SELECT * FROM question_mgmt as q WHERE id=661 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=59 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Chemical Kinetics the conversion of molecules x to y follows second

Question:

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Answer:

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as     rate = k [A]2

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k [X]2                                                    ________________________   (1)

Let [X] = a mol L-1 , then equation (1) can be written as:

Rate = k [a]2

= ka2

If the concentration of X is increased to three times, then [X] = 3a mol L-1

Now, the rate equation will be:

Rate = k [3a]2

= 9 (ka2)

Hence, the rate of formation will increase by 9 times.


< Previous Question

Next Question >

SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='9' AND ex_no!=0 AND status=1 and id!=661 ORDER BY views desc, last_viewed_on desc limit 0,10

  • Popular questions of class 12 Chemistry

SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='9' AND ex_no!=0 AND status=1 and id!=661 ORDER BY last_viewed_on desc limit 0,10

  • Recently Viewed Questions of Class 12 Chemistry

Comments

  • Answered by Ekta Mehta
  • 4 months ago

Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

  • Answered by Ekta Mehta
  • 4 months ago

Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

  • Answered by Ekta Mehta
  • 4 months ago

Taking Screenshots on your Samsung Galaxy M31s is very easy and quick.

Comment(s) on this Question


NCERT Solutions

Quick Links

Resources

Support