Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

\(2 \mathrm{NOBr} \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) .\) If \(\mathrm{NOBr}\) is dissoci- ated to the extent of \(20 \%\) at \(\mathrm{T} \mathrm{K}\) and a total pressure of 1.1 atm, then \(K_{p}\) in atm for the formation of NOBr is (a) 160 (b) \(\frac{1}{160}\) (c) \(\frac{1}{120}\) (d) 120

Step-by-Step Solution

Answer

The answer is (b) \(\frac{1}{160}\).

Step 1: Initial Concentration Setup

Let's assume the initial moles of NOBr is 1 mole. Given that NOBr dissociates 20%, it means 0.2 moles of NOBr are dissociated to form products at equilibrium.

Step 2: Determine Equilibrium Concentrations

The reaction is \(2 \mathrm{NOBr} \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g})\). Initially, assume 2x moles of NOBr. At equilibrium, 0.2*2x = 0.4x moles of NO and 0.2x moles of Br2 are formed, leaving 1.6x moles of NOBr.

Step 3: Express Total Pressure Terms

Since total pressure is 1.1 atm and we've assumed total moles after dissociation is \(2x + 0.2x + 0.2x = 1.4x\). 1.4x equals total pressure, so each component's pressure is proportionate to its number of moles.

Step 4: Calculate Partial Pressures at Equilibrium

Partial pressure of NOBr: \(\frac{1.6x}{1.4x} \cdot 1.1 = \frac{1.6}{1.4} \cdot 1.1 \approx 1.257 \text{ atm}.\) Partial pressure of NO: \(\frac{0.4x}{1.4x} \cdot 1.1 = \frac{0.4}{1.4} \cdot 1.1 \approx 0.314 \text{ atm}.\) Partial pressure of Br2: \(\frac{0.1}{1.4} \cdot 1.1 = 0.079 \text{ atm}.\)

Step 5: Apply the Equilibrium Constant Expression

The equilibrium constant expression \(K_p\) for the reaction is \[ K_p = \frac{(P_{\mathrm{NO}})^2 \cdot P_{\mathrm{Br}_2}}{(P_{\mathrm{NOBr}})^2}. \] Substituting the given partial pressures, \[ K_p = \frac{(0.314)^2 \cdot 0.079}{(1.257)^2}. \]

Step 6: Calculate the Equilibrium Constant

Calculate the expression \( K_p = \frac{(0.314)^2 \cdot 0.079}{(1.257)^2} = \frac{0.0985 \cdot 0.079}{1.5788} \approx 0.0005 .\) Thus, \(K_p \approx \frac{1}{160}.\)

Key Concepts

Dissociation

Dissociation is a fundamental concept in chemical equilibrium involving the separation of a molecule into two or more smaller molecules, ions, or atoms. In the example of the reaction \[ 2 \text{NOBr} \rightleftharpoons 2 \text{NO} (\text{g}) + \text{Br}_2 (\text{g}) \]NOBr, which is originally a single molecular entity, dissociates to form nitric oxide (NO) and bromine gas (Br\(_2\)). In this process, a certain percentage of NOBr molecules break apart and rearrange themselves into the products. Given that NOBr is dissociated to 20% at temperature T and a pressure of 1.1 atm, it implies that 20% of the initial amount of NOBr has turned into the products NO and Br\(_2\). Knowing the extent of dissociation helps us understand how far the reaction has proceeded towards equilibrium. This fraction gives insights into the concentrations of products and reactants at equilibrium.

Partial Pressure

In the context of gaseous reactions, partial pressure is an essential factor. It defines the pressure exerted by an individual gas in a mixture. When dealing with equilibrium in gaseous systems, partial pressure allows us to gauge each component's contribution to the total pressure. For the reaction given,

• The partial pressure of NOBr, NO, and Br\(_2\) can all be determined from their respective equilibrium mole fractions.
• Since the total pressure is 1.1 atm, each gas's partial pressure relies on the partial concentration of that gas to total gas concentration.

The calculation for these pressures involves using the ideal gas law concept where each component's pressure is proportional to its concentration in the system. Knowing the individual partial pressures is crucial for calculating the equilibrium constant expression \(K_p\). It allows us to express and determine how mixed gases behave under equilibrium conditions.

Equilibrium Constant Expression

The equilibrium constant expression is a key tool in understanding chemical equilibria, especially in gaseous systems. This expression is also referred to as \(K_p\) when dealing with equilibrium in terms of partial pressures. For the reaction:\[ 2 \text{NOBr} \rightleftharpoons 2 \text{NO} (\text{g}) + \text{Br}_2 (\text{g}) \]The equilibrium constant \(K_p\) is calculated using the partial pressures of the gases:\[ K_p = \frac{(P_{\text{NO}})^2 \cdot P_{\text{Br}_2}}{(P_{\text{NOBr}})^2}. \]

• The expression uses the stoichiometry of the balanced chemical equation to define the relationships between reactants and products.
• It provides a numeric value that predicts the direction of the reaction shift and the ratio of products to reactants at equilibrium.

Knowing \(K_p\) offers insights into the position of equilibrium and is crucial for calculating how the system will respond to changes in pressure, concentration, or temperature.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each reactant and product at the state of equilibrium. For a reaction at equilibrium, such as:\[ 2 \text{NOBr} \rightleftharpoons 2 \text{NO} (\text{g}) + \text{Br}_2 (\text{g}) \]

• Equilibrium concentration indicates how much of each substance exists after the reaction has reached equilibrium.
• In this case, knowing that NOBr dissociates 20% provides a basis for determining how much NO and Br\(_2\) is formed and remained at completion.

Based on the initially assumed moles of NOBr, equilibrium adjustments can be made. Calculations often start from initial amounts and take into account stoichiometric coefficients from the balanced equation. These concentrations are essential in calculating and utilizing the equilibrium constant \(K_p\) to predict how a system in equilibrium will behave when exposed to changes, like pressure shifts or additional reactants.