Question

At \(500^{\circ} \mathrm{C}\), the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g) $$ has \(K_{\text {eq }}=81\). What will be the equilibrium concentrations of the four gases if the two reactants begin with equal concentrations of \(0.50 \mathrm{M}\) ?

Short answer

At equilibrium, [SO2] = 0.05 M, [NO2] = 0.05 M, [NO] = 0.45 M, [SO3] = 0.45 M.

Step-by-step solution

Write the Equilibrium Expression

The equilibrium constant expression for the reaction: \( \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{NO}(g) + \text{SO}_3(g) \) is given by: \[ K_{eq} = \frac{[\text{NO}][\text{SO}_3]}{[\text{SO}_2][\text{NO}_2]} = 81 \]

Set Up the Initial Concentrations

Assume the initial concentrations are equal, each is 0.50 M:\( [\text{SO}_2] = [\text{NO}_2] = 0.50 \text{M} \)and initially, the products are 0:\( [\text{NO}] = [\text{SO}_3] = 0 \text{M} \).

Define the Change in Concentrations

Let the change in the concentration of SO2 and NO2 be -x. Then the change in NO and SO3 will be +x. Thus, at equilibrium:\( [\text{SO}_2] = [\text{NO}_2] = 0.50 - x \)\( [\text{NO}] = [\text{SO}_3] = x \).

Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium constant expression:\[ 81 = \frac{x \times x}{(0.50 - x)(0.50 - x)} = \frac{x^2}{(0.50 - x)^2} \] Solve for x.

Solve the Quadratic Equation

Taking the square root of both sides:\( 9 = \frac{x}{0.50 - x} \) and \( 9(0.50 - x) = x \)\( 4.5 - 9x = x \) Add 9x to both sides: \( 4.5 = 10x \)\( x = 0.45 \).

Calculate Equilibrium Concentrations

Substitute x back into the expressions:\( [\text{NO}] = [\text{SO}_3] = 0.45 \text{M} \)\( [\text{SO}_2] = [\text{NO}_2] = 0.50 - 0.45 = 0.05 \text{M} \).

Key concepts

equilibrium constant

In chemical equilibrium calculations, the equilibrium constant, denoted as \( K_{eq} \), is crucial. It's a numeric value that indicates the ratio of product concentrations to reactant concentrations at equilibrium.
For the reaction:
$$\text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{NO}(g) + \text{SO}_3(g)$$
the equilibrium constant expression is:
\[ K_{eq} = \frac{[\text{NO}][\text{SO}_3]}{[\text{SO}_2][\text{NO}_2]} \]
Here, \( K_{eq} \) is 81. This means the products (NO and SO₃) are favored at equilibrium.
To use \( K_{eq} \) in calculations, equate it to the ratio of concentrations at equilibrium. Plug known values into this equation to find unknown concentrations.

initial concentration

Initial concentration refers to how much of each substance is present at the start before any reaction has occurred.
In our example, both SO₂ and NO₂ start with 0.50 M concentrations, while NO and SO₃ start at 0 M.
These initial concentrations are crucial for setting up the 'ICE' table (Initial, Change, Equilibrium) used in equilibrium problems.
When you know initial concentrations, you can calculate the changes due to the reaction and finally, the equilibrium concentrations.

quadratic equation

Sometimes, to find equilibrium concentrations, you must solve a quadratic equation.
In our example, inserting equilibrium concentrations into the equilibrium expression leads us to solve:
\[ 81 = \frac{x^2}{(0.50 - x)^2} \]
Taking the square root of both sides, we get:
\[ 9 = \frac{x}{0.50 - x} \]
Cross-multiplying and rearranging terms gives:
\[ 4.5 - 9x = x \]
Adding \( 9x \) to both sides results in:
\[ 4.5 = 10x \]
Solving for \( x \), we find:
\[ x = 0.45 \]
This approach—often needed in equilibrium calculations—helps solve for the unknown, giving us the equilibrium concentrations.

reaction stoichiometry

Stoichiometry relates to the quantities of reactants and products in a balanced chemical reaction.
Using stoichiometry, for every mole of SO₂ and NO₂ that reacts, 1 mole of NO and SO₃ forms.
In our problem, if the change in concentration of SO₂ and NO₂ is \( -x \), the change in NO and SO₃ must be \( +x \). This gives us the stoichiometric relationship:
\[ [\text{SO}_2] = [\text{NO}_2] = 0.50 - x \]
\[ [\text{NO}] = [\text{SO}_3] = x \]
Once \( x \) is determined, these relationships help calculate the equilibrium concentrations, ensuring that the reaction's stoichiometry is maintained throughout the process.