Chapter 6, Problem 43E

(a)

Interpretation Introduction

Interpretation: Concentration of ${\text{H}}_{\text{2}}\text{O}$ , ${\text{Cl}}_2$ , and $\text{HOCl}$ at equilibrium when $\text{1}\text{.0}\text{g}$ ${\text{H}}_{\text{2}}\text{O}$ and $\text{2}\text{.00}\text{g}$ ${\text{Cl}}_{\text{2}}\text{O}$ are mixed is to be calculated.

Concept introduction:Law of mass action states that rate of any chemical reaction is proportional to product of molar concentrations of the reactants, each raised with power of its coefficient in a balanced chemical reaction.

Consider a hypothetical reversible reaction as follows:

  $\text{aA}+\text{bB}\rightleftharpoons \text{cC}+\text{dD}$

The expression used to calculate $K_{\text{eq}}$ is as follows:

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{C}\right]}_{\text{eq}}^{\text{c}}{\left[\text{D}\right]}_{\text{eq}}^{\text{d}}}{{\left[\text{A}\right]}_{\text{eq}}^{\text{a}}{\left[\text{B}\right]}_{\text{eq}}^{\text{b}}}$

Where,

• ${\text{K}}_{\text{f}}$ is the rate constant of forward reaction.
• ${\text{K}}_{\text{b}}$ is the rate constant of backward reaction.
• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.
• ${\left[\text{A}\right]}_{\text{eq}}$ and ${\left[\text{B}\right]}_{\text{eq}}$ are the concentration of reactant ‘A’ and ‘B’ respectively at equilibrium.
• ${\left[\text{C}\right]}_{\text{eq}}$ and ${\left[\text{D}\right]}_{\text{eq}}$ are the concentration of product ‘C’ and ‘D’ respectively at equilibrium..
• ‘a’ and ‘b’ are the stoichiometric coefficient of reactant ‘A’ and ‘B’ respectively.
• c’ and ‘d’ are the stoichiometric coefficient of product ‘C’ and ‘D’ respectively.

Answer to Problem 43E

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is $0.05524\text{M}$ , equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is $0.01824\text{M}$ and equilibrium concentration of $\text{HOCl}$ is $0.00952\text{M}$ .

Explanation of Solution

Given Information: $\text{1}\text{.0}\text{g}$ ${\text{H}}_{\text{2}}\text{O}$ , $\text{2}\text{.00}\text{g}$ ${\text{Cl}}_{\text{2}}\text{O}$ are introduced in a $\text{1}\text{.00}\text{L}$ container. At particular temperature rate constant of the reaction is 0.090.

The balanced chemical reaction is written as follows:

  ${\text{H}}_{\text{2}}\text{O}+{\text{Cl}}_{\text{2}}\text{O}\rightleftharpoons \text{2HOCl}$

Initial concentration of ${\text{H}}_{\text{2}}\text{O}$ is calculated as follows:

  $\begin{array}{c}{\text{Initial concentration of H}}_{\text{2}}\text{O}=\frac{\text{1}\text{g}}{\left(\text{18}\text{g}{\text{mol}}^{\text{-1}}\right)\left(\text{1}\text{L}\right)}\\ =0.06\text{M}\end{array}$

Initial concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is calculated as follows:

  $\begin{array}{c}{\text{Initial concentration of Cl}}_{\text{2}}\text{O}=\frac{2\text{g}}{\left(86.9\text{g}{\text{mol}}^{\text{-1}}\right)\left(\text{1}\text{L}\right)}\\ =0.023\text{M}\end{array}$

ICE table for above reaction is constructed as follows:

  $\begin{array}{cccccc}\text{Equation}& {\text{H}}_{\text{2}}\text{O}& +& {\text{Cl}}_{\text{2}}\text{O}& \rightleftharpoons & \text{2HOCl}\\ \text{Initial}& \text{0}\text{.06}& & \text{0}\text{.023}& & \text{0}\\ \text{Change}& -x& & -x& & +2x\\ \text{Equilibrium}& \text{0}\text{.06}-\text{x}& & \text{0}\text{.023}-x& & 2x\end{array}$

At equilibrium the equation related to the concentration of all species with the equilibrium constant is written as follows:

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{HOCl}\right]}_{\text{eq}}^{\text{2}}}{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{eq}}{\left[{\text{Cl}}_{\text{2}}\text{O}\right]}_{\text{eq}}}$

Where,

• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.
• ${\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{eq}}$ is the equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ .
• ${\left[{\text{Cl}}_{\text{2}}\text{O}\right]}_{\text{eq}}$ is the equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ .
• ${\left[\text{HOCl}\right]}_{\text{eq}}$ is the equilibrium concentration of $\text{HOCl}$ .

Equilibrium constant is 0.090.

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is $\left(\text{0}\text{.06}-\text{x}\right)\text{M}$ .

Equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is $\left(\text{0}\text{.023}-\text{x}\right)\text{M}$ .

Equilibrium concentration of $\text{HOCl}$ is $2\text{x}\text{M}$ .

Substitute the values in above equation.

  $\begin{array}{c}0.090=\frac{{\left(2\text{x}\right)}^2}{\left(\text{0}\text{.06}-\text{x}\right)\left(\text{0}\text{.023}-\text{x}\right)}\\ 3.91{\text{x}}^2+\left(7.47\times {10}^{-3}\right)\text{x}-\left(1.242\times {10}^{-4}\right)=0\end{array}$

Solve the above quadratic equation and value of $\text{x}$ obtained is $0.00476$ .

