For a first-order reaction, the half-life is constant. it depends only on the rate constant and not on the reactant concentration. it is expressed as: t1/2=0.693/k

for a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as: t1/2= 1/k[a]0
a. a certain first-order reaction (a--> products) has a rate constant of 3.30×10^-3 s^-1 at 45 degrees c. how many minutes does it take for the concentration of the reactant, [a], to drop to 6.25% of the original concentration?
b. a certain second-order reaction (b--> products) has a rate constant of 1.70×10^-3 m^-1*s^-1 at 27 degrees c and an initial half-life of 296 s. what is the concentration of the reactant b after one half-life?

A.  14 min

B . 0.995 M

Explanation:

A.

Using integrated rate law for first order kinetics as:

Where,  

is the concentration at time t

is the initial concentration

Given:

The final concentration is dropped to 6.25 % of the initial concentration. SO,

= 0.0625

k =

So,

Also, 1 s = 1/60 minutes.

So,

14 minutes it takes for the concentration of the reactant to drop to 6.25% of the original concentration.

B.

(a) Half life expression for second order kinetic is:

Where,  

is the initial concentration = ?

k is the rate constant = M⁻¹s⁻¹

Half life = 296 s

So,

Half life is the time at which the concentration of the reactant reduced to half. So, concentration remains = = 0.995 M

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)

answer; natural resources;