Question

Question #d3beb

Answer 1

46) Let initial conc. be `c M`
The change in conc. in first 20s is `0.2cM` So rate of change of conc. in first 2mins is `(0.2c)/20=0.01cM/"min"`

The change in conc. in first 30s is `0.3cM` So rate of change of conc. in first 20min is `(0.3c)/30=0.01cM/"min"`

The rate of change of conc.being independent of time interval i.e.So it is Zero order reaction.

47) We known the velocity constant of first order reaction is
`k=2.303/t*log_10(a/(a-x))`
Where a is the initial conc.of reactant x is the change in conc. after time 't'

`"Here " t =10 min and x=(90a)/100`

`k=2.303/10*log_10(a/(a-(90a)/100))=0.2303min^-1`

48 In this problem the half life or the time for 50% consumption of reactant is 0.231min.
This means when t =0.231min
`x=(50a)/100`

So Velocity constant of the first order reaction

`k=2.303/t*log_10(a/(a-x))`

#k=2.303/ 0.231*log_10(a/(a-(50a)/100))#

`k=2.303/0.231*log_10 2=3.001`