Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

A flask containing \(0.5\) atm pressure of \(\mathrm{A}_{2}(\mathrm{~g})\), some solid AB added into flask which undergoes dissociation according to \(2 \mathrm{AB}(\mathrm{s})=\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \mathrm{K}_{\mathrm{P}}=\) \(0.06 \mathrm{~atm}^{2}\). The total pressure (in atm) at equilibrium is: (a) \(0.70\) (b) \(0.6\) (c) \(0.10\) (d) None of these

Step-by-Step Solution

Answer

The total pressure at equilibrium is 0.7 atm.

Step 1: Write the Equilibrium Reaction

The dissociation reaction is given by \(2 \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{g}) + \mathrm{B}_{2}(\mathrm{g})\). Let the initial pressure of \(\mathrm{A}_{2}\) at equilibrium be \(0.5 + x\) since some dissociation occurs, creating \(x\) atm of \(\mathrm{B}_{2}\) gas.

Step 2: Define Equilibrium Pressure Conditions

At equilibrium, the pressure of \(\mathrm{A}_2\) will be \(0.5 + x\) atm and the pressure of \(\mathrm{B}_2\) will be \(x\) atm. The total pressure at equilibrium is \((0.5 + x) + x = 0.5 + 2x\).

Step 3: Apply Equilibrium Constant Expression

The equilibrium constant for the reaction in terms of pressure, \(K_P\), is given as \(0.06\). For the reaction, \(K_P = P_{\mathrm{A}_2} \times P_{\mathrm{B}_2} = (0.5 + x)(x)\). Therefore, the equation becomes \(0.06 = (0.5 + x) \times x\).

Step 4: Solve the Equilibrium Equation

Rearranging the equation \(0.06 = (0.5 + x)x\) gives a quadratic equation: \(0.5x + x^2 = 0.06\). Solving this quadratic equation, \(x^2 + 0.5x - 0.06 = 0\), using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = 0.5, c = -0.06\), we find \(x = 0.1\).

Step 5: Calculate Total Equilibrium Pressure

Now that we know \(x = 0.1\), substitute \(x\) back into the expression for total pressure: \(0.5 + 2x = 0.5 + 2(0.1) = 0.5 + 0.2 = 0.7\) atm.

Key Concepts

Dissociation Reactions

In chemistry, dissociation refers to a chemical reaction where compounds break down into smaller particles, such as atoms, ions, or simpler molecules. This is particularly common in compounds like acids or ionic compounds, where forces break the bonds holding the compound together.
A typical example, similar to the given exercise, involves equilibrium gasses. In this case, the reaction is represented as \(2 \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_2(\mathrm{g}) + \mathrm{B}_2(\mathrm{g})\). Initially, only \(\mathrm{A}_2\) gas is present, but as dissociation occurs, \(\mathrm{B}_2\) is formed.
Important points to remember include:

• Solid substances usually do not appear in the equilibrium expression since their concentration is constant.
• At equilibrium, even though reactions are occurring, the overall concentration of products and reactants stay the same.

This results in a dynamic balance that defines the system's equilibrium state.

Equilibrium Constant (Kp)

The equilibrium constant expressed in terms of pressure, known as \(K_p\), measures the relative concentrations of gases at equilibrium. It gives us insight into how far a reaction will proceed and the proportions of products to reactants in a gaseous mixture.
For the dissociation reaction \(2 \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_2(\mathrm{g}) + \mathrm{B}_2(\mathrm{g})\), \(K_p\) is calculated using the partial pressures of the gases:

• \(K_p = P_{\mathrm{A}_2} \times P_{\mathrm{B}_2}\).

In the exercise example, this results in \(K_p = (0.5 + x) \times x\), where \(x\) represents the change in pressure due to equilibrium being reached.
Knowing \(K_p\) allows chemists to predict reaction behavior, recognize possible reaction shifts, and adjust conditions to achieve desired results. Understanding and calculating \(K_p\) is crucial when dealing with reactions under equilibrium conditions.

Quadratic Equations in Chemistry

In a chemical context, quadratic equations are used when calculating equilibrium states, as seen when dealing with the pressures of gas-phase reactions. When the equilibrium constant forms a quadratic expression, solving for changes in concentration or pressure involves getting to solutions for the equation.
For instance, in the given exercise, the problem leads to the equation \(0.5x + x^2 = 0.06\), simplifying to \(x^2 + 0.5x - 0.06 = 0\). To solve it:

• Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Here, \(a = 1\), \(b = 0.5\), and \(c = -0.06\).

Solving this accurately identifies \(x\), which determines the pressure of the gases at equilibrium.
Quadratic equations present a problem-solving mechanism essential not only in chemistry but across sciences and maths, providing a pathway to understand complex interactions within chemical systems.