Answer
424.2k+ views
Hint: There are few methods of determining the rate law of a reaction. The isolation method is also a method of determining the rate law of a reaction. Try to figure out what isolation means in this context. Think about the isolated rate law of each of the reactants.
Complete step by step answer: We know that the rate law of a reaction can be determined experimentally by isolation method and some other methods. In the isolation method, we isolate one of the reactants each time keeping other reactant concentrations constant. This experiment is repeated by isolating each reactant at a time.
Consider a reaction: $aA + bB \rightarrow cC + dD $
Let us calculate the rate law for this reaction. There are 2 reactant species so we need to perform reaction 2 times isolating each reactant.
First, let us isolate A and calculate order with respect to A by keeping the concentration of B constant. Let the order of reaction with respect to A be x. Then we will get the rate of the reaction as
Rate r = ${ k }_{ 1 }\left[ A \right] ^{ x }$
Let $\left[ { B }_{ 0 } \right]$ be the constant concentration of B.
${ k }_{ 1 } = { k }\left[ { B }_{ 0 } \right] ^{ y }$ where y is the order of reaction with respect to B.
Applying log to rate expression
\[\log { \left( r \right) } =\log { \left( k \right) + } x\log { \left( \left[ A \right] \right) }\]
We plot the above equation with log r on the y-axis and log [A] values on the x-axis. We find the slope of the straight line which is nothing but the order of reaction with respect to A.
Repeating the above procedure by keeping concentration A constant we get the order with respect to B. Combining these two orders we get the rate law of reaction. Adding orders of reaction with respect to each reactant we can get the total order of the reaction.
Note: This method is easy as we require less calculation to perform at a time. We can keep the reactant’s concentration common by taking that reactant concentration very large. As we take concentration large there will not be much change in the concentration of that reactant. So, it will nearly be constant.
Complete step by step answer: We know that the rate law of a reaction can be determined experimentally by isolation method and some other methods. In the isolation method, we isolate one of the reactants each time keeping other reactant concentrations constant. This experiment is repeated by isolating each reactant at a time.
Consider a reaction: $aA + bB \rightarrow cC + dD $
Let us calculate the rate law for this reaction. There are 2 reactant species so we need to perform reaction 2 times isolating each reactant.
First, let us isolate A and calculate order with respect to A by keeping the concentration of B constant. Let the order of reaction with respect to A be x. Then we will get the rate of the reaction as
Rate r = ${ k }_{ 1 }\left[ A \right] ^{ x }$
Let $\left[ { B }_{ 0 } \right]$ be the constant concentration of B.
${ k }_{ 1 } = { k }\left[ { B }_{ 0 } \right] ^{ y }$ where y is the order of reaction with respect to B.
Applying log to rate expression
\[\log { \left( r \right) } =\log { \left( k \right) + } x\log { \left( \left[ A \right] \right) }\]
We plot the above equation with log r on the y-axis and log [A] values on the x-axis. We find the slope of the straight line which is nothing but the order of reaction with respect to A.
![seo images](https://www.vedantu.com/question-sets/9a7a3432-3977-472c-b017-d8c484e9d885364623994775698797.png)
Repeating the above procedure by keeping concentration A constant we get the order with respect to B. Combining these two orders we get the rate law of reaction. Adding orders of reaction with respect to each reactant we can get the total order of the reaction.
Note: This method is easy as we require less calculation to perform at a time. We can keep the reactant’s concentration common by taking that reactant concentration very large. As we take concentration large there will not be much change in the concentration of that reactant. So, it will nearly be constant.
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)
Which are the Top 10 Largest Countries of the World?
![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)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![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)
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)
Change the following sentences into negative and interrogative class 10 english 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)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)