Question

The rate constant for a certain reaction is k = 4.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 11.0 minutes?

Answer 1

Answer: The concentration of reactant after the given time is 0.0205 M

Step-by-step explanation:

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

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

where,

k = rate constant =
`4.50* 10^(-3)s^(-1)`

t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)

`[A_o]` = initial amount of the reactant = 0.400 M

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`4.50* 10^(-3)s^(-1)=(2.303)/(660s)\log(0.400)/([A])`

`[A]=0.0205M`

Hence, the concentration of reactant after the given time is 0.0205 M