Problem 41 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the following reaction, the equilibrium constant \(K_{c}\) at \(298 \mathrm{~K}\) is \(1.6\) \(\times 10^{17}\) $$ \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{FeS}(\mathrm{s}) $$ When equal volumes of \(0.06 \mathrm{M} \mathrm{Fe}^{2+}\) (aq) and \(0.2 \mathrm{M} \mathrm{S}^{2-}\) (aq) solutions are mixed, the equilibrium concentration of \(\mathrm{Fe}^{2+}\) (aq) is found to be \(Y \times\) \(10^{-17} \mathrm{M}\). The value of \(Y\) is

Step-by-Step Solution

Answer

The value of \(Y\) is 0.625.

Step 1: Initial Concentration Calculation

Since equal volumes of \(0.06 \, \text{M}\) \(\text{Fe}^{2+}\) and \(0.2 \, \text{M} \) \(\text{S}^{2-}\) solutions are mixed, their concentrations in the mixture become \(\frac{0.06}{2} = 0.03 \, \text{M}\) and \(\frac{0.2}{2} = 0.1 \, \text{M}\) respectively, due to dilution factor of \(\frac{1}{2}\).

Step 2: Equilibrium Expression

For the reaction, the expression for the equilibrium constant \(K_c\) is \[ K_c = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]} \] Since \(\text{FeS}\) is a solid, its activity is assumed to be 1.

Step 3: Inserting Known Values

We know \(K_c = 1.6 \times 10^{17}\). Let \(x\) be the change in concentration of \(\text{Fe}^{2+}\) and \(\text{S}^{2-}\) at equilibrium. Thus, \([\text{Fe}^{2+}] = x\) and \([\text{S}^{2-}] = 0.1 - x\). The equilibrium expression becomes: \[ 1.6 \times 10^{17} = \frac{1}{x(0.1-x)} \]

Step 4: Approximating for Small x

Given the large value of \(K_c\), the reaction lies far to the right, indicating \([\text{Fe}^{2+}]\) will be very small. Assume \(x << 0.1\), then approximately \([\text{S}^{2-}] \approx 0.1\). Substitute this in the equation: \[ 1.6 \times 10^{17} = \frac{1}{x \cdot 0.1} \]

Step 5: Solve for x

Solving \(1.6 \times 10^{17} = \frac{1}{0.1x}\) gives \(0.1x = \frac{1}{1.6 \times 10^{17}}\), so \(x = \frac{1}{1.6 \times 10^{16}} = 6.25 \times 10^{-18} \text{ M}\).

Step 6: Determine Value of Y

From the above, \([\text{Fe}^{2+}] = 6.25 \times 10^{-18} \text{ M}\) at equilibrium, which corresponds to finding \(Y\) such that \(Y \times 10^{-17} = 6.25 \times 10^{-18}\), therefore \(Y = 0.625\).

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted by \( K_c \), is a crucial component in understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium in a reaction mixture. For reactions where the state of matter includes solids or liquids, as in our example, these are excluded from the \( K_c \) expression because their activities are essentially 1.

In our given reaction, \( \text{Fe}^{2+}(\mathrm{aq}) + \text{S}^{2-}(\mathrm{aq}) \rightleftharpoons \text{FeS}(\mathrm{s}) \), the equilibrium constant \( K_c \) is given as \( 1.6 \times 10^{17} \). This exceptionally large value suggests that the reaction heavily favors the formation of the solid product, \( \text{FeS} \).

The expression derived from the equation becomes \( K_c = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]} \). Notably, this emphasizes that our interest at equilibrium is in the concentrations of the aqueous ions, since the concentration of solids like \( \text{FeS} \) remains constant and thus does not appear in the expression.

Understanding \( K_c \) allows us to predict the direction of the reaction shift when disturbances, like changes in concentration or pressure, occur.

Concentration Calculations

Calculating concentrations is essential when working with equilibrium problems. Let's explore how to calculate initial and equilibrium concentrations in our example. When you mix different concentrations of solutions, their concentration adjusts due to dilution.

In our scenario, equal volumes of \( 0.06 \, \text{M} \) \( \text{Fe}^{2+} \) and \( 0.2 \, \text{M} \) \( \text{S}^{2-} \) are mixed, resulting in halving their concentrations. Therefore, the concentrations become \( 0.03 \, \text{M} \) for \( \text{Fe}^{2+} \) and \( 0.1 \, \text{M} \) for \( \text{S}^{2-} \).

At equilibrium, calculations involve changes in concentrations denoted by \( x \). Here, it's calculated for \( \text{Fe}^{2+} \) which becomes \( x \), and for \( \text{S}^{2-} \), it becomes \( 0.1 - x \) due to their consumption in the reaction.

Approximation for small \( x \) values simplifies calculations significantly, especially when \( K_c \) is large indicating the reaction proceeds nearly to completion. This allows us to assume \( x \) is negligible in comparison to initial concentrations.

Reaction Quotient

The reaction quotient, \( Q \), is a snapshot of a reaction at any point in time, providing a way to gauge which direction a reaction will proceed to reach equilibrium. Like \( K_c \), \( Q \) is calculated using the concentrations of reactants and products, but it can be used before equilibrium is achieved.

For the reaction \( \text{Fe}^{2+} + \text{S}^{2-} \rightleftharpoons \text{FeS} \), had we required \( Q \), we would use the expression: \( Q = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]} \). When \( Q < K_c \), the forward reaction is favored, indicating a shift to produce more products until \( Q = K_c \).

In our example, we start with initial concentrations and determine the shifts towards equilibrium to find \([\text{Fe}^{2+}]\) at equilibrium. The large \( K_c \) value hints that initially \( Q < K_c \), driving the reaction towards forming more \( \text{FeS} \).

Utilizing \( Q \) effectively predicts the dynamics of reactions and assists in determining the equilibrium position given initial conditions.