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
+1 vote
506 views
in Chemical Kinetics by (36.2k points)
closed by

For a reaction A → P, the rate of the reaction doubles when the concentration of A is increased by two times.

1. Determine the order of the reaction.

2. Derive an expression to determine the 14th life period of above order reaction.

1 Answer

+1 vote
by (33.4k points)
selected by
 
Best answer

1. First order reaction.

2. 1/4th life of a first order reaction can be determined as follows.

\(k=\frac{2.303}{t}log\frac{[R]_0}{[R]}\)

When t = t/4, [R] can be taken as \(\frac{[R]_0}{4}\)

\(k=\frac{2.303}{t/4}log\,4\,;\)

\(k=\frac{2.303}{t/4}\times log\,4\)

tt/4\(\frac{2.303\,\times\,log\,4}{k}\)

\(=\frac{2.303\,\times\,0.6020}{k}\) 

\(=\frac{1.386}{k}\)

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.

...