Chapter 19, Problem 69QRT

(a)

Interpretation Introduction

Interpretation:

Oxidizing agent and reducing agent has to be identified in the given reaction.

    ${\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{Cl}}^{\text{-}}\text{(aq)}\text{+}\text{HOCl(aq)}$

Explanation of Solution

Given reaction in the problem statement is,

    ${\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{Cl}}^{\text{-}}\text{(aq)}\text{+}\text{HOCl(aq)}$

Oxidation state of chlorine in chlorine molecule is zero.  This gains an electron and gets converted into chloride ion.  In the product side, the oxidation state of chlorine in $\text{HOCl}$ is $+1$.  This means chlorine lost its electron to form ${\text{ClO}}^{\text{-}}$ ion.  Therefore, chlorine molecule acts as both oxidizing and reducing agent.

(b)

Interpretation Introduction

Interpretation:

Equilibrium constant expression for the given reaction has to be written.

Explanation of Solution

Given reaction in the problem statement is,

    ${\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{Cl}}^{\text{-}}\text{(aq)}\text{+}\text{HOCl(aq)}$

On seeing the above equation, it is found that there are different phases.  Hence, this is an heterogeneous equilibria reaction.  Therefore, for heterogeneous equilibria, the concentration symbol is used for the aqueous species and pressure symbol is used for gas-phase species.  Liquids are not considered in this expression.  Therefore, the equilibrium constant expression can be given as,

    ${\text{K}}_{\text{p}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][Cl}}^{\text{-}}\text{][HOCl]}}{{\text{P}}_{{\text{Cl}}_{\text{2}}}}$

(c)

Interpretation Introduction

Interpretation:

Concentration of $\text{HOCl}$ has to be calculated.

Explanation of Solution

Equilibrium constant is given as $2.7\times {10}^{-5}$.  Equilibrium pressure of ${\text{Cl}}_{\text{2}}$ is given as $\text{1}\text{.0}\text{atm}$.

The reaction that is given in the problem statement is,

    ${\text{Cl}}_{\text{2}}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\rightleftharpoons }{\text{H}}^{\text{+}}\text{(aq)}\text{+}{\text{Cl}}^{\text{-}}\text{(aq)}\text{+}\text{HOCl(aq)}$

Equilibrium expression for this can be given as,

    ${\text{K}}_{\text{p}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][Cl}}^{\text{-}}\text{][HOCl]}}{{\text{P}}_{{\text{Cl}}_{\text{2}}}}-------------------(1)$

Concentration of the products increases till the equilibrium is reached.

${\text{Cl}}_{\text{2}}\text{(g)}$     ${\text{HN}}_{\text{3}}\text{(aq)}$        ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$    ${\text{N}}_{\text{3}}{}^{-}\text{(aq)}$

Initial        $\text{P}\text{=}{\text{P}}_{\text{initial}}$    $\text{Conc}\text{=}\text{0}\text{M}$        $\text{Conc}\text{=}\text{0}\text{M}$    $\text{Conc}\text{=}\text{0}\text{M}$

Change    $-some$    $+x\text{M}$            $+x\text{M}$        $+x\text{M}$

Equilibrium    $\text{1}\text{.0}\text{atm}$    $x\text{M}$            $x\text{M}$        $x\text{M}$

Substituting these values in equation (1),

    $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{\text{(x)(x)(x)}}{\text{(1}\text{.0)}}\\ 2.7\times {10}^{-5}=\frac{{\text{x}}^{\text{3}}}{\text{(1}\text{.0)}}\\ x^3=2.7\times {10}^{-5}\\ x=\sqrt[3]{2.7\times {10}^{-5}}\\ =0.03\text{M}\end{array}$

Therefore, concentration of $\text{HOCl}$ is $\text{0}\text{.03}\text{M}$.