Question

(b) A first order reaction takes 40 minutes for 30% decomposition.
Calculate t1/2 for this reaction.
(Given log 1.428 = 0.1548)​

Answer 1

Answer:

Let a M be the initial concentration. After 40 minutes, the concentration is a−  

100

30a

​  

=0.70a.

k=  

t

2.303

​  

log  

[A]

[A])  

−

 

​  

 

k=  

40

2.303

​  

log  

0.70a

a

​  

 

k=8.92×10  

−3

/min

The half life period, t  

1/2

​  

=  

k

0.693

​  

=  

8.92×10  

−3

/min

0.693

​  

=77.7 min.

Explanation:

MARK AS BRAINLIEST

Answer 2

Answer:

$\Large \bf{Given} -$

t = 40min

30% of the reactants undergoes decomposition that means out of 100 particles in reactant 30 particles has been used

Now,70 particles will undergo out of 100 particles

$\Large \bf{To\:find} -$

$\Large \sf{t_\frac{1}{2}} = \Large{?}$

$\Large \bf{Answer}$

$\text{You know for a first order reaction}$ $\text{half life is -}$

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

$\text{You can directly put half life formula }$ $\text{but k is missing}$

$\text{So,now you have to bring k in order to }$ $\text{solve}$

$\text{Now you can use,}$

$\Large \sf{k} = \frac{2.303}{t} \log\frac{[R]_o}{[R]}$

$\Large \sf{k} = \frac{2.303}{40} \log\frac{100}{70}$

$\Large \sf{k} = \frac{2.303}{40} \log\frac{10}{7}$

$\Large \sf{k} = \frac{2.303}{40} \log1.428$

$\Large \sf{k} = \frac{2.303}{40}× 0.1548$

$\Large \sf{k} = \frac{0.3565}{40}$

$\large \sf{k} = 0.008913$

$\text{Now put k in half-life formula}$

$\Large \sf{t_\frac{1}{2}} = \frac{0.693}{0.008913}$

$\Large \sf{t_\frac{1}{2}} = 77.75 min$