Question

The rate constant for this first‑order reaction is 0.460 s − 1 at 400 ∘ C. A ⟶ products How long, in seconds, would it take for the concentration of A to decrease from 0.680 M to 0.370 M?

Answer 1

1.32 seconds would take for the concentration of A to decrease from 0.680 M to 0.370 M.

Step-by-step explanation:

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration

Given that:

The rate constant, k =
`0.460` s⁻¹

Initial concentration
`[A_0]` = 0.680 M

Final concentration
`[A_t]` = 0.370 M

Time = ?

Applying in the above equation, we get that:-

`0.370=0.680* e^(-0.460* t)`

`0.68e^(-0.46t)=0.37`

`68e^(-0.46t)=37`

`-0.46t=\ln \left((37)/(68)\right)`

`t=1.32\ s`

1.32 seconds would take for the concentration of A to decrease from 0.680 M to 0.370 M.