Answer
425.4k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)