Question

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.0055 M?

Answer 1

Answer: The time taken by the reaction is 84.5 seconds

Step-by-step explanation:

The equation used to calculate half life for first order kinetics:

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

where,

`t_(1/2)` = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

`k=(0.693)/(9)=0.077s^(-1)`

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])` ......(1)

where,

k = rate constant =
`0.077s^(-1)`

t = time taken for decay process = 50.7 sec

`[A_o]` = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

`0.077=(2.303)/(50.7)\log([A_o])/(0.0741)`

`[A_o]=3.67M`

Now, calculating the time taken by using equation 1:

`[A]=0.0055M`

`k=0.077s^(-1)`

`[A_o]=3.67M`

Putting values in equation 1, we get:

`0.077=(2.303)/(t)\log(3.67)/(0.0055)\\\\t=84.5s`

Hence, the time taken by the reaction is 84.5 seconds