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is calculated as follows:

  $\begin{array}{c}\text{Equilibrium concentration of}{\text{H}}_{\text{2}}\text{O}=\left(0.06-0.00476\right)\text{M}\\ =0.05524\text{M}\end{array}$

Equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is calculated as follows:

  $\begin{array}{c}\text{Equilibrium concentration of}{\text{Cl}}_{\text{2}}\text{O}=\left(0.023-0.00476\right)\text{M}\\ =0.01824\text{M}\end{array}$

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

  $\begin{array}{c}\text{Equilibrium concentration of}\text{HOCl}=0.01824\text{M}\left(2\right)\left(0.00476\right)\text{M}\\ =0.00952\text{M}\end{array}$

(b)

Interpretation Introduction

Interpretation: Concentration of ${\text{H}}_{\text{2}}\text{O}$ , ${\text{Cl}}_2$ , and $\text{HOCl}$ at equilibrium when $\text{1}\text{.0}\text{mole}$ $\text{HOCl}$ is introduced is to be calculated.

Concept introduction:Law of mass action states that rate of any chemical reaction is proportional to product of molar concentrations of the reactants, each raised with power of its coefficient in a balanced chemical reaction.

Consider a hypothetical reversible reaction as follows:

  $\text{aA}+\text{bB}\rightleftharpoons \text{cC}+\text{dD}$

The expression used to calculate $K_{\text{eq}}$ is as follows:

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{C}\right]}_{\text{eq}}^{\text{c}}{\left[\text{D}\right]}_{\text{eq}}^{\text{d}}}{{\left[\text{A}\right]}_{\text{eq}}^{\text{a}}{\left[\text{B}\right]}_{\text{eq}}^{\text{b}}}$

Where,

• ${\text{K}}_{\text{f}}$ is the rate constant of forward reaction.
• ${\text{K}}_{\text{b}}$ is the rate constant of backward reaction.
• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.
• ${\left[\text{A}\right]}_{\text{eq}}$ and ${\left[\text{B}\right]}_{\text{eq}}$ are the concentration of reactant ‘A’ and ‘B’ respectively at equilibrium.
• ${\left[\text{C}\right]}_{\text{eq}}$ and ${\left[\text{D}\right]}_{\text{eq}}$ are the concentration of product ‘C’ and ‘D’ respectively at equilibrium..
• ‘a’ and ‘b’ are the stoichiometric coefficient of reactant ‘A’ and ‘B’ respectively.
• c’ and ‘d’ are the stoichiometric coefficient of product ‘C’ and ‘D’ respectively.

Answer to Problem 43E

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is $0.22\text{M}$ , equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is $0.22\text{M}$ and equilibrium concentration of $\text{HOCl}$ is $0.065\text{M}$ .

Explanation of Solution

Given Information: $\text{1}\text{.0}\text{mol}$ of $\text{HOCl}$ is introduced in a $\text{2}\text{.00}\text{L}$ container. At particular temperature rate constant of the reaction is 0.090.

The balanced chemical reaction is written as follows:

  ${\text{H}}_{\text{2}}\text{O}+{\text{Cl}}_{\text{2}}\text{O}\rightleftharpoons \text{2HOCl}$

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

  $\begin{array}{c}\text{Initial concentration of HOCl} =\frac{\text{1}\text{mol}}{\left(\text{2}\text{L}\right)}\\ =0.5\text{M}\end{array}$

ICE table for above reaction is constructed as follows:

  $\begin{array}{cccccc}\text{Equation}& \text{2HOCl}& \rightleftharpoons & {\text{H}}_{\text{2}}\text{O}& +& {\text{Cl}}_{\text{2}}\text{O}\\ \text{Initial}& \text{0}\text{.5}& & \text{0}& & \text{0}\\ \text{Change}& -x& & +x& & +x\\ \text{Equilibrium}& \text{0}\text{.5}-2\text{x}& & x& & x\end{array}$

At equilibrium the equation related to the concentration of all species with the equilibrium constant is written as follows:

  ${\text{K}}_{\text{eq}}=\frac{{\left[\text{HOCl}\right]}_{\text{eq}}^{\text{2}}}{{\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{eq}}{\left[{\text{Cl}}_{\text{2}}\text{O}\right]}_{\text{eq}}}$

Where,

• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.
• ${\left[{\text{H}}_{\text{2}}\text{O}\right]}_{\text{eq}}$ is the equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ .
• ${\left[{\text{Cl}}_{\text{2}}\text{O}\right]}_{\text{eq}}$ is the equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ .
• ${\left[\text{HOCl}\right]}_{\text{eq}}$ is the equilibrium concentration of $\text{HOCl}$ .

Equilibrium constant is 0.090.

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is $\text{x}\text{M}$ .

Equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is $\text{x}\text{M}$ .

Equilibrium concentration of $\text{HOCl}$ is $\left(0.5-2\text{x}\right)\text{M}$ .

Substitute the values in above equation as follows:

  $\begin{array}{c}0.090=\frac{{\left(0.5-2\text{x}\right)}^2}{\left(\text{x}\right)\left(\text{x}\right)}\\ 0.3=\frac{0.5-2\text{x}}{\text{x}}\\ \text{x}=0.22\end{array}$

Equilibrium concentration of ${\text{H}}_{\text{2}}\text{O}$ is $\text{0}\text{.22}\text{M}$ and equilibrium concentration of ${\text{Cl}}_{\text{2}}\text{O}$ is $\text{0}\text{.22}\text{M}$ .

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

  $\begin{array}{c}\text{Equilibrium concentration of}\text{HOCl}=\left(0.5-\left(2\right)\left(0.22\right)\right)\text{M}\\ =0.065\text{M}\end{array}$