Chapter 16, Problem 46E

Interpretation Introduction

(a)

Interpretation:

The direction of the equilibrium on adding gaseous $\left[\text{HCl}\right]$ 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 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 writing the expression.

Answer to Problem 46E

When the gaseous $\text{HCl}$ is added then the equilibrium shifts to the left.

Explanation of Solution

The reaction is given below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$ … (1)

According to the definition of the equilibrium constant, $K_{eq}$. The $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is present in liquid state so its concentration is not considered. The equilibrium constant is written as shown below.

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

Since, gaseous $\left[\text{HCl}\right]$ is added to the solution. It contain common $\left[{\text{H}}^{\text{+}}\right]$ ions and according to the equation (2), $\left[{\text{H}}^{\text{+}}\right]$ ions is present on the product side. Therefore, it increases the concentration of $\left[{\text{H}}^{\text{+}}\right]$ ions in the solution which increases the equilibrium constant. As a result, it shifts the equilibrium to the left.

Conclusion

The equilibrium shifts to the left when the gaseous $\left[\text{HCl}\right]$ is added.

Interpretation Introduction

(b)

Interpretation:

The direction of the equilibrium on adding solid $\left[\text{NaOH}\right]$ 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 known as in equilibrium. In other words, their concentration does not vary with 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 writing the expression.

Answer to Problem 46E

When the solid $\left[\text{NaOH}\right]$ is added the equilibrium shifts toward left.

Explanation of Solution

The reaction is given below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$ … (1)

According to the definition of the equilibrium constant, $K_{eq}$. $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is present in liquid state so its concentration is not considered. The equilibrium constant is written as shown below.

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

The solid $\left[\text{NaOH}\right]$ is basically deliquescent in nature so its phase is aqueous. It also contains common $\left[{\text{OH}}^{-}\right]$ ions. Therefore, on addition of solid $\left[\text{NaOH}\right]$ the equilibrium constant value increases which shift the equilibrium towards left.

Conclusion

The equilibrium shifts toward left on adding solid $\left[\text{NaOH}\right]$.

Interpretation Introduction

(c)

Interpretation:

The direction of the equilibrium on adding liquid $\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]$ 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 known as in equilibrium. In other words, their concentration does not vary with 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 writing the expression.

Answer to Problem 46E

When the liquid $\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]$ is added then the equilibrium shifts to the left.

Explanation of Solution

The reaction is given below:

${\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)$ … (1)

According to the definition of the equilibrium constant, $K_{eq}$. $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is present in liquid state so its concentration is not considered. The equilibrium constant is written as shown below.

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

Since, liquid $\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]$ is added to the solution. It contains common $\left[{\text{H}}^{\text{+}}\right]$ ions and according to the equation (2), $\left[{\text{H}}^{\text{+}}\right]$ ions is present on the product side. Therefore, it increases the concentration $\left[{\text{H}}^{\text{+}}\right]$ ions in the solution which increases the equiibrium constant. As a result, it shifts the equilibrium to the left.

Conclusion

The equilibrium shifts toward left on adding liquid $\left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right]$.

Interpretation Introduction

(d)

Interpretation:

The direction of the equilibrium on adding solid $\left[\text{NaF}\right]$ 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 known as 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 writing the expression.

Answer to Problem 46E

When the solid $\left[\text{NaF}\right]$ is added then there is no shift in the equilibrium.

Explanation of Solution

The reaction is given below:

${\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$ … (1)

According to the definition of the equilibrium constant, $K_{eq}$. $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is present in liquid so its concentration is not considered. The equilibrium constant is written as shown below.

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

From equation (2) there is no solid $\left[\text{NaF}\right]$ compound present. Therefore, there is no effect of solid $\left[\text{NaF}\right]$ on the equilibrium direction.

Conclusion

There is no shift in the equilibrium of the reaction when the solid $\left[\text{NaF}\right]$ is added.