Question

A reaction has a rate constant of 0.0642 sec-1; how long will it take (in minutes) until 0.479 mol/L of the compound is left, if there was 0.502 mol/L at the start? (give answer to 3 decimal places)?

Answer 1

`t=0.012\ min`

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 = 0.0642 s⁻¹

Initial concentration
`[A_0]` = 0.502 mol/L

Final concentration
`[A_t]` = 0.479 mol/L

Time = ?

Applying in the above equation, we get that:-

`0.479=0.502e^(-0.0642* t)`

`e^(-0.0642t)=(0.479)/(0.502)`

`t=-(\ln \left((0.479)/(0.502)\right))/(0.0642)`

`t=0.731\ sec`

Also, 1 sec = 1/60 min

So,
`t=0.012\ min`