Problem 14 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The equilibrium constant at \(298 \mathrm{~K}\) for a reaction \(A+B \rightleftharpoons C+D\) is 100 . If the initial concentration of all the four species were \(1 \mathrm{M}\) each, then equilibrium concentration of \(D\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) will be : (a) \(1.818\) (b) \(1.182\) (c) \(0.182\) (d) \(0.818\)

Step-by-Step Solution

Answer

The equilibrium concentration of \(D\) is 1.818 mol L\(^{-1}\).

Step 1: Write the equilibrium expression

For the reaction \(A + B \rightleftharpoons C + D\), the equilibrium constant expression \(K_{eq}\) is given by:\[ K_{eq} = \frac{[C][D]}{[A][B]} \]We are given that \(K_{eq} = 100\).

Step 2: Define change in concentration

Let the change in concentration of \(D\) at equilibrium be \(x\). As the reaction moves toward equilibrium, the concentrations of \(A\) and \(B\) each decrease by \(x\), while the concentrations of \(C\) and \(D\) each increase by \(x\). So at equilibrium:- \([A] = 1 - x\)- \([B] = 1 - x\)- \([C] = 1 + x\)- \([D] = 1 + x\)

Step 3: Substitute into equilibrium expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ 100 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} \]\[ 100 = \frac{(1 + x)^2}{(1 - x)^2} \]

Step 4: Simplify the equation

Take the square root of both sides to deal with squared terms:\[ \sqrt{100} = \frac{1 + x}{1 - x} \]\[ 10 = \frac{1 + x}{1 - x} \]

Step 5: Solve for x

Solve the equation \(10 = \frac{1 + x}{1 - x}\):Multiply both sides by \((1 - x)\):\[ 10(1 - x) = 1 + x \]Simplify and solve for \(x\):\[ 10 - 10x = 1 + x \]\[ 10 - 1 = 10x + x \]\[ 9 = 11x \]\[ x = \frac{9}{11} \approx 0.818 \]

Step 6: Determine equilibrium concentration of D

The equilibrium concentration of \(D\) is given by \(1 + x\). Since \(x = 0.818\):\[ [D]_{equilibrium} = 1 + 0.818 = 1.818 \]

Key Concepts

chemical_equilibrium

Chemical equilibrium refers to the state where a chemical reaction and its reverse occur at equal rates, meaning the concentrations of the reactants and products remain constant over time. This state doesn't mean that the reactants and products are equal in concentration, but rather that the speed of transformation from reactants to products and from products back to reactants is balanced.

In this equilibrium state, for a reaction such as \( A + B \rightleftharpoons C + D \), both the forward reaction \( A + B \to C + D \) and the backward reaction \( C + D \to A + B \) continue to happen, but they don't cause changes in the concentrations. Therefore, by observing the constant concentration values, one can confirm the system is in equilibrium. The equilibrium state is crucial in many chemical processes, industrial applications, and even biological systems, providing a balance that maintains consistency despite continuing processes.

reaction_concentration

Reaction concentration in the context of chemical equilibrium refers to the amounts of each reactant and product present in a system when it has reached equilibrium. Initially, concentrations are set, and as the reaction moves toward equilibrium, these concentrations shift according to the reaction dynamics.

For the given reaction \( A + B \rightleftharpoons C + D \), we started with all species at a concentration of \( 1 \text{ M} \). As the reaction proceeds, these amounts shift: \( A \) and \( B \) decrease by some value \( x \), while \( C \) and \( D \) increase by that same value \( x \). It’s important to track these changes as they inform how to calculate the equilibrium concentrations using the equilibrium constant.

Thus, understanding reaction concentration involves anticipating how the quantities of substances will adjust and reach a stable state consistent with the equilibrium condition set by the equilibrium constant.

equilibrium_expression

The equilibrium expression is a mathematical equation that represents the relations between the concentrations of reactants and products at equilibrium in a reversible chemical reaction. For the reaction \( A + B \rightleftharpoons C + D \), the equilibrium constant \( K_{eq} \) is defined by:

• \[ K_{eq} = \frac{[C][D]}{[A][B]} \]

This equation captures the ratio of the product of the concentrations of the products \([C][D]\) to the product of the concentrations of the reactants \([A][B]\), each raised to the power of their stoichiometric coefficients.

In practical terms, it signifies that at equilibrium, the ratio of concentrations is maintained as per the equilibrium constant, here being 100. By inserting the equilibrium concentrations into this expression, we can solve for unknown values, as shown in the problem above where solving \( x \) helps determine each substance's equilibrium concentration. Understanding the equilibrium expression allows chemists to predict how shifts (like adding a reactant or changing the temperature) can affect the system, providing insights into reaction behavior and control.