Chapter 17, Problem 17.82P

(a)

Interpretation Introduction

Interpretation:

The given reactions are

For Nickel, the sulfide ore is roasted in air:

  ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{NiO}}_{\text{(s)}}{\text{+ SO}}_{\text{2(g)}}$

The metal oxide is reduced by the ${\text{H}}_{\text{2}}$ in water gas (${\text{CO+H}}_{\text{2}}$) to impure $\text{Ni}$:

  ${\text{NiO}}_{\text{(s)}}{\text{+ H}}_{\text{2(g)}}\rightleftharpoons {\text{Ni}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

The $\text{CO}$ in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl which is subsequently decomposed to the metal.

  ${\text{Ni}}_{\text{(s)}}{\text{+ CO}}_{\text{(g)}}\rightleftharpoons {\text{Ni(CO)}}_{\text{4(g)}}$

The given three steps have to be balanced and overall balanced equation for the conversion of ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2}}{\text{ to Ni(CO)}}_{\text{4}}$ has to be obtained.

Explanation of Solution

The given reactions can be balanced as

  ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{NiO}}_{\text{(s)}}{\text{+ SO}}_{\text{2(g)}}$

Balancing the Nickel in the equation:

Three atoms of nickel are present in reactant side and only one atom of nickel is present in product side.  To balance the nickel on both sides, co-efficient $\text{''2''}$ is added to ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2}}$ on reactant side and co-efficient $\text{''6''}$ is added to $\text{NiO}$ on product side.  The balanced equation for nickel atom is obtained as

  ${\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{6NiO}}_{\text{(s)}}{\text{+ SO}}_{\text{2(g)}}$

Balancing the Sulfur in the equation:

Four atoms of sulfur are present in reactant side and only one atom of sulfur is present in product side.  To balance the sulfur on both sides, co-efficient $\text{''4''}$ is added to ${\text{SO}}_{\text{2}}$ on product side.  The balanced equation for sulfur atom is obtained as

  ${\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{6NiO}}_{\text{(s)}}{\text{+ 4SO}}_{\text{2(g)}}$

Balancing the Oxygen in the equation:

Two atoms of oxygen are present in reactant side and fourteen atoms of oxygen are present in product side.  To balance the oxygen on both sides, co-efficient $\text{''7''}$ is added to ${\text{O}}_{\text{2}}$ on reactant side.  The balanced equation for oxygen atom is obtained as

  ${\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ 7O}}_{\text{2(g)}}\rightleftharpoons {\text{6NiO}}_{\text{(s)}}{\text{+ 4SO}}_{\text{2(g)}}$        (1)

The second reaction is a balanced equation.

  ${\text{NiO}}_{\text{(s)}}{\text{+ H}}_{\text{2(g)}}\rightleftharpoons {\text{Ni}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}$        (2)

The third reaction can be balanced as

  ${\text{Ni}}_{\text{(s)}}{\text{+ CO}}_{\text{(g)}}\rightleftharpoons {\text{Ni(CO)}}_{\text{4(g)}}$

In the above equation, nickel is balanced and $\text{CO}$ group has to be balanced.

Balancing the $\text{CO}$ group:

Only one $\text{CO}$ group is present in reactant side whereas four $\text{CO}$ groups are present in product side.  To balance, co-efficient $\text{''4''}$ is added to the $\text{CO}$ group on reactant side.  The balanced equation is

  ${\text{Ni}}_{\text{(s)}}{\text{+ 4CO}}_{\text{(g)}}\rightleftharpoons {\text{Ni(CO)}}_{\text{4(g)}}$        (3)

Overall balanced equation for the conversion of ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2}}{\text{ to Ni(CO)}}_{\text{4}}$:

Equation 1:

  ${\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ 7O}}_{\text{2(g)}}\rightleftharpoons {\text{6NiO}}_{\text{(s)}}{\text{+ 4SO}}_{\text{2(g)}}$

Equation 2:

  ${\text{NiO}}_{\text{(s)}}{\text{+ H}}_{\text{2(g)}}\rightleftharpoons {\text{Ni}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

In order to get the overall equation of conversion of ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2}}{\text{ to Ni(CO)}}_{\text{4}}$, the above equation has to be multiplied by $\text{''6''}$.  The equation will be

  ${\text{6NiO}}_{\text{(s)}}{\text{+ 6H}}_{\text{2(g)}}\rightleftharpoons {\text{6Ni}}_{\text{(s)}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Equation 3:

