Q92CP - Chemistry: Molecular Nature Of Matter And Change

Question

Use each of the following reaction quotients to write the balanced equation:

           

 

                 $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{Q}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{CO}}_{\mathbf{2}}{\mathbf{]}}^{\mathbf{2}}\mathbf{\right[}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{\mathbf{]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{\right]}\mathbf{ }\mathbf{[}{\mathbf{O}}_{\mathbf{2}}{\mathbf{]}}^{\mathbf{3}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }\mathbf{Q}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{NH}}_{\mathbf{3}}{\mathbf{]}}^{\mathbf{4}}\mathbf{\right[}{\mathbf{O}}_{\mathbf{2}}{\mathbf{]}}^{\mathbf{7}}}{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}{\mathbf{]}}^{\mathbf{4}}\mathbf{\right[}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}{\mathbf{]}}^{\mathbf{6}}}$

Step-by-Step Solution

Answer

(a) For reaction quotient , $\mathrm{Q}=\frac{\left[{\mathrm{CO}}_2{]}^2\right[{\mathrm{H}}_2\mathrm{O}{]}^2)}{\left[{\mathrm{C}}_2{\mathrm{H}}_4\right][{\mathrm{O}}_2{]}^3}$ The balanced equation is:      ${\mathrm{C}}_2{\mathrm{H}}_{4\left(\mathrm{g}\right)}+3{\mathrm{O}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 4{\mathrm{NH}}_{3\left(\mathrm{g}\right)}+7{\mathrm{O}}_{2\left(\mathrm{g}\right)}$

(b) For reaction quotient , $\mathrm{Q}=\frac{\left[{\mathrm{NH}}_3{]}^4\right[{\mathrm{O}}_2{]}^2)}{{\left[{\mathrm{NO}}_2\right]}^4[{\mathrm{H}}_2\mathrm{O}{]}^6}$ The balanced equation is:$4{\mathrm{NO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons 4{\mathrm{NH}}_{3\left(\mathrm{g}\right)}+7{\mathrm{O}}_{2\left(\mathrm{g}\right)}$

Step 1: Write the balanced equation for Q = [ N H 3 ] 4 [ O 2 ] 2 ) [ N O 2 ] 4 [ H 2 O ] 6

For any given reaction;

${\mathrm{A}}_{\left(\mathrm{g}\right)}+2{\mathrm{B}}_{\left(\mathrm{g}\right)}3{\mathrm{C}}_{\left(\mathrm{g}\right)}+4{\mathrm{D}}_{\left(\mathrm{g}\right)}$

Reaction quotient $=\frac{\left[\mathrm{Product}\right]}{\left[\mathrm{Reactant}\right]}$

$\therefore {\mathrm{Q}}_{\mathrm{C}}=\frac{\left[\mathrm{C}{]}^3\right[\mathrm{D}{]}^4}{\left[\mathrm{A}\right][\mathrm{B}{]}^2}$

Hence in the reaction, numerator is product and denominator is reactant. Powers are the coefficients

(a) Thus, for reaction quotient,

 $\mathrm{Q}=\frac{\left[{\mathrm{CO}}_2{]}^2\right[{\mathrm{H}}_2\mathrm{O}{]}^2}{\left(\right[{\mathrm{C}}_2{\mathrm{H}}_4\left]\right[{\mathrm{O}}_2{]}^3)}$

The balanced equation is

${\mathrm{C}}_2{\mathrm{H}}_{4\left(\mathrm{g}\right)}+3{\mathrm{O}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 4{\mathrm{NH}}_{3\left(\mathrm{g}\right)}+7{\mathrm{O}}_{2\left(\mathrm{g}\right)}$             

Step 2: Write the balanced equation for Q = [ NH 3 ] 4 [ O 2 ] 2 ) [ NO 2 ] 4 [ H 2 O ] 6

 

(b) For reaction quotient,

 $\mathrm{Q}=\frac{\left[{\mathrm{NH}}_3{]}^4\right[{\mathrm{O}}_2{]}^2)}{{\left[{\mathrm{NO}}_2\right]}^4[{\mathrm{H}}_2\mathrm{O}{]}^6}$

The balanced equation is:

$4{\mathrm{NO}}_{2\left(\mathrm{g}\right)}+6{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons 4{\mathrm{NH}}_{3\left(\mathrm{g}\right)}+7{\mathrm{O}}_{2\left(\mathrm{g}\right)}$