Question

A first order is 20% complete in 10 mins. Calcutta (i) the specific rate constant, (ii) the time taken for the reactions to go to 75% completion. [In 10 = 2.3, In 2 = 0.7]

Answer 1

$K=\frac{2.303}{t}\log \frac{R_o}{R}$

as $r\times n$ is $20$%completed so,remaining conc. is $80$%.

here,$t=10$min

$\left[R_o\right]=100$

$\left[R\right]=80$

$K=\frac{2.303}{10}\log \frac{100}{80}$

$=0.2303\log \frac{5}{4}$

$=0.0223$

For $75$% completion,remaining conc.is $25$%

$t=\frac{2.303}{K}\log \frac{R_0}{R}$

$=\frac{2.303}{0.0223}\log \frac{100}{25}$

$t=62.18$