# Question d3beb

Jul 24, 2016

46) Let initial conc. be $c M$
The change in conc. in first 20s is $0.2 c M$ So rate of change of conc. in first 2mins is $\frac{0.2 c}{20} = 0.01 c \frac{M}{\text{min}}$

The change in conc. in first 30s is $0.3 c M$ So rate of change of conc. in first 20min is $\frac{0.3 c}{30} = 0.01 c \frac{M}{\text{min}}$

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
$k = \frac{2.303}{t} \cdot {\log}_{10} \left(\frac{a}{a - x}\right)$
Where a is the initial conc.of reactant x is the change in conc. after time 't'

$\text{Here } t = 10 \min \mathmr{and} x = \frac{90 a}{100}$

$k = \frac{2.303}{10} \cdot {\log}_{10} \left(\frac{a}{a - \frac{90 a}{100}}\right) = 0.2303 {\min}^{-} 1$

48 In this problem the half life or the time for 50% consumption of reactant is 0.231min.
This means when t =0.231min
$x = \frac{50 a}{100}$

So Velocity constant of the first order reaction

$k = \frac{2.303}{t} \cdot {\log}_{10} \left(\frac{a}{a - x}\right)$

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

$k = \frac{2.303}{0.231} \cdot {\log}_{10} 2 = 3.001$