is written as:\[K_C = \frac{[ ext{SO}_3][ ext{NO}]}{[ ext{SO}_2][ ext{NO}_2]}\]
This expression is derived from the reaction stoichiometry and shows how the equilibrium constant defines the extent to which reactions proceed.
In practice, you replace the placeholders with actual concentrations at equilibrium to verify consistency with the given \( K_C \).
Joining this with known values, the equilibrium expression becomes a powerful tool for calculating unknown concentrations in the reaction mixture, which makes it an essential aspect of chemical equilibrium studies.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. This transformation can either be reversible or irreversible.
For reversible reactions, such as

• \( ext{SO}_2 + ext{NO}_2 \rightleftharpoons ext{SO}_3 + ext{NO} \),

both forward and reverse reactions occur, eventually reaching a state of equilibrium.
At equilibrium, although both reactions continue to proceed, their rates are equal, yielding no net change in the concentrations of reactants and products.
Understanding the chemical behavior of reacting elements, the conditions under which they react, and how they reach equilibrium is vital.
This knowledge aids in predicting the outcomes of reactions under various conditions, helping chemists design processes that maximize desired yields efficiently.