
For the reaction ${N_2}\, + \,3{H_2}\, \to \,2N{H_3}$ . The rate of change of concentration for hydrogen is $0.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$ . The rate of change of concentration ammonia is:
A.$ - 0.2\, \times 10{\,^{ - 4\,}}$
B.$0.2\, \times 10{\,^{ - 4\,}}$
C.$1\, \times 10{\,^{ - 4\,}}$
D.$0.3 \times 10{\,^{ - 4\,}}$
Answer
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Hint: 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 $0.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$ , we have to calculate the rate of change of ammonia.
Now, we have the given reaction
${N_2}\, + \,3{H_2}\, \to \,2N{H_3}$
It is given the rate of change of hydrogen so we can write it as,
$ - \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,
$ - \dfrac{1}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}}\, = \,\dfrac{1}{2}\,\dfrac{{d\left[ {N{H_3}} \right]}}{{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.
$\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - \dfrac{2}{3}\, \times \,0.3\, \times \,{10^{ - 4}}$
On solving the above equation,
$\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$
Hence, the rate of change of concentration of ammonia is $ - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$ .
So, the correct option is $A$.
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.
Complete step-by-step answer: In the given question it is given that the rate of change of hydrogen is $0.3\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$ , we have to calculate the rate of change of ammonia.
Now, we have the given reaction
${N_2}\, + \,3{H_2}\, \to \,2N{H_3}$
It is given the rate of change of hydrogen so we can write it as,
$ - \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,
$ - \dfrac{1}{3}\dfrac{{d\left[ {{H_2}} \right]}}{{dt}}\, = \,\dfrac{1}{2}\,\dfrac{{d\left[ {N{H_3}} \right]}}{{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.
$\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - \dfrac{2}{3}\, \times \,0.3\, \times \,{10^{ - 4}}$
On solving the above equation,
$\dfrac{{d\left[ {N{H_3}} \right]}}{{dt}} = \, - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$
Hence, the rate of change of concentration of ammonia is $ - 0.2\,\, \times \,{10^{ - 4}}\,m{s^{ - 1}}$ .
So, the correct option is $A$.
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.
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