Chapter 13, Problem 13.26CP

Consider the interconversion of A molecules (red spheres) and B molecules (blue spheres) according to the reaction A ⇌ B. Each of the following series of pictures represents a separate experiment in which time increases from left to right:

1. (a) Which of the experiments has resulted in an equilibrium state?
2. (b) What is the value of the equilibrium constant K_c for the reaction A ⇌ B?
3. (c) Explain why you calculate K_c without knowing the volume of the reaction vessel.

Interpretation Introduction

Interpretation:

From the given experiments, the one resulted in equilibrium state has to be identified.

Concept introduction:

Chemical equilibrium:

It is defined as a state in which concentrations of reactants and products are same over a period of time. An equilibrium mixture is a mixture in which reactants and products are in equilibrium.

Answer to Problem 13.26CP

Experiment one and three are in equilibrium state.

Explanation of Solution

Reason correct options:

Given,

Figure 1

In experiment one and three, the number of red spheres A and blue spheres B are same over a period of time. In both experiments third and fourth box have same number of A and B. Hence, experiment one and three are in equilibrium state.

Reason incorrect options:

In experiment two, the number of A and B are not same over a period of time. Hence, experiment two is not in equilibrium.

Interpretation Introduction

Interpretation:

The equilibrium constant for given reaction has to be calculated.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Answer to Problem 13.26CP

The value of equilibrium constant is $1.5$

Explanation of Solution

Given,

$\text{A}\underset{}{\overset{}{\rightleftharpoons }}\text{B}$

The equilibrium constant is calculated by substituting concentrations of reactants and products in equilibrium constant expression.

${\text{K}}_{\text{c}}\text{=}\frac{\text{[B]}}{\text{[A]}}\text{=}\frac{\text{6}}{\text{4}}\text{=}\text{1}\text{.5}$

Therefore, the value of equilibrium constant is $1.5$

Interpretation Introduction

Interpretation:

The reason for calculating equilibrium constant without knowing volume of the reaction vessel has to be explained.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.

Explanation of Solution

At equilibrium the number of molecules on reactant and product side remains same and volume gets cancelled while substituting in equilibrium constant expression. Hence, equilibrium constant ${\text{K}}_{\text{c}}$ can be calculated without knowing volume term.