Question

For the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) \(\mathrm{g}\) \(\mathrm{g}\) If we start with 2 mol. \(\mathrm{SO}_{2}\) and \(1 \mathrm{~mol} . \mathrm{O}_{2}\) in \(1 \mathrm{~L}\) flask, the mixture needs \(0.4 \mathrm{~mol} \mathrm{MnO}_{4}^{-}\) in acidic medium for the complete oxidation of \(\mathrm{SO}_{2}\). The value of \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(1 / 2\) (b) 2 (c) 1 (d) \(0.6\)

Short answer

The value of \(K_c\) is 2 (option b).

Step-by-step solution

Identify the limiting reactant

For the reaction \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), identify the limiting reactant. Start with 2 mol of \(\mathrm{SO}_{2}\) and 1 mol of \(\mathrm{O}_{2}\). According to the stoichiometry, you need 1 mol of \(\mathrm{O}_{2}\) for every 2 mol of \(\mathrm{SO}_{2}\). Therefore, neither reactant is in excess, but instead, they will react completely with each other.

Calculate moles of SO2 oxidized

The reaction of \(\mathrm{SO}_{2}\) with \(\mathrm{MnO}_4^-\) in acidic medium is given as \(5 \mathrm{SO}_{2} + 2 \mathrm{MnO}_{4}^{-} + 2\mathrm{H}_2\mathrm{O} \rightarrow 5\mathrm{SO}_4^{2-} + 2\mathrm{Mn}^{2+} + 4\mathrm{H}^+\). Given, 0.4 mol \(\mathrm{MnO}_4^-\) is used, so \(\frac{5}{2} \times 0.4 = 1\) mol of \(\mathrm{SO}_{2}\) is oxidized.

Determine Moles of SO3 formed

Initially, we had 2 moles of \(\mathrm{SO}_{2}\) and after oxidation, 1 mol of \(\mathrm{SO}_{2}\) remains to reach equilibrium. Therefore, 1 mol of \(\mathrm{SO}_{2}\) gets converted to \(\mathrm{SO}_{3}\), resulting in the formation of 1 mol of \(\mathrm{SO}_{3}\).

Calculate equilibrium concentrations

At equilibrium, 1 mol of \(\mathrm{SO}_{2}\) produces 1 mol of \(\mathrm{SO}_{3}\). With the 1 L volume, the final concentrations are: \([\mathrm{SO}_{2}] = 1 \, \text{M}, \; [\mathrm{O}_2] = 0.5 \, \text{M}, \; [\mathrm{SO}_3] = 1 \, \text{M}\).

Calculate equilibrium constant \(K_c\)

Use the expression for \(K_c\) for the reaction: \(K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 \times [\mathrm{O}_2]}\). Plug in the equilibrium concentrations: \(K_c = \frac{1^2}{1^2 \times 0.5} = 2\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time.
This occurs when the forward and reverse reactions proceed at the same rate. Understanding chemical equilibrium entails acknowledging that it's not a static state but a dynamic balance.
The equilibrium constant, denoted as \( K_c \) for reactions in solution, is a vital concept and it's defined based on the concentrations of the products and reactants at equilibrium.

• The formula for \( K_c \) for the reaction \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \) is \( K_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 \times [\mathrm{O}_2]} \).
• This ratio remains constant as long as temperature remains unchanged, irrespective of the initial concentrations, given the reaction has reached equilibrium.

By calculating \( K_c \), we can predict the extent of a reaction and whether products or reactants are favored at equilibrium.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions.
It is based on the conservation of mass where the total mass of the reactants must equal the total mass of the products.
In the given reaction, \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), stoichiometry dictates that two moles of sulfur dioxide react with one mole of oxygen to produce two moles of sulfur trioxide.

• Using stoichiometry allows us to gauge the amount of each substance that will react or be produced.
• It's crucial for determining the equilibrium concentrations and the equilibrium constant.

Understanding stoichiometry is essential for any calculations involving the amounts of reactants or products in a chemical equation.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product formed.
For the reaction \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), we start with 2 moles of \(\mathrm{SO}_2\) and 1 mole of \(\mathrm{O}_2\).
The stoichiometry of the reaction indicates the amount of \(\mathrm{O}_2\) needed to completely react with \(\mathrm{SO}_2\).

• In our example, 1 mole of \(\mathrm{O}_2\) reacts completely with 2 moles of \(\mathrm{SO}_2\), so neither is in excess.
• Understanding which reactant is limiting is key to calculating how much product can be formed and helps in determining equilibrium composition.

Identifying the limiting reactant helps in determining how far a reaction can proceed toward equilibrium.

Oxidation Reaction

An oxidation reaction is a chemical process where a substance loses electrons.
In the context of our exercise, \(\mathrm{SO}_2\) is oxidized in the presence of \(\mathrm{MnO}_4^-\) ions in acidic medium, resulting in \(\mathrm{SO}_4^{2-}\) ions.
This step is vital in understanding how much \(\mathrm{SO}_2\) is used up by another chemical entity.

• The reaction \(5 \mathrm{SO}_2 + 2 \mathrm{MnO}_4^- + 2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{SO}_4^{2-} + 2 \mathrm{Mn}^{2+} + 4 \mathrm{H}^+\) shows the conversion process.
• 0.4 moles of \(\mathrm{MnO}_4^-\) are used, indicating that it results in the oxidation of 1 mole of \(\mathrm{SO}_2\).

Oxidation reactions are integral in altering reactant concentrations, impacting the equilibrium positions in chemical equations.