Question

prove that "The time required for 99% completion of as First order reaction is double the time require for 90% completion​

Answer 1

Explanation:

k=2.303 log[Ao]/[A] /t

t90%=2.303 log100/10 /k

=2.303 log 10^1/k

=2.303×1/k

=2.303/k

t99%=2.303 log 100/1 /k

=2.303 log10^2 / k

=2.303×2 /k

=4.606/k

t99%=2t90%