Question

At some temperature, a gaseous mixture in a 1.00 - \(\mathrm{L}\) vessel originally contained \(1.00 \mathrm{~mol} \mathrm{SO}_{2}\) and \(5.00 \mathrm{~mol} \mathrm{O}_{2}\). When equilibrium was reached, \(77.8 \%\) of the \(\mathrm{SO}_{2}\) had been converted to \(\mathrm{SO}_{3}\). Calculate the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for this reaction at this temperature.

Short answer

The equilibrium constant \( K_c \) is approximately 52.06.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction we're considering is the formation of sulfur trioxide from sulfur dioxide and oxygen. The balanced equation is:\[ 2 ext{SO}_2 (g) + ext{O}_2 (g) ightleftharpoons 2 ext{SO}_3 (g) \]

Determine the Initial Moles and Changes

Initially, we have 1.00 mol of \( \text{SO}_2 \) and 5.00 mol of \( \text{O}_2 \). No \( \text{SO}_3 \) is present initially. At equilibrium, 77.8% of \( \text{SO}_2 \) is converted to \( \text{SO}_3 \).Thus, moles of \( \text{SO}_2 \) converted: \( 0.778 \times 1.00 = 0.778 \text{ mol} \).So, moles of \( \text{SO}_3 \) formed = 0.778 mol (since 2 moles \( \text{SO}_2 \) form 2 moles \( \text{SO}_3 \)).

Calculate the Moles at Equilibrium

The moles of \( \text{SO}_2 \) at equilibrium = Initial moles - Converted moles = \( 1.00 - 0.778 = 0.222 \text{ mol} \).The moles of \( \text{O}_2 \) at equilibrium = Initial moles - \( \frac{1}{2} \times \text{moles of } \text{SO}_2 \text{ consumed} \).The moles of \( \text{O}_2 \) consumed = \( \frac{1}{2} \times 0.778 = 0.389 \text{ mol} \).Thus, moles of \( \text{O}_2 \) at equilibrium = \( 5.00 - 0.389 = 4.611 \text{ mol} \).

Calculate the Concentrations at Equilibrium

Since the total volume of the container is 1.00 L, the concentrations at equilibrium will be the same as the moles at equilibrium:- \( [\text{SO}_2] = \frac{0.222 \text{ mol}}{1.00 \text{ L}} = 0.222 \text{ M} \)- \( [\text{O}_2] = \frac{4.611 \text{ mol}}{1.00 \text{ L}} = 4.611 \text{ M} \)- \( [\text{SO}_3] = \frac{0.778 \text{ mol}}{1.00 \text{ L}} = 0.778 \text{ M} \)

Write the Expression for Kc

The equilibrium constant for the reaction \( K_c \) is given by the expression:\[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} \]

Calculate the Value of Kc

Substitute the equilibrium concentrations into the expression for \( K_c \):\[ K_c = \frac{(0.778)^2}{(0.222)^2 \, (4.611)} \]Calculating this gives:\[ K_c = \frac{0.605284}{0.011623764} \approx 52.06 \]

Key concepts

Chemical Equilibrium

When we talk about chemical equilibrium, we are discussing the point in a chemical reaction where the forward and reverse reactions occur at the same rate. This means that the concentrations of the reactants and products remain constant over time.
In our example, the conversion of sulfur dioxide (\(\text{SO}_2\)) to sulfur trioxide (\(\text{SO}_3\)) has reached a state of equilibrium. At this point, both the forward reaction (formation of \(\text{SO}_3\)) and reverse reaction (decomposition back to \(\text{SO}_2\) and oxygen) happen simultaneously, maintaining a balance.
The equilibrium constant, \(K_c\), is a crucial value that reflects the ratio of the concentrations of products to reactants at equilibrium. It does not depend on the initial concentrations but is influenced by temperature. This constant helps us predict the position of equilibrium and the extent of the reaction.
A larger \(K_c\) value suggests that, at equilibrium, the products are favored, common in reactions that go nearly to completion. Conversely, a smaller \(K_c\) points to reactant favor, meaning the reaction does not proceed far before reaching equilibrium.

Reaction Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's built on the law of conservation of mass, where the quantities of each element remain constant during the reaction.
In the reaction \(2\text{SO}_2 + \text{O}_2 \leftrightarrow 2\text{SO}_3\), stoichiometry helps us track the conversion of \(\text{SO}_2\) to \(\text{SO}_3\). According to the balanced equation, two moles of \(\text{SO}_2\) react with one mole of \(\text{O}_2\) to produce two moles of \(\text{SO}_3\).
While 77.8% of \(\text{SO}_2\) converts to \(\text{SO}_3\), stoichiometry ensures we use precise coefficients from the balanced equation to determine the changes in moles of other species. Knowing this is vital to calculating the exact concentrations needed in equilibrium expressions.

Concentration Calculations

Understanding concentration is pivotal for calculating equilibrium constants. Concentrations are typically expressed in molarity (\(M\)), which is moles of solute per liter of solution.
For this problem, we used the initial moles of reactants to determine their concentrations at equilibrium. This involved calculating the number of moles converted during the reaction and then determining the remaining and formed moles at equilibrium.
For example, we began with 1.00 mol of \(\text{SO}_2\) and 5.00 mol of \(\text{O}_2\), in a 1-liter vessel:

• 0.778 mol of \(\text{SO}_2\) was converted to 0.778 mol of \(\text{SO}_3\)
• Remaining \(\text{SO}_2\): 0.222 mol
• Remaining \(\text{O}_2\): 4.611 mol

Since the mixture was in a 1.00-liter vessel, these mole values directly translate to molarity, enabling us to plug into the \(K_c\) expression, as seen in step 6: \[ K_c = \frac{(0.778)^2}{(0.222)^2 \times 4.611}\]