Chapter 6, Problem 37E

a)

Interpretation Introduction

Interpretation:Whether system of $1.0\text{ L}$ flask with $1.0\text{ mol HOCl}$ , $0.10{\text{ mol Cl}}_2\text{O}$ and $0.10{\text{ mol H}}_2\text{O}$ is at equilibrium or not should be determined. Also, shift in system direction if it is not at equilibrium should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, reactants and products have constant concentration. Equilibrium constant is denoted by $K$ .

Reaction quotient is determined with help of initial concentrations of reactants and products. It is denoted by $Q$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q$ and $K$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q$ and $K$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q$ is greater than $K$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q$ is lessthan $K$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Given Information: Value of $\left[\text{HOCl}\right]$ is $1.0M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.10M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.10M$ .

Concentration of $\text{HOCl}$ is calculated as follows:

  $\begin{array}{c}\left[\text{HOCl}\right]=\frac{1.0\text{ mol}}{1.0\text{ L}}\\ =1.0M\end{array}$

Concentration of ${\text{Cl}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}_2\text{O}\right]=\frac{0.10\text{ mol}}{1.0\text{ L}}\\ =0.10M\end{array}$

Concentration of ${\text{H}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\frac{0.10\text{ mol}}{1.0\text{ L}}\\ =0.10M\end{array}$

Given chemical reaction is as follows:

  ${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\leftrightarrow 2\text{HOCl}\left(g\right)$

Expression for $Q$ is as follows:

  $Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}$

Where,

• $Q$ isreaction quotient.
• $\left[\text{HOCl}\right]$ is concentration of $\text{HOCl}$ .
• $\left[{\text{Cl}}_2\text{O}\right]$ is concentration of ${\text{Cl}}_2\text{O}$ .
• $\left[{\text{H}}_2\text{O}\right]$ is concentration of ${\text{H}}_2\text{O}$ .

Value of $\left[\text{HOCl}\right]$ is $1.0M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.10M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.10M$ .

Substitute the values in above equation.

  $\begin{array}{c}Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}\\ =\frac{{\left(1.0M\right)}^2}{\left(0.10M\right)\left(0.10M\right)}\\ =100\end{array}$

But value of $K$ is 0.0900. Since value of $Q$ is very much greater than that of $K$ , shift in left direction is observed. This means consumption of products will be more that leads to more formation of reactants and no equilibrium is attained.

b)

Interpretation Introduction

Interpretation: Whether system of $2.0\text{ L}$ flask with $0.084\text{ mol HOCl}$ , $0.080{\text{ mol Cl}}_2\text{O}$ and $0.98{\text{ mol H}}_2\text{O}$ is at equilibrium or not should be determined. Also, shift in system direction if it is not at equilibrium should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, reactants and products have constant concentration. Equilibrium constant is denoted by $K$ .

Reaction quotient is determined with help of initial concentrations of reactants and products. It is denoted by $Q$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q$ and $K$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q$ and $K$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q$ is greater than $K$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q$ is less than $K$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Given Information: Value of $\left[\text{HOCl}\right]$ is $0.042M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.040M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.49M$ .

Concentration of $\text{HOCl}$ is calculated as follows:

  $\begin{array}{c}\left[\text{HOCl}\right]=\frac{0.084\text{ mol}}{2.0\text{ L}}\\ =0.042M\end{array}$

Concentration of ${\text{Cl}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}_2\text{O}\right]=\frac{0.080\text{ mol}}{2.0\text{ L}}\\ =0.040M\end{array}$

Concentration of ${\text{H}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\frac{0.98\text{ mol}}{2.0\text{ L}}\\ =0.49M\end{array}$

Given chemical reaction is as follows:

  ${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2\text{HOCl}\left(g\right)$

Expression for $Q$ is as follows:

  $Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}$

Where,

• $Q$ is reaction quotient.
• $\left[\text{HOCl}\right]$ is concentration of $\text{HOCl}$ .
• $\left[{\text{Cl}}_2\text{O}\right]$ is concentration of ${\text{Cl}}_2\text{O}$ .
• $\left[{\text{H}}_2\text{O}\right]$ is concentration of ${\text{H}}_2\text{O}$ .