  ${\text{Ni}}_{\text{(s)}}{\text{+ 4CO}}_{\text{(g)}}\rightleftharpoons {\text{Ni(CO)}}_{\text{4(g)}}$

In order to get the overall equation of conversion of ${\text{Ni}}_{\text{3}}{\text{S}}_{\text{2}}{\text{ to Ni(CO)}}_{\text{4}}$, the above equation has to be multiplied by $\text{''6''}$.  The equation will be

  ${\text{6Ni}}_{\text{(s)}}{\text{+ 24CO}}_{\text{(g)}}\rightleftharpoons {\text{6Ni(CO)}}_{\text{4(g)}}$

Overall reaction:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ 7O}}_{\text{2(g)}}\rightleftharpoons \bcancel{{\text{6NiO}}_{\text{(s)}}}{\text{+ 4SO}}_{\text{2(g)}}\\ \\ \bcancel{{\text{6NiO}}_{\text{(s)}}}{\text{+ 6H}}_{\text{2(g)}}\rightleftharpoons \bcancel{{\text{6Ni}}_{\text{(s)}}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\\ \\ \bcancel{{\text{6Ni}}_{\text{(s)}}}{\text{+ 24CO}}_{\text{(g)}}\rightleftharpoons {\text{ 6Ni(CO)}}_{\text{4(g)}}\end{array}}\\ \\ {\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ 7O}}_{\text{2(g)}}{\text{+ 6H}}_{\text{2(g)}}{\text{+ 24CO}}_{\text{(g)}}\rightleftharpoons {\text{ 4SO}}_{\text{2(g)}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{+ 6Ni(CO)}}_{\text{4(g)}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The overall $Q_{\text{c}}$ is the product of the $Q_{\text{c}}$’s for the individual reactions has to be proved.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by $Q$.  For a reversible reaction of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{ to NO}}_{\text{2}}$, the reaction quotient is represented as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2(g)}}\\ \\ Q_{\text{c}}\text{ = }\frac{{\left({\text{NO}}_{\text{2}}\right)}^{\text{2}}}{\left({\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\right)}\end{array}$

Explanation of Solution

The obtained overall reaction is

  ${\text{2Ni}}_{\text{3}}{\text{S}}_{\text{2(s)}}{\text{+ 7O}}_{\text{2(g)}}{\text{+ 6H}}_{\text{2(g)}}{\text{+ 24CO}}_{\text{(g)}}\rightleftharpoons {\text{ 4SO}}_{\text{2(g)}}{\text{+ 6H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{+ 6Ni(CO)}}_{\text{4(g)}}$

In reaction quotient, solids are not included in the expression.

Reaction quotient expression can be written by taking the products in nominator and reactants in denominator.

$Q_{\text{c}}$ of overall reaction:

  $Q_{\text{c}}\text{ = }\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{4}}{{\text{[H}}_{\text{2}}\text{O]}}^{\text{6}}{{\text{[Ni(CO)}}_{\text{4}}\text{]}}^{\text{6}}}{{{\text{[O}}_{\text{2}}\text{]}}^{\text{7}}{{\text{[H}}_{\text{2}}\text{]}}^{\text{6}}{\text{[CO]}}^{\text{24}}}$

$Q_{\text{c}}$ of individual reactions:

  $\begin{array}{l}{Q}_{\text{1}}\text{×}{Q}_{\text{2}}\text{×}{Q}_{\text{3}}\text{=}\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{4}}}{{{\text{[O}}_{\text{2}}\text{]}}^{\text{7}}}\text{×}\frac{{{\text{[H}}_{\text{2}}\text{O]}}^{\text{6}}}{{{\text{[H}}_{\text{2}}\text{]}}^{\text{6}}}\text{×}\frac{{{\text{[Ni(CO)}}_{\text{4}}\text{]}}^{\text{6}}}{{\text{[CO]}}^{\text{24}}}\\ \\ \text{=}\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{4}}{{\text{[H}}_{\text{2}}\text{O]}}^{\text{6}}{{\text{[Ni(CO)}}_{\text{4}}\text{]}}^{\text{6}}}{{{\text{[O}}_{\text{2}}\text{]}}^{\text{7}}{{\text{[H}}_{\text{2}}\text{]}}^{\text{6}}{\text{[CO]}}^{\text{24}}}\end{array}$

Hence, the overall $Q_{\text{c}}$ is the product of the $Q_{\text{c}}$’s for the individual reactions.