Question

9. The rate of a first order, reaction at 20 min is
0.55 mol L-'minand 0.055 mol L-' min-1 at 40 min
after initiation. The half life of the reaction in minutes
is?
(a) 3
(b) 6
(c) 60
(d) 20​

Answer 1

Answer:

sorry bro weak in chemistry

Answer 2

Answer:

The half life period is equal to 6minutes for the given first order reaction.

Explanation:

Given: $Rate_{1}=0.55molL^{-1}min^{-1}$ at time $t_{1}=20min$

and, $Rate_{2}=0.055molL^{-1}min^{-1}$  at time $t_{2}=40min$

We know for the first order expression of rate law:

$Rate= k[A]$

Therefore , $0.55=k[A]_{20}$                                    .................(1)

and, $0.055=k[A]_{40}$                                             .................(2)

Divide the equation (1) by eq.(2):

$\frac{[A]_{20}}{[A]_{40}}= \frac{0.55}{0.055} =10$

The expression for rate constant at time 20 min:

$k=\frac{2.303}{t}log\frac{[A]_{20}}{[A]_{40}}$

$k=\frac{2.303}{20}log10$

$k=0.1152min^{-1}$

The half life period for first order reaction is given by:

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Therefore, $t_{\frac{1}{2} }=\frac{0.693}{0.1152}$$t_{\frac{1}{2} }=6 min$

Therefore, the half life period is six minutes.