Problem 206 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The equilibrium concentration of D is \(1.818 \text{ M}\).

Step 1: Write Down the Equilibrium Expression

For the reaction \( \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \), the equilibrium constant expression is given by:\[K = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\]We know \( K = 100 \) and all initial concentrations \([\mathrm{A}]_0 = [\mathrm{B}]_0 = [\mathrm{C}]_0 = [\mathrm{D}]_0 = 1\, \text{M}\).

Step 2: Determine Change in Concentration

Let's assume the change in concentration at equilibrium is \(x\) for \(\mathrm{A}\) and \(\mathrm{B}\). This means that for \(\mathrm{C}\) and \(\mathrm{D}\), the concentration will increase by \(x\).At equilibrium:\[[\mathrm{A}] = 1 - x, \quad [\mathrm{B}] = 1 - x, \quad [\mathrm{C}] = 1 + x, \quad [\mathrm{D}] = 1 + x\]

Step 3: Substitute into Equilibrium Expression

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

Step 4: Solve the Equation for x

We solve the equation \((1+x)^2 = 100(1-x)^2\) by expanding both sides:\[1 + 2x + x^2 = 100(1 - 2x + x^2)\]which simplifies to:\[1 + 2x + x^2 = 100 - 200x + 100x^2\]Rearranging gives:\[0 = 99x^2 - 202x + 99\]Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=99, \ b=-202, \ c=99\):\[x = \frac{202 \pm \sqrt{202^2 - 4 \cdot 99 \cdot 99}}{2 \cdot 99}\]\[x = \frac{202 \pm \sqrt{40804 - 39204}}{198}\]\[x = \frac{202 \pm \sqrt{1600}}{198}\]\[x = \frac{202 \pm 40}{198}\]\[x = \frac{242}{198} \quad \text{or} \quad x = \frac{162}{198}\]Giving approximate \(x = 1.22\ \text{for} \ \frac{242}{198} \), not valid, as concentrations can't exceed initial concentration, thus we use \(x = 0.818\ \text{for} \ \frac{162}{198} \).

Step 5: Calculate Equilibrium Concentration of D

Using \(x = 0.818\):\[[\mathrm{D}] = 1 + x = 1 + 0.818 = 1.818 \text{ M}\]

Key Concepts

Equilibrium Constant

Imagine you're observing a party where some guests are constantly moving between two rooms, and yet, at any given moment, it seems like the same number of guests are in each room. This is similar to the concept of chemical equilibrium in chemistry, where reactions don't stop but reach a state where they balance each other out. In our example reaction \( \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \), the equilibrium constant \( K \) is a measure of this balance. The value of \( K \) indicates the extent to which a reaction proceeds to the right (products) or to the left (reactants) at equilibrium. The equilibrium constant expression for the given reaction is: \[ K = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \] Here, \([\mathrm{C}]\), \([\mathrm{D}]\), \([\mathrm{A}]\), and \([\mathrm{B}]\) represent the molar concentrations of the substances involved.

• If \( K \) is large (\( K \gg 1 \)), the products dominate at equilibrium.
• If \( K \) is small (\( K \ll 1 \)), there is a higher concentration of reactants at equilibrium.
• A value of \( K \approx 1 \) suggests that both reactants and products are present in significant amounts.

In this exercise, \( K = 100 \), a relatively large value, suggests that products \( \mathrm{C} \) and \( \mathrm{D} \) are favored at equilibrium.

Quadratic Equation in Chemistry

Solving chemical equilibrium problems often involves handling quadratic equations. These mathematical expressions are key when determining concentrations of reactants and products at equilibrium. In our reaction, we let changes in concentration be represented by \( x \). This gives us equilibrium concentrations: - \([\mathrm{A}] = 1 - x\)- \([\mathrm{B}] = 1 - x\)- \([\mathrm{C}] = 1 + x\)- \([\mathrm{D}] = 1 + x\)Substituting these into the equilibrium equation results in a quadratic form which must then be solved to find \( x \). This becomes: \[ 100 = \frac{(1+x)^2}{(1-x)^2} \] When simplified and rearranged, this leads to: \[ 99x^2 - 202x + 99 = 0 \]

Using the Quadratic Formula

Every quadratic equation \( ax^2 + bx + c = 0 \) can be solved using the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \( a = 99 \), \( b = -202 \), and \( c = 99 \). Applying these, we calculate \( x \) and interpret them to respect chemical principles, ensuring physical feasibility (e.g., concentrations can't be negative or exceed initial values). Here, \( x = 0.818 \) is the valid solution.

Reaction Concentration Changes

Tracking how the concentrations of substances change in a reaction reaching equilibrium is vital for understanding dynamic chemical processes. As a reaction approaches equilibrium, concentrations adjust to balance the forward and reverse reaction rates. In our particular setup starting with \(1 \text{ M}\) each of \( \mathrm{A}, \mathrm{B}, \mathrm{C}, \) and \( \mathrm{D} \): - Reactants \( \mathrm{A} \) and \( \mathrm{B} \) concentrations decrease by \( x \).- Products \( \mathrm{C} \) and \( \mathrm{D} \) concentrations increase by \( x \).

Why Changes Matter

Understanding these changes helps chemists predict how the concentration of any specific substance at equilibrium affects the others. As \( x \) was determined to be 0.818, the increase in concentration for \( \mathrm{D} \), is significant given the equilibrium constant of 100. Therefore, at equilibrium, the concentration of \( \mathrm{D} \) becomes \( 1 + 0.818 = 1.818 \text{ M} \).This process involves conceptualizing how concentrations evolve over time and aids in the mastery of equilibrium problems in chemistry.