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For ${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$ the rate of disappearance of ${{H}_{2}}$ is 0.01 M/min. The amount of $N{{H}_{3}}$ formed at that instance will be:
[A] -0.02 mol
[B] Zero
[C] 0.1132 g
[D] 0.17 g

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Last updated date: 18th Jun 2024
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Answer
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Hint: To solve this question, remember that rate of appearance or disappearance of the product or reactant is given by the change in its concentration in time T. Also, remember that the rate of disappearance of reactants is equal to the rate of formation of each product.

Complete answer:
In the given question, we are given the reaction of formation of ammonia from nitrogen and hydrogen. So firstly, lets we can write down the given reaction as-
     ${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$
We know that we can find the rate of a reaction by equating the rate of disappearance of its products and reactants. And, we can write the rate of disappearance or appearance as - $\dfrac{\Delta C}{\Delta T}$ where, C is the concentration of the product or the reactant in time T.
From the given balanced reaction, we can write that the rate of reaction is- $-\dfrac{1}{3}\dfrac{d\left[ {{H}_{2}} \right]}{dT}=-\dfrac{d\left[ {{N}_{2}} \right]}{dT}=\dfrac{1}{2}\dfrac{d\left[ N{{H}_{3}} \right]}{dT}$
The rate of appearance of the ammonia is equal to the rate of disappearance of hydrogen and nitrogen and we’ve used the balanced equation to obtain the correct number of moles.
The appearance rate is positive because it is formed and that of disappearance is negative.
Now, we can see that in the question the rate of disappearance of ${{H}_{2}}$ is given to us.
Therefore, we can write that $\dfrac{1}{3}\dfrac{d\left[ {{H}_{2}} \right]}{dT}=-0.01M/\min $
Now, we know that 1 minute = 60 sec.
Therefore, the rate of disappearance of hydrogen = $\dfrac{1}{3}\dfrac{d\left[ {{H}_{2}} \right]}{dT}=-0.01M/\min =-1.667\times {{10}^{-4}}M/\sec $
Now, we have to find the amount of $N{{H}_{3}}$ at that instance. Therefore, we can write that-
$\begin{align}
  & -\dfrac{1}{3}\dfrac{d\left[ {{H}_{2}} \right]}{dT}=\dfrac{1}{2}\dfrac{d\left[ N{{H}_{3}} \right]}{dT} \\
 & Or,-1.667\times {{10}^{-4}}M/\sec =\dfrac{1}{2}\dfrac{d\left[ N{{H}_{3}} \right]}{dT} \\
 & Or,\dfrac{d\left[ N{{H}_{3}} \right]}{dT}=-2\times 1.667\times {{10}^{-4}}M/\sec =-3.334\times {{10}^{-4}}M/\sec \\
\end{align}$
Therefore, we can say that rate of appearance of ammonia is $-3.334\times {{10}^{-4}}M/\sec $ which means there are $3.334\times {{10}^{-4}}$ moles present at that instance.
If we want to convert it to M/sec, we can write that, amount of ammonia formed = $-3.334\times {{10}^{-4}}M/\sec \times 60\sec =-0.02$

Therefore, the correct answer is option [A] -0.02 mol.

Note:
The rate of a reaction depends upon several factors like the nature of the reactants, presence of catalyst, temperature at which the reaction is taking place and also on the presence of catalyst. The rate of a reaction increases upon increasing the temperature according to the equation given by Arrhenius. However, the rate constant rapidly increases with the increase in temperature but at a very high temperature, the rate of increase falls off eventually.