Question

For the reaction \[{\text{4N}}{{\text{H}}_{\text{3}}}{\text{ + 5}}{{\text{O}}_{\text{2}}} \to \,{\text{4NO}}\,{\text{ + }}\,{\text{6}}{{\text{H}}_{\text{2}}}{\text{O}}\], the rate of reaction with respect to \[{\text{N}}{{\text{H}}_{\text{3}}}\] is \[2 \times {10^{ - 3}}\,{\text{M}}{{\text{s}}^{ - 1}}\]. Find the rate of the reaction with respect to oxygen in \[\,{\text{M}}{{\text{s}}^{ - 1}}\].

Answer 1

Hint:The rate is used to determine the speed of the reaction. Chemical Kinetics is used to determine the rates of the carious chemical reaction.The rate of the reaction is given in terms of the change in concentration of species that are either reactants or products. The unit of the rate of reaction is the unit the concentration per time.

Complete solution:
Here, the reaction given is as follows:
\[{\text{4N}}{{\text{H}}_{\text{3}}}{\text{ + 5}}{{\text{O}}_{\text{2}}} \to \,{\text{4NO}}\,{\text{ + }}\,{\text{6}}{{\text{H}}_{\text{2}}}{\text{O}}\]
For this reaction the rate can be written as follows:
 {\text{rate}}\,{\text{of}}\,{\text{reaction = decrease}}\,{\text{in}}\,{\text{concentration}}\,{\text{of}}\,{\text{reactants = }} \\
  {\text{increase}}\,{\text{in}}\,{\text{concentration}}\,{\text{of}}\,{\text{products}} \\
\[{\text{rate = - }}\dfrac{1}{4} \times \dfrac{{d\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}}{{dt}}{\text{ = - }}\dfrac{1}{5} \times \dfrac{{d\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{dt}}{\text{ = + }}\dfrac{1}{4} \times \dfrac{{d\left[ {{\text{NO}}} \right]}}{{dt}} = {\text{ + }}\dfrac{1}{5} \times \dfrac{{d\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}{{dt}}\]
Here, the rate is expressed per mole.
From this equation, we can equate the rate of ammonia and oxygen to determine the rate of oxygen.
\[{\text{ - }}\dfrac{1}{4} \times \dfrac{{d\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}}{{dt}}{\text{ = - }}\dfrac{1}{5} \times \dfrac{{d\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{dt}}\]
Now, rearrange the equation to determine the rate of oxygen.
\[{\text{ - }}\dfrac{{d\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{dt}}{\text{ = - }}\dfrac{5}{4} \times \dfrac{{d\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}}{{dt}}\]
Here, substitutes the value of the rate concerning ammonia as \[2 \times {10^{ - 3}}\,{\text{M}}{{\text{s}}^{ - 1}}\].
\[{\text{ - }}\dfrac{{d\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{dt}}{\text{ = - }}\dfrac{5}{4} \times 2 \times {10^{ - 3}}\,{\text{M}}{{\text{s}}^{ - 1}}\]
\[{\text{ - }}\dfrac{{d\left[ {{{\text{O}}_{\text{2}}}} \right]}}{{dt}}{\text{ = }}2.5 \times {10^{ - 3}}\,{\text{M}}{{\text{s}}^{ - 1}}\]
Thus, the rate of reaction concerning oxygen obtained is \[2.5 \times {10^{ - 3}}\,{\text{M}}{{\text{s}}^{ - 1}}\].

Note:As the reaction proceeds the concentration of the reactants decreases and simultaneously concentration of products increases. Therefore, the rate of reaction is given in terms of the decrease in the concentration of reactants or an increase in the concentration of products.
The rate is expressed per mole hence while determining the rate of species it is divide by the coefficients.