Question

\(\mathrm{K}_{\mathrm{c}}\) for \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) is 10 at \(25^{\circ} \mathrm{C}\). If a con- tainer contains \(1,2,3,4\) moles per litre of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and D, respectively at \(25^{\circ} \mathbf{C}\) the reaction shall (1) Procecd from left to right (2) Procecd from right to left (3) Be at cquilibrium (4) Nonc

Short answer

The reaction will proceed from left to right.

Step-by-step solution

Understand Given Information

Given equilibrium constant \(\text{K}_c\) for the reaction \( \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \) is 10 at \( 25^{\text{o}} \text{C}.\)

Initial Concentrations

The initial concentrations are given: \[ [\text{A}] = 1 \, \text{mol/L}, \, [\text{B}] = 2 \, \text{mol/L}, \, [\text{C}] = 3 \, \text{mol/L}, \, [\text{D}] = 4 \, \text{mol/L}. \]

Calculate Reaction Quotient \( Q \)

The reaction quotient is calculated as follows: \[ Q = \frac{[\text{C}] \times [\text{D}]}{[\text{A}] \times [\text{B}]} = \frac{3 \times 4}{1 \times 2} = 6 \]

Compare \( Q \) with \( \text{K}_c \)

Compare the value of \( Q \) with the equilibrium constant \( \text{K}_c \): \[ Q = 6 < \text{K}_c = 10 \]

Determine Reaction Direction

Since \( Q < \text{K}_c \), the reaction will proceed from left to right to reach equilibrium.

Key concepts

Reaction Quotient

The reaction quotient, often denoted as \( Q \), is a measure used to determine the direction in which a chemical reaction will proceed to reach equilibrium. You can calculate \( Q \) using the concentrations of the reactants and products at any point in time using the formula: \[ Q = \frac{[\text{products}]}{[\text{reactants}]} \] This is similar to the equilibrium constant \( K \), but \( Q \) can be used at any stage of the reaction, not just at equilibrium. In the given exercise, we calculated \( Q \) for the reaction \( A + B \rightleftharpoons C + D \) using the initial concentrations:

• \( Q = \frac{[C] \times [D]}{[A] \times [B]} = \frac{3 \times 4}{1 \times 2} = 6 \).

Comparing this value to the equilibrium constant helps in predicting the reaction's direction.

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), provides a snapshot of the ratio of product concentrations to reactant concentrations when a reaction is at equilibrium. Each chemical reaction has its unique \( K_c \) value, which remains constant at a given temperature. In the exercise, we know that \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \) is 10 at 25°C. This signifies that, at equilibrium, the ratio \[ K_c = \frac{[C] \times [D]}{[A] \times [B]} = 10 \] must hold. The equilibrium constant tells us about the position of equilibrium and whether a reaction mixture will have more products or reactants at equilibrium. A large \( K_c \) suggests a product-favored reaction, while a small one suggests reactant-favored. Comparing this to the calculated \( Q \) allows us to predict the shift needed to reach equilibrium.

Reaction Direction

Determining the direction of a reaction is essential for understanding how a system reaches equilibrium. By comparing \( Q \) to \( K_c \), we can predict this direction:

• If \( Q < K_c \), the reaction will proceed from left to right (forming more products).
• If \( Q > K_c \), the reaction will proceed from right to left (forming more reactants).
• If \( Q = K_c \), the reaction is already at equilibrium.

In our specific problem, since \( Q = 6 < K_c = 10 \), the reaction will move from left to right, favoring the production of \( C \) and \( D \), hence moving towards equilibrium. Knowing the direction is crucial for controlling reactions in chemistry and industrial processes.