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A first-order reaction has a specific reaction rate of $\text{1}{{\text{0}}^{-3}}\text{ }{{\text{s}}^{1-}}$. How much time will it take for 10gm of this to reduce to 2.5 gm?

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Hint: The first-order reaction depends only on one reactant whereas Zero order reaction does not depend on any reacting species. The formula of the first-order reaction is $\text{k = }\dfrac{2.303}{\text{t}}\log \dfrac{{{\text{A}}_{\text{O}}}}{\text{A}}$.

Complete answer:
-To find the time that will reduce the concentration from 10 g to 2.5 g it is given that specific reaction rate or k = ${{10}^{-3}}$, initial concentration is $({{\text{A}}_{\text{O}}}\text{) = 10g}$ and the final concentration is $(\text{A) = 2}\text{.5g}$.
-So, we can apply the formula of the first order that is:
$\text{k = }\dfrac{2.303}{\text{t}}\log \dfrac{{{\text{A}}_{\text{O}}}}{\text{A}}$
-It can also be written as:
$\text{t = }\dfrac{2.303}{\text{k}}\log \dfrac{{{\text{A}}_{\text{O}}}}{\text{A}}$
$\text{t = }\dfrac{2.303}{{{10}^{-3}}}\log \dfrac{10}{2.5}$
$\text{t = 2030 }\cdot \text{ log4}$
-Now, the value of log4 from the log table is 0.6021, so by applying it in the above equation, we will get:
$\text{t = 2303 }\cdot \text{ 0}\text{.6021}$
$\text{t = 1386}\text{.6 sec}$
-Therefore, a total of 1386.6 sec time will be taken by the reactant to reduce up to 2.5g.

Additional Information:
-The physical significance of k is: It represents the fraction of the reactant decomposed per unit time of the constant concentration.
-The common formula of rate equation for all the orders except for n=1 will be: ${{\text{k}}_{\text{n}}}\text{ = }\dfrac{1}{t\left( n-1 \right)}\left( \dfrac{1}{{{\left( a-x \right)}^{n-1}}}-\dfrac{1}{{{a}^{n-1}}} \right)$
-Zero-order reactions occur under special conditions and are very uncommon.
They generally occur in the heterogeneous system.

Note: Students should not get confused between specific reaction rate and rate constant. The specific reaction rate is also a rate constant but it is a rate of reaction in which the specific conditions are applied.


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