Answer
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Hint: The rate of disappearance is the rate of that particular chemical concentration going down. This means the chemical reactant is getting consumed in the reaction. One can use any reaction to prove the rate of disappearance of ammonia.
Complete step by step answer:
1) First of all we will learn about the concept of the rate of disappearance. The rate of disappearance can be elaborated as the rate at which a reactant gets disappeared that gets consumed during the reaction progress.
2) Now to calculate the rate of disappearance of ammonia let us consider a reaction of ammonia with Sulphur dioxide as below,
$4N{H_3}_{(g)} + S{O_2}_{(g)}\xrightarrow{{}}4N{O_{(g)}} + 6{H_2}{O_{(g)}}$
In the above reaction when four moles of ammonia are reacted with one mole of sulfur dioxide, the reactions yield products as four moles of nitric oxide and six moles of water.
3) In the above reaction the ammonia gets consumed in the reaction to form the products. This means the ammonia disappears in the reaction. Now to calculate the rate of disappearance of ammonia let us first write a rate equation for the given reaction as below,
Rate of reaction,
$\dfrac{{ - d\left[ {N{H_3}} \right]}}{{dt}} \times \dfrac{1}{4} = \dfrac{1}{4} \times \dfrac{{d\left[ {NO} \right]}}{{dt}}$
Now by canceling the common value $\dfrac{1}{4}$ on both sides we get the above equation as,
$\dfrac{{ - d\left[ {N{H_3}} \right]}}{{dt}} = \dfrac{{d\left[ {NO} \right]}}{{dt}}$
4) Now from the above equation, we can say that the rate of formation of nitric oxide is equal to the rate of disappearance of the ammonia. The rate of formation of nitric oxide is $3 \cdot 2 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{{\text{S}}^{ - 1}}$
5) As the rate of disappearance of ammonia is equal to the rate of formation of nitric oxide we can say that the rate of disappearance of ammonia is $ - 3 \cdot 2 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{{\text{S}}^{ - 1}}$.
Note:
The rate of disappearance value is always negative as the concentration of reactant gets consumed in the reaction. The rate of appearance value is always positive as the product is formed in the reaction.
Complete step by step answer:
1) First of all we will learn about the concept of the rate of disappearance. The rate of disappearance can be elaborated as the rate at which a reactant gets disappeared that gets consumed during the reaction progress.
2) Now to calculate the rate of disappearance of ammonia let us consider a reaction of ammonia with Sulphur dioxide as below,
$4N{H_3}_{(g)} + S{O_2}_{(g)}\xrightarrow{{}}4N{O_{(g)}} + 6{H_2}{O_{(g)}}$
In the above reaction when four moles of ammonia are reacted with one mole of sulfur dioxide, the reactions yield products as four moles of nitric oxide and six moles of water.
3) In the above reaction the ammonia gets consumed in the reaction to form the products. This means the ammonia disappears in the reaction. Now to calculate the rate of disappearance of ammonia let us first write a rate equation for the given reaction as below,
Rate of reaction,
$\dfrac{{ - d\left[ {N{H_3}} \right]}}{{dt}} \times \dfrac{1}{4} = \dfrac{1}{4} \times \dfrac{{d\left[ {NO} \right]}}{{dt}}$
Now by canceling the common value $\dfrac{1}{4}$ on both sides we get the above equation as,
$\dfrac{{ - d\left[ {N{H_3}} \right]}}{{dt}} = \dfrac{{d\left[ {NO} \right]}}{{dt}}$
4) Now from the above equation, we can say that the rate of formation of nitric oxide is equal to the rate of disappearance of the ammonia. The rate of formation of nitric oxide is $3 \cdot 2 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{{\text{S}}^{ - 1}}$
5) As the rate of disappearance of ammonia is equal to the rate of formation of nitric oxide we can say that the rate of disappearance of ammonia is $ - 3 \cdot 2 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{{\text{S}}^{ - 1}}$.
Note:
The rate of disappearance value is always negative as the concentration of reactant gets consumed in the reaction. The rate of appearance value is always positive as the product is formed in the reaction.
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