Question

A first-order reaction has a half-life of 29.8 s . How long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Answer 1

Answer : The time taken for the concentration of the reactant in the reaction to fall to one-fourth of its initial value is, 12.4 seconds.

Explanation :

Half-life = 29.8 s

First we have to calculate the rate constant, we use the formula :

`k=(0.693)/(t_(1/2))`

`k=(0.693)/(29.8s)`

`k=0.0232s^(-1)`

Now we have to calculate the time passed.

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

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`0.0232s^(-1)`

t = time passed by the sample = ?

a = let initial amount of the reactant = x M

a - x = amount left after decay process =
`x-(1)/(4)* (x)=(3)/(4)* (X)=(3x)/(4)M`

Now put all the given values in above equation, we get

`t=(2.303)/(0.0232s^(-1))\log(x)/(((3x)/(4)))`

`t=12.4s`

Therefore, the time taken for the concentration of the reactant in the reaction to fall to one-fourth of its initial value is, 12.4 seconds.