Value of $\left[\text{HOCl}\right]$ is $0.042M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.040M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.49M$ .

Substitute the values in above equation.

  $\begin{array}{c}Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}\\ =\frac{{\left(0.042M\right)}^2}{\left(0.040M\right)\left(0.49M\right)}\\ =0.09\end{array}$

But value of $K$ is 0.0900. Since value of $Q$ is equal to that of $K$ , equilibrium is achieved and no shift in any direction is observed.

c)

Interpretation Introduction

Interpretation: Whether system of $3.0\text{ L}$ flask with $0.25\text{ mol HOCl}$ , $0.0010{\text{ mol Cl}}_2\text{O}$ and $0.56{\text{ mol H}}_2\text{O}$ is at equilibrium or not should be determined. Also, shift in system direction if it is not at equilibrium should be determined.

Concept introduction:Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, reactants and products have constant concentration. Equilibrium constant is denoted by $K$ .

Reaction quotient is determined with help of initial concentrations of reactants and products. It is denoted by $Q$ . If system is not in equilibrium, direction of shift is determined by comparison of $Q$ and $K$ . Below mentioned are three cases to determine shift of reaction direction.

1. If $Q$ and $K$ are equal, equilibrium is attained and no shift in direction is observed.

2. If $Q$ is greater than $K$ , shift in left direction is observed. This results in more consumption of products and formation of reactants.

3. If $Q$ is less than $K$ , shift in right direction is observed. This results in more consumption of reactants and formation of products.

Explanation of Solution

Given Information: Value of $\left[\text{HOCl}\right]$ is $0.083M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.00033M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.1867M$ .

Concentration of $\text{HOCl}$ is calculated as follows:

  $\begin{array}{c}\left[\text{HOCl}\right]=\frac{0.25\text{ mol}}{3.0\text{ L}}\\ =0.083M\end{array}$

Concentration of ${\text{Cl}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}_2\text{O}\right]=\frac{0.0010\text{ mol}}{3.0\text{ L}}\\ =0.00033M\end{array}$

Concentration of ${\text{H}}_2\text{O}$ is calculated as follows:

  $\begin{array}{c}\left[{\text{H}}_2\text{O}\right]=\frac{0.56\text{ mol}}{3.0\text{ L}}\\ =0.1867M\end{array}$

Given chemical reaction is as follows:

  ${\text{H}}_{\text{2}}\text{O}\left(g\right)+{\text{Cl}}_{\text{2}}\text{O}\left(g\right)\rightleftharpoons 2\text{HOCl}\left(g\right)$

Expression for $Q$ is as follows:

  $Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}$

Where,

• $Q$ is reaction quotient.
• $\left[\text{HOCl}\right]$ is concentration of $\text{HOCl}$ .
• $\left[{\text{Cl}}_2\text{O}\right]$ is concentration of ${\text{Cl}}_2\text{O}$ .
• $\left[{\text{H}}_2\text{O}\right]$ is concentration of ${\text{H}}_2\text{O}$ .

Value of $\left[\text{HOCl}\right]$ is $0.083M$ .

Value of $\left[{\text{Cl}}_2\text{O}\right]$ is $0.00033M$ .

Value of $\left[{\text{H}}_2\text{O}\right]$ is $0.1867M$ .

Substitute the values in above equation.

  $\begin{array}{c}Q=\frac{{\left[\text{HOCl}\right]}^2}{\left[{\text{Cl}}_2\text{O}\right]\left[{\text{H}}_2\text{O}\right]}\\ =\frac{{\left(0.083M\right)}^2}{\left(0.00033M\right)\left(0.1867M\right)}\\ =111.82\end{array}$

But value of $K$ is 0.0900. Since value of $Q$ is very much greater than that of $K$ , shift in left direction is observed. This means consumption of products will be more that leads to more formation of reactants and no equilibrium is attained.