Question

The rate of first order reaction is 0.04mol^-1 at 10 min. and 0.03 at 20 min. after initiation
Find half lie of the reaction​

Answer 1

K= 2.303/t2-t1 log (rate)1/. (rate)2

= 2.303/20-10 log 0.04/0.03

=2.878×10^-2min^-1

T1/2=0.693/2.878×10^-2

=1444.7

Answer 2

U have to just put the values into the formulae for the first order eqn. that is:

K = (2.303/t2 – t1)log(Q1/Q2)

=(2.303/(10)llog(0.04/0.03)

After solving,

K = 0.02876 /min

thalf = 0.693/k = > 24 min approx.

Hope it's Helpful for you.......