Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
3.2k views
in Equilibrium by (48.1k points)

Explain law of mass Action by taking the following example,
CH3COOH + C2H2OH ➝ CH3COOC2H5 + H2O

1 Answer

+1 vote
by (49.4k points)
selected by
 
Best answer

Law of Mass Action:

According to this law, “the rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time”. Molar concentration is called active mass. Active mass is the number of moles dissolved in one litre of solution.
eg. CH3COOH + C2H5OH ➝ CH3COOC2 H5 + H2O
According to law of mass action,
Rate of forward reaction a [CH3COOH] [C2H5OH]
= K1 [CH3COOH] [C2H5OH]
When K2 = rate constant for backward reaction At equilibrium,
Rate of backward reaction a [CH3COOC2H5] [H2O]
= K2 [CH3COOC2H5] [H2O]

Where K= rate constant for backward reaction At equilibrium,
Rate of forward reaction = Rate of backward reaction K2 [CH3COOH] [C2H6OH] – K2 [CH3COOC2H5] [H2O]

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...