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) \(2 \mathrm{~atm}\) (d) \(1.8 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The value of \( \mathrm{K} \) is \( 1.8 \mathrm{~atm} \).

Step 1: Write the Reaction

The given problem involves the reaction of carbon dioxide with graphite to form carbon monoxide. The chemical equation for this reaction is: \[ \text{C (graphite)} + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \] At equilibrium, the partial pressures will be affected and defined by the initial conditions and the extent of reaction.

Step 2: Define Initial and Change in Pressures

Initially, the pressure of \( \mathrm{CO}_2 \) is \(0.5 \mathrm{~atm}\), and that of \( \mathrm{CO} \) is \(0\). Let \( x \) be the change in pressure as \( \mathrm{CO}_2 \) is converted to \( \mathrm{CO} \). At equilibrium, the pressure of \( \mathrm{CO}_2 \) will be \((0.5 - x) \mathrm{~atm}\) and \( \mathrm{CO} \) will be \(2x \mathrm{~atm}\).

Step 3: Use Total Equilibrium Pressure

We are given that the total pressure at equilibrium is \(0.8 \mathrm{~atm}\). Therefore, the equation for the total pressure at equilibrium is: \[ (0.5 - x) + 2x = 0.8 \] Simplify this equation to solve for \( x \).

Step 4: Solve for x

Solving the equation from Step 3: \[ 0.5 - x + 2x = 0.8 \] \[ 0.5 + x = 0.8 \] \[ x = 0.3 \mathrm{~atm} \]

Step 5: Determine Equilibrium Partial Pressures

With \( x = 0.3 \mathrm{~atm} \), calculate the equilibrium partial pressures:- For \( \mathrm{CO}_2 \): \( 0.5 - x = 0.5 - 0.3 = 0.2 \mathrm{~atm} \)- For \( \mathrm{CO} \): \( 2x = 2 imes 0.3 = 0.6 \mathrm{~atm} \)

Step 6: Write Equilibrium Expression

The equilibrium constant \( \mathrm{K} \) for the reaction is given by: \[ \mathrm{K} = \frac{(\text{Partial pressure of } \mathrm{CO})^2}{\text{Partial pressure of } \mathrm{CO}_2} \] Substitute the values: \[ \mathrm{K} = \frac{(0.6)^2}{0.2} \]

Step 7: Calculate K

Finally, calculate the value of \( \mathrm{K} \): \[ \mathrm{K} = \frac{0.36}{0.2} = 1.8 \mathrm{~atm} \]

Key Concepts

Equilibrium Constant (K)

The equilibrium constant (K) measures the ratio of the concentration of products to reactants at equilibrium in a chemical reaction. It tells us how far a reaction will proceed at a certain temperature. In our problem, we calculate the equilibrium constant using partial pressures because we are dealing with gases. The formula for calculating K is:

• For a reaction \[ aA + bB \rightleftharpoons cC + dD\]we use this expression: \[K = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\]

By using this formula, we can determine the value of K based on the pressures of carbon monoxide (CO) and carbon dioxide (CO₂). Knowing K helps understand how much of the reactants are converted to products in a reaction at equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. In our exercise, we need to monitor how the partial pressures of CO₂ and CO change as the reaction reaches equilibrium. Initially, the partial pressure of CO₂ is 0.5 atm and CO is 0 atm. As the reaction progresses:

• The pressure of CO₂ decreases as it's consumed, represented by \((0.5 - x)\) atm, where \(x\) is the change in pressure.
• CO, on the other hand, is formed, and its partial pressure becomes \(2x\) atm, considering the stoichiometry of the reaction.

The total pressure at equilibrium is given as 0.8 atm, allowing us to solve for \(x\). By understanding these pressure changes, we can determine how the constituents of the reaction behave as equilibrium is reached.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical equation. It allows us to predict the amounts of products formed from given reactants. In our example, we consider the balanced equation:\[ \text{C (graphite)} + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)\]The stoichiometry shows that for every mole of CO₂ reacted, two moles of CO are produced. This relationship guides the calculation of partial pressures:

• For every decrease in CO₂, the amount of CO formed is double, hence the factor \(2x\).
• This stoichiometric insight is crucial for setting up the equation of partial pressures and solving for the equilibrium constant K.

Understanding stoichiometry helps predict how a change in one component affects the entire system, ensuring proper balance in chemical reactions.