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 \(\mathrm{D}\left(\mathrm{in} \mathrm{mol} \mathrm{L}^{-1}\right)\) will be: (a) \(0.818\) (b) \(1.818\) (c) \(1.182\) (d) \(0.182\)

Step-by-Step Solution

Answer

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

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction \(\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\) is given by:\[K_c = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\]We are given that \(K_c = 100\).

Step 2: Initial Concentrations and Change

Initially, all species \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) have a concentration of \(1 \text{ M}\). Let the change in concentration at equilibrium be \(x\) for products \(\mathrm{C}, \mathrm{D}\) and \(-x\) for reactants \(\mathrm{A}, \mathrm{B}\).

Step 3: Write Equilibrium Concentrations

At equilibrium:- \([\mathrm{A}] = 1 - x\)- \([\mathrm{B}] = 1 - x\)- \([\mathrm{C}] = 1 + x\)- \([\mathrm{D}] = 1 + x\)

Step 4: Plug into Equilibrium Expression

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

Step 5: Solve the Equation

To find \(x\), solve the equation:\[\frac{(1 + x)^2}{(1 - x)^2} = 100\]Taking the square root of both sides, we get:\[\frac{1+x}{1-x} = 10\]Solve for \(x\):\[1+x = 10(1-x)\]\[1+x = 10 - 10x\]\[11x = 9\]\[ x = \frac{9}{11}\]

Step 6: Calculate Equilibrium Concentration of D

Substitute \(x = \frac{9}{11}\) into \([\mathrm{D}] = 1 + x\):\[[\mathrm{D}] = 1 + \frac{9}{11} = \frac{11}{11} + \frac{9}{11} = \frac{20}{11} = 1.818\]

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented by the symbol \(K_c\), is a vital aspect of understanding chemical reactions at equilibrium. It quantifies the ratio of the concentrations of products to the concentrations of reactants at equilibrium. For the general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression \(K_c\) is given by:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]This expression presumes that all concentrations are measured in moles per liter (mol/L) and that the reaction has reached a state where the rates of the forward and reverse reactions are equal. An equilibrium constant with a larger value implies a greater concentration of products compared to reactants at equilibrium. If \(K_c\) is 100, for example, like in the given exercise, it signifies that products are favored at equilibrium compared to the reactants. It is important to note that the equilibrium constant is dependent on temperature and alterations in temperature can change its value.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of reactants and products in a reversible chemical reaction when the forward and reverse reactions occur at the same rate. At this time, the concentration of each substance remains constant. In the exercise you're dealing with, we have initial concentrations for all species as \(1 \text{ M}\). As the system reaches equilibrium, the concentrations change, and we denote these changes by \( x \) for products \( \mathrm{C} \) and \( \mathrm{D} \), and \(-x\) for reactants \( \mathrm{A} \) and \( \mathrm{B} \).The equilibrium concentrations are found by adjusting the initial concentrations by these changes:- For \([\mathrm{A}]\) and \([\mathrm{B}]\), the equilibrium concentrations are \(1 - x\).- For \([\mathrm{C}]\) and \([\mathrm{D}]\), it is \(1 + x\).These calculations become practical in solving for an unknown when using the equilibrium expression, substituting these values. Solving for \( x \) allows us to find the equilibrium concentration of any specific component like \( [\mathrm{D}] \), as was required in the solution provided.

Reaction Quotient

The reaction quotient, represented by \(Q_c\), is similar to the equilibrium constant in that it expresses the ratio of product concentrations to reactant concentrations. However, while \(K_c\) is calculated only at equilibrium, \(Q_c\) can be calculated at any point during the reaction to predict the direction the reaction will proceed to reach equilibrium.The expression for \(Q_c\) is:\[Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]After determining \(Q_c\), it's compared to \(K_c\):

• If \(Q_c = K_c\), the system is at equilibrium.
• If \(Q_c > K_c\), there are more products than needed for equilibrium, so the reaction will proceed in the reverse direction to form more reactants.
• If \(Q_c < K_c\), more products need to form, so the reaction moves forward to produce more products.

In the context of this exercise, once the equilibrium concentrations are calculated, they would give the value for \(Q_c\), which should equal \(K_c\) at equilibrium, confirming that our calculations are consistent and correct. The reaction quotient is an indispensable tool in predicting and monitoring the shifts in a reaction toward equilibrium.