Question

show that in a first order reaction time required for completion of 99.9%is 10 times of half life of the reaction ​

Answer 1

Answer:-

Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

b) for completion of 99.9 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-99.9}$

$t=\frac{2.303}{k}\log\frac{100}{100-99.9}$

$t=\frac{2.303}{k}\times 3$

$t_{99.9}=\frac{6.90}{k}$

$\frac{t_{99.9}}{t_{\frac{1}{2}}}=\frac{\frac{6.90}{k}}{\frac{0.69}{k}}}$

$\frac{t_{99.9}}{t_{\frac{1}{2}}}=10$

Thus it is proved that time take for completion of 99.9% for reaction is 10 times of that of half life.

Answer 2

Answer:

first order reaction t= 2.303/k *log R•/R