Question

A certain first-order reaction (a→products) has a rate constant of 9.60×10−3 s−1 at 45 ∘c. how many minutes does it take for the concentration of the reactant, [a], to drop to 6.25% of the original concentration?

Answer 1

In 4.81 minutes the concentration of reactant will drop to 6.25% of the original concentration.

Step-by-step explanation:

Integrated rate law of first order kinetic is given as:

`\log [A]=\log [A_o]-(kt)/(2.303)`

Where:

`[A_o]`= Initial concentration of reactant

[A] =concentration left after time t .

k = Rate constant

We are given :

`k=9.60* 10^(-3) s^(-1)`

`[A_o]=x`

`[A]=6.25% of x =0.0625x`

Time taken during the reaction = t=?

`\log [0.0625 x]=\log[x]-(9.60* 10^(-3) s^(-1)\tmes t)/(2.303)`

t = 288.86 seconds = 4.81 minutes

1 min = 60 seconds

In 4.81 minutes the concentration of reactant will drop to 6.25% of the original concentration.

Answer 2

The integrated rate law expression for a first order reaction is

`ln([A_(0)])/([A_(t)])=kt`

where

[A0]=100

[At]=6.25

[6.25% of 100 = 6.25]

k = 9.60X10⁻³s⁻¹

Putting values

`ln(100)/(6.25)=9.6X10^(-3)t`

taking log of 100/6.25

100/6.25 = 16

ln(16) = 2.7726

Time = 2.7726 / 0.0096 = 288.81 seconds