Chapter 16, Problem 42E

Interpretation Introduction

(a)

Interpretation:

The direction of the equilibrium with the increase in the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

The equilibrium shifts to left with the increase in the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant, $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

In this case, $\left[{\text{Cd}}^{\text{2+}}\right]$ is present on the product side or right side. When its concentration is increased, it increases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards left.

Conclusion

The equilibrium shifts to left if the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$ is increased.

Interpretation Introduction

(b)

Interpretation:

The direction of the equilibrium with the increase in the concentration of $\left[{\text{S}}^{\text{2}-}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

The equilibrium shifts to left with the increase in the concentration of $\left[{\text{S}}^{\text{2}-}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

In this case, $\left[{\text{S}}^{\text{2}-}\right]$ is present on the product side or right side. When its concentration is increased, it increases the equilibrium constant. Therefore, it in order to maintain the equilibrium, the reaction will shift towards left.

Conclusion

The equilibrium shifts to left if the concentration of $\left[{\text{S}}^{\text{2}-}\right]$ is increased.

Interpretation Introduction

(c)

Interpretation:

The direction of the equilibrium with the decrease in the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

The equilibrium shifts to right with the decrease in the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

In this case, $\left[{\text{Cd}}^{\text{2+}}\right]$ ions are present on the product side or right side. When its concentration is decreased, it decreases the equilibrium constant. Therefore, in order to maintain the equilibrium, the reaction will shift towards right.

Conclusion

The equilibrium shifts to right if the concentration of $\left[{\text{Cd}}^{\text{2+}}\right]$ is decreased.

Interpretation Introduction

(d)

Interpretation:

The direction of the equilibrium with the decrease in the concentration of $\left[{\text{S}}^{\text{2}-}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

When the concentration of $\left[{\text{S}}^{\text{2}-}\right]$ is decreased the equilibrium shift toward right.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

In this case, $\left[{\text{S}}^{\text{2}-}\right]$ is present on the product side or right side. When its concentration is decreased, it decreases the equilibrium constant. in order to maintain the equilibrium, the reaction will shift towards right.

Conclusion

The equilibrium shifts to right if the concentration of $\left[{\text{S}}^{\text{2}-}\right]$ is decreased.

Interpretation Introduction

(e)

Interpretation:

The direction of the equilibrium with the addition of solid $\text{CdS}$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

It does not affect the equilibrium direction if solid $\text{CdS}$ is added.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

Since $\left[\text{CdS}\right]$ is not present in the equilibrium constant equation. Therefore, the addition of solid $\left[\text{CdS}\right]$ does not cause any change in the value of the equilibrium constant value or no change in the direction of equilibrium.

Conclusion

If solid $\text{CdS}$ is added then it does not cause any change in the direction of the equilibrium.

Interpretation Introduction

(f)

Interpretation:

The direction of the equilibrium with the addition of solid $\left[\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right]$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

When the solid $\left[\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right]$ is added the equilibrium shift towards left.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

The solid $\left[\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right]$ is basically deliquescent in nature so its phase is aqueous. It also contain $\left[{\text{Cd}}^{2+}\right]$ ions. Therefore, the addition of solid $\left[\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right]$ increases the equilibrium constant value which shifts the equilibrium towards left.

Conclusion

The direction of equilibrium is left if solid $\left[\text{Cd}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right]$ is added.

Interpretation Introduction

(g)

Interpretation:

The direction of the equilibrium with the addition of solid ${\text{NaNO}}_3$ is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

It does not affect the equilibrium direction if solid ${\text{NaNo}}_3$ is added.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

From equation (2), there is no affect on the equilibrium direction. When the solid $\left[{\text{NaNO}}_3\right]$ is added.

Conclusion

It does not cause any change in the equilibrium direction if solid ${\text{NaNO}}_3$ is added.

reaction.

Interpretation Introduction

(h)

Interpretation:

The direction of the equilibrium with the addition of ${\text{H}}^{\text{+}}$ ions is to be stated.

Concept introduction:

In a chemical reaction, when the rate of both forward and reverse reaction are equal. Then the chemical reaction is in equilibrium. In other words, their concentration does not vary with the time change. They are reversible in nature.

It is represented as $\rightleftarrows$ between the chemical reaction. It is denoted as $K_{eq}$. It is known as equilibrium constant. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients as shown below.

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Also, the concentration of liquid and solid is not considered in the equilibrium constant expression.

Answer to Problem 42E

It does not affect the equilibrium direction if ${\text{H}}^{\text{+}}$ ions are added.

Explanation of Solution

According to the Le Chatelier’s principle, the change in concentration, volume, pressure and temperature affect the equilibrium of the reaction.

The reaction is given below.

$\text{CdS}\left(s\right)\rightleftarrows {\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2}-}\left(aq\right)$…(1)

According to the definition of the equilibrium constant $K_{eq}$. The $\left[\text{CdS}\right]$ molecule is present in solid state so its concentration is not considered. The equilibrium constant is written as shown below.

$K_{eq}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{2-}\right]$…(2)

When $\left[{\text{H}}^{+}\right]$ ions are added, it only increases the $\left[{\text{H}}^{+}\right]$ ions in the reaction. Although it is not present in the equation (2) so it has no affect on equilibrium.

Conclusion

It does not cause any change in the direction of equilibrium if ${\text{H}}^{\text{+}}$ ions are added.