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 \(2 \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \mathrm{K}_{\mathrm{P}}=\) \(0.06 \mathrm{~atm}^{2} .\) The total pressure (in \(\mathrm{atm}\) ) 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 reaction given is \(2\ \mathrm{AB}(\mathrm{s}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{g})\). This is a dissociation reaction of solid \(\mathrm{AB}\) into two gas components \(\mathrm{A}_2(\mathrm{g})\) and \(\mathrm{B}_2(\mathrm{g})\).

Step 2: Initial Conditions

Initially, there is \(0.5\) atm of \(\mathrm{A}_2(\mathrm{g})\) present in the flask. The pressure of \(\mathrm{B}_2(\mathrm{g})\) is initially \(0\) atm since it only forms from the dissociation of \(\mathrm{AB}\).

Step 3: Set Up the Equilibrium Expression

At equilibrium, additional \(x\) atm of \(\mathrm{A}_2(\mathrm{g})\) and \(x\) atm of \(\mathrm{B}_2(\mathrm{g})\) will be formed from the dissociation of \(\mathrm{AB}\). Thus, the total pressure of \(\mathrm{A}_2(\mathrm{g})\) becomes \(0.5 + x\) atm, and that of \(\mathrm{B}_2(\mathrm{g})\) is \(x\) atm. The equilibrium constant expression for the pressure is \(K_P = P_{\mathrm{A}_2} \cdot P_{\mathrm{B}_2} = 0.06\ \text{atm}^2\).

Step 4: Substitute into the Equilibrium Expression

Using the equilibrium constant expression, \(K_P = (0.5 + x) \cdot x = 0.06\). Expand this to get the quadratic equation: \(0.5x + x^2 = 0.06\).

Step 5: Solve the Quadratic Equation

The quadratic equation is \(x^2 + 0.5x - 0.06 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = 0.5\), \(c = -0.06\), we find the value of \(x\). Calculate \(b^2 - 4ac = (0.5)^2 - 4 \times 1 \times (-0.06) = 0.49\). Then, \(x = \frac{-0.5 \pm 0.7}{2}\). This gives \(x = 0.1\) as the valid solution since pressure cannot be negative.

Step 6: Find Total Pressure at Equilibrium

The total pressure at equilibrium is the sum of the partial pressures of \(\mathrm{A}_2(\mathrm{g})\) and \(\mathrm{B}_2(\mathrm{g})\), which is \(0.5 + x + x = 0.5 + 2x = 0.5 + 2(0.1) = 0.7\).

Key Concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, denoted as \(K_P\) in terms of partial pressure, is a key factor. It tells us about the position of the equilibrium for the reaction when it involves gases. For a given reaction, \(K_P\) is derived from the ratio of the partial pressures of the products to those of the reactants, each raised to the power of their stoichiometric coefficients.
For instance, in the dissociation reaction \(2\ \mathrm{AB}(s) \rightleftharpoons \mathrm{A}_2(g) + \mathrm{B}_2(g)\), the equilibrium constant expression can be written as:

• \(K_P = P_{\mathrm{A}_2} \cdot P_{\mathrm{B}_2}\)

Since solids like \(\mathrm{AB}\) do not have a partial pressure, they do not appear in the \(K_P\) expression. Understanding how the equilibrium constant applies allows us to calculate the concentration of gases at equilibrium, giving insight into the system's behavior under established conditions.
A high \(K_P\) indicates a reaction that heavily favors products, whereas a low \(K_P\) suggests reactants are favored at equilibrium.

Partial Pressure

Partial pressure is a term you'll often encounter when dealing with gases, especially in equilibrium reactions. It represents the pressure that a single type of gas in a mixture would exert if it occupied the entire volume on its own. In mixed gases, each gas contributes to the total pressure in proportion to its amount.
In the given reaction, initially, we have \(0.5\) atm of \(\mathrm{A}_2(g)\) in the flask, but no \(\mathrm{B}_2(g)\) due to the absence of dissociation initially. At equilibrium, as the reaction proceeds, some of the solid \(\mathrm{AB}\) dissociates to form \(\mathrm{A}_2(g)\) and \(\mathrm{B}_2(g)\). Additional \(x\) atm is contributed by both \(\mathrm{A}_2(g)\) and \(\mathrm{B}_2(g)\), leading to partial pressures of \(0.5 + x\) atm for \(\mathrm{A}_2(g)\) and \(x\) atm for \(\mathrm{B}_2(g)\).

• Total equilibrium pressure is the sum of these partial pressures.

Understanding this concept is fundamental to calculating the equilibrium state of a gaseous reaction.

Quadratic Equation

Quadratic equations pop up often in the analysis of chemical equilibriums, particularly when dealing with the equilibrium constant in a reaction. The quadratic equation we derived in this problem is the result of setting the \(K_P\) expression and rearranging it into a standard form.
The general form is:

• \(ax^2 + bx + c = 0\)

For this exercise, we ended up with the equation \(x^2 + 0.5x - 0.06 = 0\). To solve this, the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

is used, substituting \(a = 1\), \(b = 0.5\), and \(c = -0.06\). This approach helps find the values of \(x\) that satisfy the equilibrium condition. Typically, only the positive root is meaningful since we are dealing with physical quantities like pressure that cannot be negative.

Dissociation Reaction

A dissociation reaction involves a compound breaking down into two or more components, usually into simpler molecules or atoms. In this exercise, the solid \(\mathrm{AB}\) dissociates into two gaseous forms: \(\mathrm{A}_2(g)\) and \(\mathrm{B}_2(g)\).
Dissociation reactions can often alter the pressure and concentration of the reacting system, as observed with an increase in the number of gas molecules.
This transformation can be particularly significant in closed systems, like the flask in our problem, where the initial amounts of reactants change over time until equilibrium is reached. This reaction type highlights the dynamic nature of chemical equilibriums, where reactants and products continually interconvert, even at equilibrium. Understanding the characteristics of dissociation reactions helps in predicting and calculating the conditions at equilibrium, such as pressure and concentration.