Show that for a first order reaction, time required for completion of 99% of reaction is twice the time required for complrtion of 90% of reaction.

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

Time = \frac{2.303}{K}log \frac{initial concentration}{final concentration}

for 99% completion of reaction:

T_{1}= \frac{2.303}{K}\times \log \frac{100}{100-99}=\frac{2.303}{K}\times \log \frac{100}{1}= \frac{2 \times 2.303}{K}

for 90% completion of reaction:

T_{2}= \frac{2.303}{K}\times \log \frac{100}{100-90}=\frac{2.303}{K}\times \log 10= \frac{1 \times 2.303}{K}

\therefore T_{1} = 2\times T_{2}

Preparation Products

Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions