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-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.

-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.