Problem 202 - A Complete Resource Book in Chemistry for JEE Main

Question

A vessel at \(1000 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) with a pressure of \(0.5 \mathrm{~atm}\). Some of the \(\mathrm{CO}_{2}\) is converted into \(\mathrm{CO}\) on the addition of graphite. If the total pressure at equilibrium is \(0.8 \mathrm{~atm}\), the value of \(\mathrm{K}\) is: (a) \(3.6 \mathrm{~atm}\) (b) \(1 \mathrm{~atm}\) (c) 2 atm (d) \(1.8 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K \) is \( 1.8 \text{ atm} \). (Option d)

Step 1: Understanding the Reaction

The conversion of \( \mathrm{CO}_2 \) to \( \mathrm{CO} \) in the presence of graphite can be represented by the reaction: \[ \mathrm{C} + \mathrm{CO}_2 \rightleftharpoons 2 \mathrm{CO} \] At equilibrium, we need to determine the equilibrium constant \( K \) using the given pressure information.

Step 2: Initial and Change in Pressure

Initially, the partial pressure of \( \mathrm{CO}_2 \) is \( 0.5 \text{ atm} \). Let's assume \( x \) atm of \( \mathrm{CO}_2 \) reacts at equilibrium. Thus, the change in pressure for \( \mathrm{CO}_2 \) is \(-x\) and for \( \mathrm{CO} \) it is \(+2x\) because two moles of \( \mathrm{CO} \) are produced for each mole of \( \mathrm{CO}_2 \) reacted.

Step 3: Write the Expression for Total Pressure at Equilibrium

At equilibrium, the total pressure is given as \( 0.8 \text{ atm} \). Therefore, \[ 0.5 - x + 2x = 0.8 \] Simplifying this gives: \[ 0.5 + x = 0.8 \] Solving for \( x \) gives: \[ x = 0.3 \text{ atm} \]

Step 4: Calculate the Equilibrium Pressures

Now, we calculate the equilibrium pressures of \( \mathrm{CO}_2 \) and \( \mathrm{CO} \): - The pressure of \( \mathrm{CO}_2 \) at equilibrium is \( 0.5 - 0.3 = 0.2 \text{ atm} \). - The pressure of \( \mathrm{CO} \) at equilibrium is \( 2 \times 0.3 = 0.6 \text{ atm} \).

Step 5: Calculate the Equilibrium Constant

The equilibrium constant \( K \) in terms of pressure for the reaction \( \mathrm{C} + \mathrm{CO}_2 \rightleftharpoons 2 \mathrm{CO} \) is given by: \[ K = \frac{{(P_{\mathrm{CO}})^2}}{{P_{\mathrm{CO}_2}}} \] Substitute the equilibrium pressures: \[ K = \frac{{(0.6)^2}}{{0.2}} = \frac{0.36}{0.2} = 1.8 \text{ atm} \]

Step 6: Summary Step: Identify the Correct Answer

The equilibrium constant \( K \) is calculated to be \( 1.8 \text{ atm} \). Therefore, the correct answer is option (d).

Key Concepts

Chemical Equilibrium

In chemical reactions, equilibrium represents the state where the rate of the forward reaction is equal to the rate of the backward reaction. This means that the concentrations or pressures of reactants and products remain constant over time. In the context of this exercise, we explore the equilibrium between carbon dioxide (\( \mathrm{CO}_2 \)) and carbon monoxide (\( \mathrm{CO} \)) as part of a reaction where graphite also participates. The reaction can be represented as \( \mathrm{C} + \mathrm{CO}_2 \rightleftharpoons 2 \mathrm{CO} \).

At equilibrium, no net change in the amounts of each gas is observed, even though both reactions continue to occur. It's important to distinguish between the concept of dynamic equilibrium and a state where reactions cease entirely. Although the concentrations remain stable at equilibrium, it indicates a balance of reaction rates rather than no activity at all.

Understanding equilibrium is crucial in many areas of chemistry, allowing scientists to predict how changes in conditions can affect reaction yields and product formation. In turn, this knowledge can be applied to optimize industrial chemical processes, laboratory experiments, and environmental management.

Pressure Calculations

Pressure calculations are a fundamental part of understanding gas-phase reactions, particularly when dealing with equilibria. For gaseous reactions like this one, the pressures of each substance can be used similarly to concentrations in liquid-phase reactions.

When conducting pressure calculations for the reaction \( \mathrm{C} + \mathrm{CO}_2 \rightleftharpoons 2 \mathrm{CO} \), we begin by knowing the initial and total equilibrium pressures. The initial pressure of \( \mathrm{CO}_2 \) is 0.5 atm, which changes as it participates in the reaction. By introducing the variable \( x \) to represent the change in pressure caused by the conversion of \( \mathrm{CO}_2 \) to \( \mathrm{CO} \), we account for these changes.

In this exercise, the total pressure at equilibrium is given as 0.8 atm. By setting up the equation \( 0.5 - x + 2x = 0.8 \) and solving for \( x \), we determine that 0.3 atm of \( \mathrm{CO}_2 \) has reacted, allowing us to deduce the respective pressures of each substance at equilibrium. Mastery of pressure calculations is essential for solving many chemical problems efficiently and accurately.

Equilibrium Pressure

The equilibrium pressures of gases provide valuable information about the state of a reaction. In our exercise, the equilibrium pressure of \( \mathrm{CO}_2 \) was calculated to be 0.2 atm, and \( \mathrm{CO} \) reached a pressure of 0.6 atm at equilibrium.

Equilibrium pressures can tell us much about how far the reaction has proceeded from its initial state, as well as the relative stability of reactants and products under given conditions. By knowing the equilibrium pressures and the total pressure, we can compute the equilibrium constant, \( K_p \), which quantifies the ratio of products to reactants when the system is at equilibrium (in terms of pressure).

In practical applications, controlling equilibrium pressures through temperature, volume, or other variables can manipulate the position of equilibrium to favor the production of desired products. This idea is the foundation of techniques used in chemical manufacturing and industrial processes. By understanding and predicting these pressures, chemists can efficiently design reactions for maximum efficacy.

Reaction Stoichiometry

Reaction stoichiometry is the balanced relationship between reactants and products in a chemical reaction, based on the concept of moles. In this exercise, the stoichiometry of the reaction \( \mathrm{C} + \mathrm{CO}_2 \rightleftharpoons 2 \mathrm{CO} \) highlights that one mole of \( \mathrm{CO}_2 \) converts into two moles of \( \mathrm{CO} \).

This stoichiometric relationship is crucial in calculating how much pressure change occurs when a particular amount of \( \mathrm{CO}_2 \) reacts. As 0.3 atm of \( \mathrm{CO}_2 \) was converted, it led to the increase in pressure of \( \mathrm{CO} \) by 0.6 atm (because the stoichiometry indicates that twice as many moles of \( \mathrm{CO} \) are produced per mole of \( \mathrm{CO}_2 \) that reacts).

Stoichiometry allows chemists to predict the outcomes of reactions and calculate exact amounts needed or produced, essential in laboratory settings as well as industrial chemical processes. It simplifies the calculation of reactants required and products formed, which is a cornerstone principle in preparing any chemical reaction.