Question

In a catalytic conversion of ${N_2}$ to $N{H_3}$ by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is $40 \times {10^{ - 3}}{\text{ }}mol{\text{ }}{L^{ - 1}}{\text{ }}{s^{ - 1}}$ . If there is no side reaction, the rate of the reaction as expressed in terms of hydrogen is:
(in $mol{\text{ }}{L^{ - 1}}{\text{ }}{s^{ - 1}}$)
\[
  A.{\text{ }}60 \times {10^{ - 3}} \\
  B.{\text{ }}20 \times {10^{ - 3}} \\
  C.{\text{ }}1.200 \\
  D.{\text{ }}10.3 \times {10^{ - 3}} \\
 \]

Answer 1

Hint: In order to deal with this question first we will define the term rate of reaction further we will write a balanced equation and according to it we will calculate the required rate of reaction.
Complete step-by-step answer:

Formula used- \[{\text{Rate of reaction}} = \dfrac{{\Delta conc}}{{\Delta time}}\]
The reaction rate or response time is the frequency at which a chemical reaction occurs. We describe reaction rate as the velocity of transforming reactants into products. Level of reaction will differ considerably.
Balanced reaction is as following:
${N_2} + 3{H_2} \to 2N{H_3}$
Rate of reaction can be expressed as:
\[\dfrac{{\Delta conc}}{{\Delta time}} = \dfrac{{ - d\left( {{N_2}} \right)}}{{dt}} = \dfrac{{ - 1}}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = \dfrac{1}{2}\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}}\]
Since, we have to express it in terms of hydrogen also we know the rate of reaction in terms of ammonia, so we will take second and third part of the above equation to find the result.
So, we write it as:
\[\dfrac{{ - 1}}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = \dfrac{1}{2}\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}}\]
Now let us substitute the values:
\[
  \because \dfrac{{ - 1}}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = \dfrac{1}{2}\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} \\
   \Rightarrow - \dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = \dfrac{3}{2}\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} \\
   \Rightarrow - \dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = \dfrac{3}{2} \times 40 \times {10^{ - 3}}{\text{ }}mol{\text{ }}{L^{ - 1}}{\text{ }}{s^{ - 1}} \\
   \Rightarrow - \dfrac{{d\left[ {{H_2}} \right]}}{{dt}} = 60 \times {10^{ - 3}}{\text{ }}mol{\text{ }}{L^{ - 1}}{\text{ }}{s^{ - 1}} \\
 \]
Hence, the rate of the reaction as expressed in terms of hydrogen is \[60 \times {10^{ - 3}}{\text{ }}mol{\text{ }}{L^{ - 1}}{\text{ }}{s^{ - 1}}\]
So, the correct answer is option A.

Additional information-
Ammonia, also known as $N{H_3}$ , is a colorless gas consisting of nitrogen and hydrogen atoms, with a distinct odour. It is created in the human body and nature, naturally in water, soil, and air, including in small molecules of bacteria. Ammonia and the ammonium ion are important components of metabolic processes in human health.
Ammonia is also used in the manufacturing of fabrics, explosives, textiles, pesticides, dyes and other chemicals as a refrigerant agent, for purification of water sources. It's used in many applications for domestic and industrial cleaning.

Note- The response rate is the frequency at which a chemical reaction occurs. When a reaction has a low rate, this means that the molecules interact at a slower pace than a high-rate reaction. Factors that affect the reaction rates of chemical reactions include reactant concentration, temperature, physical state and dispersion of reactants, solvent, and availability of a catalyst.