Question

For the reaction: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the initial concentration of \(\mathrm{A}\) and \(\mathrm{B}\) is equal, but the equilibrium concentration of \(\mathrm{C}\) is twice that of equilibrium concentration of A. The equilibrium constant is (a) 4 (b) 9 (c) \(1 / 4\) (d) \(1 / 9\)

Short answer

The equilibrium constant (K_{eq}) for the reaction is 4 (option a).

Step-by-step solution

Identify Initial Conditions

We are given that initially, the concentrations of reactants A and B are equal. Let's denote this initial concentration as 'a'.

Set up Equilibrium Expression

The equilibrium expression for the reaction is written as: \[ K_{eq} = \frac{[C][D]}{[A][B]} \]

Define Changes in Concentration at Equilibrium

If 'x' is the change in concentration of A and B that reacts to achieve equilibrium, then at equilibrium, the concentration of A and B would be '[A] = 'a - x' and [B] = 'a - x' respectively. Since the equilibrium concentration of C is twice that of A at equilibrium, [C] = '2x', and by stoichiometry [D] = 'x'.

Write the Equilibrium Constant Expression with Variables

Substitute the equilibrium concentrations of reactants and products in terms of x into the equilibrium expression to get: \[ K_{eq} = \frac{(2x)(x)}{(a-x)(a-x)} \]

Apply the Given Relationship between Concentrations

Since [C] (which is '2x') is twice that of equilibrium concentration of A ('a - x'), we write '2x = 2(a - x)'. Simplifying this, we get '2x = 2a - 2x', which leads to '4x = 2a'. Thus, 'x = a/2'.

Substitute the Value of 'x' into the Equilibrium Expression

Substituting 'x = a/2' into the expression for K gives: \[ K_{eq} = \frac{(2\frac{a}{2})(\frac{a}{2})}{(a-\frac{a}{2})(a-\frac{a}{2})} \] Simplifying, we have: \[ K_{eq} = \frac{a^2}{(\frac{a}{2})^2} = \frac{a^2}{\frac{a^2}{4}} = 4 \]

Determine the Numerical Value of the Equilibrium Constant

Since the specific initial concentrations are not required to determine the equilibrium constant, we can see that the mathematical relationship results in K_{eq} being equal to 4.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for interpreting reactions that can proceed in both the forward and reverse directions. Equilibrium is reached when the rates of the forward and reverse reactions are equal, resulting in no further net change in the concentrations of the reactants and products. However, this doesn't mean the amounts of reactants and products are equal, but that their concentrations no longer change over time.

In our exercise, chemical equilibrium is observed in the reaction \( \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \). Here, the concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) initially are the same, and when the system reaches equilibrium, the concentration of \( \mathrm{C} \) is twice that of \( \mathrm{A} \). These conditions define the equilibrium state for this particular reaction.

Equation Stoichiometry

When calculating the equilibrium constant, it is essential to consider the stoichiometry of the equation, which describes the quantitative relationship between reactants and products in a chemical reaction. In stoichiometric calculations, we often use the mole ratio obtained from the balanced chemical equation to convert between amounts of reactants and products.

In our scenario, for every mole of \( \mathrm{A} \) and \( \mathrm{B} \) that reacts, one mole of \( \mathrm{C} \) and \( \mathrm{D} \) is produced. This 1:1:1:1 stoichiometry is critical in setting up the equilibrium expressions correctly. It allows us to express changes in reactant and product concentrations in terms of a single variable, 'x', simplifying the calculation of the equilibrium constant.

Concentration Changes in Reactions

In chemical reactions, concentration changes are what drive the system to equilibrium. As the reaction proceeds, reactants are consumed and products are formed until their rates of formation and consumption become equal. The change in concentration of each species is proportional to their stoichiometry in the balanced equation.

In the given problem, the change in concentration of \( \mathrm{A} \) and \( \mathrm{B} \) is denoted as 'x'. This value of 'x' reflects the amount of \( \mathrm{A} \) and \( \mathrm{B} \) that reacts to form \( \mathrm{C} \) and \( \mathrm{D} \) when the reaction achieves equilibrium. Since the concentration of \( \mathrm{C} \) at equilibrium is given to be twice that of \( \mathrm{A} \) at equilibrium, we set up the relationship \( 2x = 2(a - x) \) to find 'x'. Understanding how these concentration changes relate is vital for solving equilibrium constant problems.