Question

If the rate for a first order reaction is k the time required for the completion of 99% of the reaction

Answer 1

K=2.303/tlog10^2

K=2.303/tx2log10

t=2.303/kx2

t=4.606/k

Answer 2

Answer:

Explanation:

The 1st order

t = $\frac{2.303}{k} log (\frac{a}{a-x} )$

$\frac{2.303}{k} log (\frac{100}{100-99} )$

$\frac{2.303}{k} log 10^{2}$

$\frac{2.303}{k} * 2 * log 10$

$\frac{2.303}{k} = \frac{4.606}{k}$