Question

Consider this reaction:

Answer 1

Answer:   0.0345 sec

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[H_3PO_4]^2$

k= rate constant = $46.6s^{-1}$

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  

for completion of 20 % of reaction

$t=\frac{2.303}{46.6}\log\frac{0.660}{\frac{20}{100}\times 0.660}$

$t=\frac{2.303}{46.6}\log\frac{0.660}{0.132}$

$t=0.0345sec$

The time taken for the concentration of $H_3PO_4$ to decrease to 20% to its natural value is 0.0345 sec