Chapter 6, Problem 6.61P

Interpretation Introduction

(a)

Interpretation:

Whether the given reversible reaction is at equilibrium or not needs to be determined.

Concept introduction:

For a reversible reaction, equilibrium constant ( $K_{eq}$ ) is defined as the ratio of the product of the concentrations of the products to the product of concentrations of reactants at equilibrium. The coefficients of each chemical component in a balanced equation should be taken as the power of its concentration in the equation.

  $\begin{array}{l}aA+bB\rightleftarrows cC+dD\\ \text{equilibrium constant = }{K}_{eq}=\frac{{\left[C\right]}^c\times {\left[D\right]}^d}{{\left[A\right]}^a\times {\left[B\right]}^b}\end{array}$

By comparing the magnitude of $K$ of a reaction at a given instance with its equilibrium constant ( $K_{eq}$ ), the progress of a reaction can be determined.

  $\begin{array}{l}\text{When }K=K_{eq},\text{ reaction is at equilibrium}\\ \text{When }K>K_{eq},\text{ the backward reaction is in progress to achieve equilibrium}\\ \text{When }K<K_{eq},\text{ the forward reaction is in progress to achieve equilibrium}\end{array}$

Answer to Problem 6.61P

The system is not at equilibrium.

Explanation of Solution

For the given reaction,

  $\begin{array}{l}\text{A+B}\rightleftarrows \text{C+D}\\ \text{equilibrium constant = }{K}_{eq}\text{=}\frac{\left[\text{C}\right]\text{×}\left[\text{D}\right]}{\left[\text{A}\right]\text{×}\left[\text{B}\right]}\end{array}$

Let's calculate the $K$ of the above reaction instance:

  $\begin{array}{l}\text{A+B}\rightleftarrows \text{C+D}\\ \text{No of A atoms in the system: 5}\\ \text{No of B atoms in the system: 6}\\ \text{No of C atoms in the system: 1}\\ \text{No of D atoms in the system: 2}\end{array}$

Therefore,

  $K\text{ = }\frac{\left[\text{C}\right]\text{×}\left[\text{D}\right]}{\left[\text{A}\right]\text{×}\left[\text{B}\right]}=\frac{1\times 2}{5\times 6}=0.067$

K of the reaction at the given instance is smaller than $K_{eq}$ value ( $K<K_{eq}$ ), thus the reaction is not at equilibrium.

Interpretation Introduction

(a)

Interpretation:

If the reaction is not in the equilibrium, the direction of reaction needs to be explained.

Concept introduction:

For a reversible reaction, equilibrium constant ( $K_{eq}$ ) is defined as the ratio of the product of the concentrations of the products to the product of concentrations of reactants at equilibrium. The coefficients of each chemical component in a balanced equation should be taken as the power of its concentration in the equation.

  $\begin{array}{l}aA+bB\rightleftarrows cC+dD\\ \text{equilibrium constant = }{K}_{eq}=\frac{{\left[C\right]}^c\times {\left[D\right]}^d}{{\left[A\right]}^a\times {\left[B\right]}^b}\end{array}$

By comparing the magnitude of $K$ of a reaction at a given instance with its equilibrium constant ( $K_{eq}$ ), the progress of a reaction can be determined.

  $\begin{array}{l}\text{When }K=K_{eq},\text{ reaction is at equilibrium}\\ \text{When }K>K_{eq},\text{ the backward reaction is in progress to achieve equilibrium}\\ \text{When }K<K_{eq},\text{ the forward reaction is in progress to achieve equilibrium}\end{array}$

Answer to Problem 6.61P

The reaction should proceed to the right (forward reaction) to achieve the equilibrium

Explanation of Solution

For the given reaction,

  $\begin{array}{l}\text{A+B}\rightleftarrows \text{C+D}\\ \text{equilibrium constant = }{K}_{eq}\text{=}\frac{\left[\text{C}\right]\text{×}\left[\text{D}\right]}{\left[\text{A}\right]\text{×}\left[\text{B}\right]}\end{array}$

Here,

  $K\text{ = }\frac{\left[\text{C}\right]\text{×}\left[\text{D}\right]}{\left[\text{A}\right]\text{×}\left[\text{B}\right]}=\frac{1\times 2}{5\times 6}=0.067$

Thus, ( $K<K_{eq}$ ), the forward reaction is in progress to achieve the equilibrium. In other words, more and more products should be formed in order to reach $K=K_{eq}=4$