Question

If the rate of decomposition of ammonia, \({\bf{N}}{{\bf{H}}_{\bf{3}}}\), at 1150 K is \(2.10 \times 1{0^{ - 6}}mol/L/s\), what is the rate of production of nitrogen and hydrogen?

Short answer

The rate of production of nitrogen and hydrogen \(1.05 \times 1{0^{ - 6}}mol/L/s\), \({{\bf{N}}_{\bf{2}}}\) and \(3.15 \times 1{0^{ - 6}}mol/L/s\), \({{\bf{H}}_{\bf{2}}}\).

Step-by-step solution

Rate of a Reaction

The rate of the reaction is the rate of the reaction or how fast or slow reaction can happen. The reaction involves the reactant which are bound to give product at a certain period of time.

Explanation

One may determine after using the stoichiometry of the reaction,

\(\begin{aligned} - 1/2 \Delta \left( {N{H_3}} \right)/\Delta t = 1/3 \Delta \left( {{N_2}} \right)/\Delta t = \Delta \left( {{H_2}} \right)/\Delta t,\\\begin{aligned}{{20}{l}} {Rate of disappearance N{H_3} = - 1/2 \Delta \left( {N{H_3}} \right) / \Delta t}\\{Rate of Formation of {N_2} = 1/3 \Delta \left( {{N_2}} \right) / \Delta t}\\{Rate of formation of {H_2} = \Delta \left( {{H_2}} \right) / \Delta t}\end{aligned}\end{aligned}\)

Therefore:

Rate of Production of \({{\bf{N}}_{\bf{2}}}\)

\(\begin{aligned}1/2 \times 2.10 \times 1{0^{ - 6}}mol {\left( {Ls} \right)^{ - 1}} = \Delta ({N_2})/\Delta t\\\Delta ({N_2})/\Delta t = 1.05 \times 1{0^{ - 6}}mol {\left( {Ls} \right)^{ - 1}}\end{aligned}\)

Rate of production of \({{\bf{H}}_{\bf{2}}}\)

\(\begin{aligned}\Delta ({H_2})/\Delta t = 3 \times 2.10 \times 1{0^{ - 6}}mol {\left( {Ls} \right)^{ - 1}}/ 2\\\Delta ({H_2})/\Delta t = 3.10 1{0^{ - 6}}mol {\left( {Ls} \right)^{ - 1}}\end{aligned}\)