Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

In an equilibrium reaction, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=\) \(2 \mathrm{SO}_{3}(\mathrm{~g})\), the partial pressure \(\mathrm{SO}_{2}, \mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) are \(0.662,0.101\) and \(0.331\) atm respectively. What should be the partial pressure of oxygen if the equilibrium concentration of \(\mathrm{SO}_{3}\) and \(\mathrm{SO}_{2}\) becomes equal? (a) \(0.4 \mathrm{~atm}^{-1}\) (b) \(0.6 \mathrm{~atm}^{-1}\) (c) \(0.12 \mathrm{~atm}^{-1}\) (d) \(0.8 \mathrm{~atm}^{-1}\)

Step-by-Step Solution

Answer

None of the options correctly match the calculated partial pressure, which is 0.18375 atm.

Step 1: Understand the Reaction Equation

The equilibrium reaction is given as \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\). We start by noting the stoichiometry in the balanced equation, which indicates the ratio of reactants to products.

Step 2: Initial Partial Pressures

Initially, the partial pressures given are \( P_{\mathrm{SO}_{2}} = 0.662 \) atm, \( P_{\mathrm{O}_{2}} = 0.101 \) atm, and \( P_{\mathrm{SO}_{3}} = 0.331 \) atm.

Step 3: Equilibrium Condition

At equilibrium, we want the partial pressures of \(\mathrm{SO}_{3}\) and \(\mathrm{SO}_{2}\) to be equal. Hence, let \(P_{\mathrm{SO}_{2}}' = P_{\mathrm{SO}_{3}}'\), and both will become equal to a common pressure \(x\).

Step 4: Change in Pressures

Since \(\mathrm{SO}_{3}\) and \(\mathrm{SO}_{2}\) should have equal partial pressures, let the change in \(\mathrm{SO}_{3}\) be \(+y\) and in \(\mathrm{SO}_{2}\) be \(+y\). The change in \(\mathrm{O}_{2}\) will be \(+\frac{y}{2}\) due to stoichiometry.

Step 5: Equilibrium Equation Setup

Using these changes, we set up the equation for equilibrium: \[ P_{\mathrm{SO}_{3}}' = 0.331 + y, \] \[ P_{\mathrm{SO}_{2}}' = 0.662 - y, \] and \[ P_{\mathrm{SO}_{3}}' = P_{\mathrm{SO}_{2}}' = x, \] inferring that \(0.331 + y = 0.662 - y\).

Step 6: Solve for Change \(y\)

Solving \(0.331 + y = 0.662 - y\), we get: \[ 2y = 0.662 - 0.331 \] \[ 2y = 0.331 \] \[ y = 0.1655 \]. Therefore, \(x = 0.331 + 0.1655 = 0.4965\) atm because both \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) should be equal.

Step 7: Calculate New \(P_{\mathrm{O}_{2}}'\)

The partial pressure of \(\mathrm{O}_{2}\) will increase due to the reverse reaction. Thus, the new \(P_{\mathrm{O}_{2}}' = 0.101 + \frac{0.1655}{2} = 0.18375\) atm.

Step 8: Choose the Correct Option

Comparing the calculated partial pressure of \(\mathrm{O}_{2}\), which is 0.18375 atm, with the provided options, we notice the options are not accurate for the units (atm instead of atm\(^{-1}\)). None of the given answers correctly match the solution.

Key Concepts

Understanding Stoichiometry in Equilibrium Reactions

Stoichiometry is a foundational concept in chemistry, where it refers to the precise calculation of reactants and products in a chemical reaction. In an equilibrium reaction, such as \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\), stoichiometry is essential to understanding how the quantities of substances relate to one another. The coefficients in the balanced chemical equation demonstrate the stoichiometric relationships: two moles of \(\mathrm{SO}_{2}\) react with one mole of \(\mathrm{O}_{2}\) to produce two moles of \(\mathrm{SO}_{3}\).

When analyzing equilibrium systems, stoichiometry allows us to set up the correct relationships between changes in reactants and products. In the given problem, as the partial pressures adjust to reach equilibrium with equal pressures, this stoichiometric balance guides the calculations needed to determine how each pressure changes. Understanding these relationships is key to successfully solving equilibrium problems and predicting how reactions proceed.

To practice, try the following steps:

• Identify the coefficients that indicate the mole ratio of reactants to products.
• Use the mole ratio to calculate the changes in quantities when certain conditions are set, such as equal partial pressures.

Partial Pressure in Gas Reactions

Partial pressure is an essential concept in the study of gases. It refers to the pressure exerted by an individual gas in a mixture of gases. For a system at equilibrium, partial pressures play a crucial role in describing the state of the system. In the exercise involving the reaction \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\), each gas has a defined partial pressure that must be considered to fully understand the equilibrium state.

To solve problems involving partial pressure:

• Identify the initial partial pressures of all gases involved. These are provided or can be calculated from given data.
• Understand how partial pressures affect each other through stoichiometry, impacting the equilibrium conditions.

Partial pressures can also change as equilibrium shifts. For instance, as the reaction progresses, partial pressures adjust in response to the stoichiometry, as we saw in the provided exercise. Grasping how these changes occur involves applying these principles to determine new equilibrium conditions.

Remember, the sum of partial pressures of all gases in a mixture will always equal the total pressure of the system.

Effective Chemistry Problem-Solving

Chemistry problem-solving skills are crucial for tackling equilibrium reactions and other complex chemical scenarios. It requires a nuanced understanding of various concepts, including stoichiometry and partial pressures, and the ability to interlink these ideas to solve problems effectively.

Here are some strategies to enhance your chemistry problem-solving:

• Read the problem carefully and identify what is known and what needs to be discovered.
• Write out relevant equations and setup relations using stoichiometric coefficients.
• Calculate step-by-step, checking your logic at each stage of the process.
• Verify your final answer with given options or through logical reasoning.

Chemistry problems often have a defined set of steps that guide you from the problem statement to the solution. Emphasizing a clear understanding of these steps ensures that each part of a problem is addressed and resolved effectively. This methodical approach not only leads to the correct answer but also reinforces your grasp of chemical principles. By integrating these techniques, solving chemical problems becomes a more manageable and even enjoyable task.