Question

In a reaction $ {N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right) $ the rate of appearance of $ N{H_3} $ is $ 2.5 \times {10^{ - 4}}mol{L^{ - 1}}{\sec ^{ - 1}} $ . The rate of reaction and rate of disappearance of $ {H_2} $ will be:
(A) $ 3.75 \times {10^{ - 4}} $ , $ 1.25 \times {10^{ - 4}} $
(B) $ 1.25 \times {10^{ - 4}} $ , $ 2.5 \times {10^{ - 4}} $
(C) $ 1.25 \times {10^{ - 4}} $ , $ 3.75 \times {10^{ - 4}} $
(D) $ 5.0 \times {10^{ - 4}} $ , $ 3.75 \times {10^{ - 4}} $

Answer 1

Hint: We need to find the rate of the reaction that means with what speed the reaction is carried out. . And also the rate of disappearance of hydrogen during the reaction. According to the equating the rate of the nitrogen , hydrogen, ammonia will be same and given as:
 $ - \dfrac{{d\left[ {{N_2}} \right]}}{{dt}} = - \dfrac{{d\left[ {{H_2}} \right]}}{{3dt}} = \dfrac{{d\left[ {N{H_3}} \right]}}{{2dt}} $

Complete step by step solution
Here we are given that the reaction is carried out between the nitrogen and hydrogen and it further leads to the formation of the product. Here we are given the rate of appearance of the product that means the rate with which the product appeared after the reaction. Further we need to find the rate of the reaction that means with what speed the reaction is carried out. And also we need to find the rate of disappearance of hydrogen during the reaction.
Firstly we will write the reaction which was carried out in the form of the equation:
 $ {N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right) $
Also we are given with the rate of appearance of the product: $ 2.5 \times {10^{ - 4}}mol{L^{ - 1}}{\sec ^{ - 1}} $
Now we need to find the rate of the reaction which can be calculated as:
According to the equating the rate of the nitrogen , hydrogen, ammonia will be same and given as:
 $ - \dfrac{{d\left[ {{N_2}} \right]}}{{dt}} = - \dfrac{{d\left[ {{H_2}} \right]}}{{3dt}} = \dfrac{{d\left[ {N{H_3}} \right]}}{{2dt}} $
And we are given the rate the of ammonia which is equal to its appearance= $ 2.5 \times {10^{ - 4}}mol{L^{ - 1}}{\sec ^{ - 1}} $
this should be further divided by 2 and we will get $ 1.25 \times {10^{ - 4}} $
so the rate of the reaction is $ 1.25 \times {10^{ - 4}} $
then we need to calculate the rate of disappearance of the hydrogen
as we already know that the three rates are equal so we will now the rate of ammonia is $ 2.5 \times {10^{ - 4}}mol{L^{ - 1}}{\sec ^{ - 1}} $ then ,
 $ \dfrac{d}{{dt}}{H_2} = \dfrac{{3d\left[ {N{H_3}} \right]}}{{2dt}} $
This is according to the above written equation.
 $ \dfrac{3}{2} \times 2.5 \times {10^{ - 4}} = 3.75 \times {10^{ - 4}} $
Hence, the option C is correct.

Note
We should keep in mind here that the rates of all the three will be equal but the only thing that needs to be remembered is that the coefficient in front of each molecule should be considered and the coefficient value should be considered and adjusted during calculations.