Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is (a) \(6.5\) (b) \(7.0\) (c) \(8.0\) (d) \(2.0\)

Step-by-Step Solution

Answer

The total number of moles at equilibrium is 7. Hence, the answer is (b) 7.0.

Step 1: Understand the Reaction

The chemical reaction for the formation of \(\mathrm{SO}_3\) from \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) is: \[ 2 \mathrm{SO}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{SO}_3 \]. We start with 4 moles of \(\mathrm{SO}_2\) and 4 moles of \(\mathrm{O}_2\).

Step 2: Determine Moles of \(\mathrm{O}_2\) Reacted

Since 25% of \(\mathrm{O}_2\) is used up, we calculate 25% of 4 moles: \[ \text{Moles of } \mathrm{O}_2 \text{ reacted} = \frac{25}{100} \times 4 = 1 \text{ mole}. \]

Step 3: Calculate Moles of Reacted \(\mathrm{SO}_2\) and Produced \(\mathrm{SO}_3\)

The stoichiometry of the reaction shows that 1 mole of \(\mathrm{O}_2\) reacts with 2 moles of \(\mathrm{SO}_2\) to produce 2 moles of \(\mathrm{SO}_3\). Thus, 2 moles of \(\mathrm{SO}_2\) react, forming 2 moles of \(\mathrm{SO}_3\).

Step 4: Calculate Remaining Moles of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\)

Subtract the moles reacted from the initial amounts: \[\text{Remaining moles of } \mathrm{SO}_2 = 4 - 2 = 2 \text{ moles},\]\[\text{Remaining moles of } \mathrm{O}_2 = 4 - 1 = 3 \text{ moles}.\]

Step 5: Calculate Total Moles at Equilibrium

Add up all the moles present at equilibrium: \[\text{Total moles at equilibrium} = 2 \text{ (SO}_2\text{)} + 3 \text{ (O}_2\text{)} + 2 \text{ (SO}_3\text{)} = 7 \text{ moles}.\]

Step 6: Choose Correct Option

Based on the calculation, the total number of moles at equilibrium is 7 moles, which corresponds to option (b).

Key Concepts

Moles calculation

Calculating moles involves determining the quantity of a substance needed or produced in a chemical reaction. In this exercise, we start with a known amount: 4 moles each of sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)). Moles are a convenient unit because they relate directly to the subatomic scale of atoms and molecules, allowing us to balance chemical reactions accurately.
To understand how many moles react, we calculate the specific percentage given. For oxygen, 25% of 4 moles is used, so 1 mole of \(\mathrm{O}_2\) reacts. This simple calculation allows us to further analyze what happens to the other reactant and product, determining overall changes in the system.

Stoichiometry

Stoichiometry is the method by which chemists relate the amounts of substances in a reaction. It uses the balanced chemical equation: for this reaction, \(2 \mathrm{SO}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{SO}_3\). This equation tells us the proportion of reactants to products.
- So, when 1 mole of \(\mathrm{O}_2\) reacts, it requires 2 moles of \(\mathrm{SO}_2\).- And, in return, this reaction produces 2 moles of sulfur trioxide (\(\mathrm{SO}_3\)).
The stoichiometry dictates that no matter how much substance you start with, the ratios remain constant. This principle is foundational for planning reactions in labs and industry to ensure efficiency and expected outcomes.

Reaction yield

Reaction yield is the amount of product formed in a chemical reaction. It can be influenced by several factors, including temperature, pressure, and the presence of a catalyst. In ideal conditions, the yield is calculated based on the stoichiometric ratios from the balanced equation.
In our scenario, we calculated the formation of 2 moles of \(\mathrm{SO}_3\) based on the reaction of 1 mole of \(\mathrm{O}_2\). This yield not only depends on stoichiometry but also on what occurs at equilibrium. Hence, understanding yield helps in predicting how much of the reactants convert to products under specific conditions.

Equilibrium state

When a reaction reaches equilibrium, it means the rates of the forward reaction and the reverse reaction are equal. This leads to constant concentrations of reactants and products over time. In our example, we initially had 4 moles each of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\), but equilibrium shifts the amounts.
- At equilibrium, 1 mole of \(\mathrm{O}_2\) has reacted, leaving 3 moles.- Similarly, \(\mathrm{SO}_2\) decreases by 2 moles, resulting in 2 moles.- The production of \(\mathrm{SO}_3\) yields 2 moles.
So at equilibrium, all of these moles add up to 7. Understanding equilibrium helps better grasp reaction dynamics and optimize conditions for desired outcomes, crucial for both academic and industrial chemistry.