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 of all gases at equilibrium is 7.0; the answer is (b).

Step 1: Write the Balanced Chemical Equation

The reaction between sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)) to form sulfur trioxide (\(\mathrm{SO}_3\)) is represented by the balanced chemical equation: \[ 2\,\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2\,\mathrm{SO}_{3(g)} \].

Step 2: Initial Moles Provided

Initially, there are 4 moles of \(\mathrm{SO}_2\) and 4 moles of \(\mathrm{O}_2\).

Step 3: Calculate Moles of \(\mathrm{O}_2\) Used

Given that \(25\%\) of \(\mathrm{O}_2\) is used up, calculate the moles used: \[ \text{Moles of } \mathrm{O}_2 \text{ used} = 0.25 \times 4 = 1 \text{ mole} \].

Step 4: Determine the Change in Moles of Reactants and Products

From the reaction equation, consuming 1 mole of \(\mathrm{O}_2\) results in the formation of 2 moles of \(\mathrm{SO}_3\). As \(\mathrm{O}_2\) is consumed, \(\mathrm{SO}_2\) is used in the ratio 2:1. Therefore, \(2\) moles of \(\mathrm{SO}_2\) are consumed, and \(2\) moles of \(\mathrm{SO}_3\) are produced.

Step 5: Calculate Remaining Moles at Equilibrium

At equilibrium:- Moles of \(\mathrm{SO}_2\) = 4 - 2 = 2 - Moles of \(\mathrm{O}_2\) = 4 - 1 = 3 - Moles of \(\mathrm{SO}_3\) = 0 + 2 = 2

Step 6: Sum the Moles to Find Total at Equilibrium

Add the moles of all gases at equilibrium: \[ 2 \text{ (from \(\mathrm{SO}_2\))} + 3 \text{ (from \(\mathrm{O}_2\))} + 2 \text{ (from \(\mathrm{SO}_3\))} = 7 \text{ moles} \].

Key Concepts

Moles Calculation

Understanding moles is essential in chemistry for quantifying the amount of a substance. In this exercise, we're dealing with moles of gases, specifically sulfur dioxide \(\mathrm{SO}_2\), oxygen \(\mathrm{O}_2\), and sulfur trioxide \(\mathrm{SO}_3\). The initial condition provided is that there are 4 moles each of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\). Moles essentially represent the quantity of molecules, with 1 mole being equal to Avogadro's number: approximately \(6.022 \times 10^{23}\) molecules.

Calculating moles at equilibrium involves understanding how many moles react. When \(25\%\) of \(\mathrm{O}_2\) is consumed, it corresponds to \(0.25\) times the initial moles of \(\mathrm{O}_2\). Thus, \(1\) mole of \(\mathrm{O}_2\) is used up. This calculation helps us understand the extent of the reaction and is crucial for predicting how much \(\mathrm{SO}_3\) forms and how much of the reactants remain.

• Initial moles of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) are both 4.
• \(25\%\) consumption of \(\mathrm{O}_2\) means \(0.25 \times 4 = 1\) mole is used.
• The stoichiometry of the reaction guides how these changes affect the amounts of each substance.

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding the relationship between reactants and products. The balanced equation for the reaction of sulfur dioxide with oxygen to form sulfur trioxide is given by: \[2\,\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2\,\mathrm{SO}_{3(g)}\]
This equation tells us several key pieces of information. Firstly, it shows the stoichiometric coefficients: 2 moles of \(\mathrm{SO}_2\) react with 1 mole of \(\mathrm{O}_2\) to produce 2 moles of \(\mathrm{SO}_3\). These coefficients are vital for knowing the proportion in which substances react and are produced.

Having a balanced equation ensures we fulfill the law of conservation of mass, meaning that the total mass of reactants equals the total mass of products. These coefficients are used to calculate how much of each substance will react and how much product will be formed. In our exercise, knowing that 1 mole of \(\mathrm{O}_2\) results in 2 moles of \(\mathrm{SO}_3\) is crucial.

• The equation provides stoichiometric ratios: 2:1:2 for \(\mathrm{SO}_2 : \mathrm{O}_2 : \mathrm{SO}_3\).
• It ensures mass and molecules are conserved throughout the reaction.
• The equation is the foundation for further moles calculation and eventual determination of equilibrium state.

Sulfur Trioxide Formation

Sulfur trioxide \(\mathrm{SO}_3\) is formed through a chemical reaction between sulfur dioxide \(\mathrm{SO}_2\) and oxygen \(\mathrm{O}_2\). This reaction exemplifies a classic equilibrium process where reactants and products coexist. Understanding how \(\mathrm{SO}_3\) forms requires grasping the relationship depicted in the balanced equation: \[2\,\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2\,\mathrm{SO}_{3(g)}\]
In this context, for each mole of \(\mathrm{O}_2\) consumed, 2 moles of \(\mathrm{SO}_3\) are produced. The equilibrium point is reached when there's no further net change in the amounts of reactants and products, even though reactions continue to occur.

This example highlights how the formation of \(\mathrm{SO}_3\) impacts the total moles present in the system. Specifically, when \(1\) mole of \(\mathrm{O}_2\) is consumed, it leads to the formation of \(2\) moles of \(\mathrm{SO}_3\), increasing the number of product moles and thus altering the balance of the gaseous mixture.

• \(\mathrm{SO}_3\) formation is directly proportional to the amount of \(\mathrm{O}_2\) consumed.
• Reactant consumption and product formation are tightly linked through stoichiometry.
• The formation of \(2\) moles of \(\mathrm{SO}_3\) confirms the effectiveness of the reaction's stoichiometry.