Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

In an experiment (reactant $\rightarrow$ product) we observe that if we increase the reactant to twice then time taken in the concentration of reaction to drop to $\large\frac{1}{2}$ of its value is reduced to $\large\frac{1}{4}$ then order is


Can you answer this question?

1 Answer

0 votes
$t_{\large\frac{1}{2}}\propto [A]^{1-n}$
$t_{\large\frac{1}{2}}=P [A]^{1-n}$
A = Rate of concentration of the reaction
t = Time taken
P=constant of proportionality
Hence (d) is the correct answer.
answered Dec 13, 2013 by sreemathi.v
edited Dec 27, 2014 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App