Chapter 12, Problem 98QRT

(a)

Interpretation Introduction

Interpretation:

In the formation of mustard gas from ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$, the partial pressure of each gases has to be calculated after equilibrium is established.

Concept Introduction:

Ideal gas equation:

The ideal gas equation is given by,

  $\text{P}\text{V}\text{=}\text{n}\text{R}\text{T}$

Where,

  $\text{P}$ is the pressure,

  $\text{V}$ is the volume,

  $\text{T}$ is the temperature,

  $\text{R}$ is molar gas constant,

  $n$ is the number of moles.

Explanation of Solution

Given information,

Temperature is $\text{20}\text{.0}{}^{\text{o}}\text{C}\text{or}\text{293}\text{.2}\text{K}$, volume of flask is $\text{5}\text{.00}\text{L}$, amount of ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ are 0.258 and $\text{0}\text{.592}\text{mol}$ respectively.

  Formation of mustard gas from ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$,

  ${\text{SCl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}\left(\text{g}\right)}$

  After equilibrium reached, the amount of mustard gas present is $\text{0}\text{.0349}\text{mol}$.

Calculate the equilibrium concentrations of ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$,

  $\begin{array}{l}\left[{\text{SCl}}_{\text{2}}\right]\text{=}\frac{\text{0}\text{.258}\text{mol}{\text{SCl}}_{\text{2}}}{\text{5}\text{.00}\text{L}}\text{=}\text{0}\text{.0516}\text{mol}\text{/}\text{L}\\ \\ \left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\text{=}\frac{\text{0}\text{.592}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{5}\text{.00}\text{L}}\text{=}\text{0}\text{.118}\text{mol}\text{/}\text{L}\end{array}$

Construct ICE table for the above reaction,

  $\begin{array}{l}\underset{\_}{{\text{SCl}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}\left(\text{g}\right)}}\\ \text{Initial}\text{conc}\text{.}\text{(M):}\text{0}\text{.0516}\text{0}\text{.118}\text{0}\\ \text{Change}\text{in}\text{conc}\text{.}\text{(M):}\text{-x}\text{-2x}\text{+}\text{x}\\ \text{Equilibrium}\text{conc}\text{(M):}\text{0}\text{.0516-x}\text{0}\text{.118}\text{-}\text{2x}\text{x}\end{array}$

At equilibrium,

  $\begin{array}{l}\text{x}\text{=}\left[\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}\right]\text{=}\text{0}\text{.0349}\text{mol}\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}\text{/5}\text{.00}\text{L}\text{=}\text{0}\text{.00698}\text{mol/L}\\ \\ \left[{\text{SCl}}_{\text{2}}\right]\text{=}\text{0}\text{.0516}\text{-}\text{0}\text{.00698}\text{=}\text{0}\text{.0446}\text{mol/L}\\ \\ \left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\text{=}\text{0}\text{.118}\text{-}\text{2}\left(\text{0}\text{.00698}\right)\text{=}\text{0}\text{.104}\text{mol/L}\end{array}$

Calculate the partial pressures by using ideal gas equation,

  $\begin{array}{l}\text{P}\text{=}\left(\text{n/V}\right)\text{RT}\\ \\ {\text{P}}_{{\text{SCl}}_{\text{2}}}\text{=}\left(\text{0}\text{.0446}\text{mol/L}\right)\left(\text{0}\text{.08206}\frac{\text{atm}\text{.L}}{\text{K}\text{.mol}}\right)\left(\text{293}\text{.2}\text{K}\right)\text{=}\text{1}\text{.07}\text{atm}\end{array}$

  $\begin{array}{l}\text{P}\text{=}\left(\text{n/V}\right)\text{RT}\\ \\ {\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\text{=}\left(\text{0}\text{.104}\text{mol/L}\right)\left(\text{0}\text{.08206}\frac{\text{atm}\text{.L}}{\text{K}\text{.mol}}\right)\left(\text{293}\text{.2}\text{K}\right)\text{=}\text{2}\text{.50}\text{atm}\end{array}$

  $\begin{array}{l}\text{P}\text{=}\left(\text{n/V}\right)\text{RT}\\ \\ {\text{P}}_{\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}}\text{=}\left(\text{0}\text{.00698}\text{mol/L}\right)\left(\text{0}\text{.08206}\frac{\text{atm}\text{.L}}{\text{K}\text{.mol}}\right)\left(\text{293}\text{.2}\text{K}\right)\text{=}\text{0}\text{.168}\text{atm}\end{array}$

(b)

Interpretation Introduction

Interpretation:

In the formation of mustard gas from ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$, the value of ${\text{K}}_{\text{C}}$ has to be calculated at $\text{20}\text{.0}{}^{\text{o}}\text{C}$.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

    $\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

    ${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

    ${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

    ${\text{K}}_{\text{c}}$ is the equilibrium constant.

Explanation of Solution

Given information,

Temperature is $\text{20}\text{.0}{}^{\text{o}}\text{C}\text{or}\text{293}\text{.2}\text{K}$, volume of flask is $\text{5}\text{.00}\text{L}$, amount of ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ are 0.258 and $\text{0}\text{.592}\text{mol}$ respectively.

  Formation of mustard gas from ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$,

  ${\text{SCl}}_{\text{2}\left(\text{g}\right)}\text{+}\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}\left(\text{g}\right)}$

  After equilibrium reached, the amount of mustard gas present is $\text{0}\text{.0349}\text{mol}$.

Calculate the equilibrium concentrations of ${\text{SCl}}_{\text{2}}\text{and}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$,

  $\begin{array}{l}\left[{\text{SCl}}_{\text{2}}\right]\text{=}\frac{\text{0}\text{.258}\text{mol}{\text{SCl}}_{\text{2}}}{\text{5}\text{.00}\text{L}}\text{=}\text{0}\text{.0516}\text{mol}\text{/}\text{L}\\ \\ \left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\text{=}\frac{\text{0}\text{.592}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{5}\text{.00}\text{L}}\text{=}\text{0}\text{.118}\text{mol}\text{/}\text{L}\end{array}$

Construct ICE table for the above reaction,

  $\begin{array}{l}\underset{\_}{{\text{SCl}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}\text{2}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}\left(\text{g}\right)}}\\ \text{Initial}\text{conc}\text{.}\text{(M):}\text{0}\text{.0516}\text{0}\text{.118}\text{0}\\ \text{Change}\text{in}\text{conc}\text{.}\text{(M):}\text{-x}\text{-2x}\text{+}\text{x}\\ \text{Equilibrium}\text{conc}\text{(M):}\text{0}\text{.0516-x}\text{0}\text{.118}\text{-}\text{2x}\text{x}\end{array}$

At equilibrium,

  $\begin{array}{l}\text{x}\text{=}\left[\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}\right]\text{=}\text{0}\text{.0349}\text{mol}\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}\text{/5}\text{.00}\text{L}\text{=}\text{0}\text{.00698}\text{mol/L}\\ \\ \left[{\text{SCl}}_{\text{2}}\right]\text{=}\text{0}\text{.0516}\text{-}\text{0}\text{.00698}\text{=}\text{0}\text{.0446}\text{mol/L}\\ \\ \left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\text{=}\text{0}\text{.118}\text{-}\text{2}\left(\text{0}\text{.00698}\right)\text{=}\text{0}\text{.104}\text{mol/L}\end{array}$

Calculate the value of ${\text{K}}_{\text{C}}$,

  $\begin{array}{l}{\text{K}}_{\text{C}}\text{=}\frac{\left[\text{S}{\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Cl}\right)}_{\text{2}}\right]}{\left[{\text{SCl}}_{\text{2}}\right]{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]}^{\text{2}}}\\ \\ \text{=}\frac{\text{0}\text{.00698}}{\left(\text{0}\text{.0446}\right){\left(\text{0}\text{.104}\right)}^{\text{2}}}\text{=}\text{14}\text{.5}\end{array}$

Therefore, for the given reaction ${\text{K}}_{\text{C}}\text{=}\text{14}\text{.5}$.