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Question: For the reaction \({N_2}\, + \,3{H_2}\, \to \,2N{H_3}\) . The rate of change of concentration for hy...

For the reaction N2 + 3H2 → 2NH3{N_2}\, + \,3{H_2}\, \to \,2N{H_3} . The rate of change of concentration for hydrogen is 0.3 × 10−4 ms−10.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}} . The rate of change of concentration ammonia is:
A.−0.2 ×10 −4  - 0.2\, \times 10{\,^{ - 4\,}}
B.0.2 ×10 −4 0.2\, \times 10{\,^{ - 4\,}}
C.1 ×10 −4 1\, \times 10{\,^{ - 4\,}}
D.0.3×10 −4 0.3 \times 10{\,^{ - 4\,}}

Explanation

Solution

The rate of change of concentration of ammonia can be calculated by the concept that the rate of change of concentration of a reactant is equal to the rate of change of concentration of a product. For example in this reaction the rate of change of concentration of hydrogen is equal to the rate of change of concentration of ammonia.

Complete step-by-step answer: In the given question it is given that the rate of change of hydrogen is 0.3 × 10−4 ms−10.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}} , we have to calculate the rate of change of ammonia.
Now, we have the given reaction
N2 + 3H2 → 2NH3{N_2}\, + \,3{H_2}\, \to \,2N{H_3}
It is given the rate of change of hydrogen so we can write it as,
−d[H]dt = 0.3 × 10−4 ms−1- \dfrac{{d\left[ H \right]}}{{dt}}\, = \,0.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}}
We know the rate of change of concentration of a reactant is equal to the rate of change of concentration of a product divided by their stoichiometry coefficient. So we can write the equations as,
−13d[H2]dt = 12 d[NH3]dt- \dfrac{1}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}}\, = \,\dfrac{1}{2}\,\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}}
d[NH3]dt= −23d[H2]dt\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - \dfrac{2}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}}
By putting the value of rate of change of concentration of hydrogen, we get the value of rate of change of concentration of ammonia.
d[NH3]dt= −23 × 0.3 × 10−4\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - \dfrac{2}{3}\, \times \,0.3\, \times \,{10^{ - 4}}
On solving the above equation,
d[NH3]dt= −0.2  × 10−4 ms−1\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}}
Hence, the rate of change of concentration of ammonia is −0.2  × 10−4 ms−1 - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}} .

So, the correct option is AA.

Note: As we know that speed of an automobile is expressed in terms of change in the position or distance covered by it in a certain period of time. Similarly, the speed of reaction or the rate of reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of; the rate of decrease in concentration of any of one of the reactants or the rate of increase in concentration of any of the product.