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

The integrated rate law expression for a first order reaction is

$ln\frac{[A_{0}]}{[A_{t}]}=kt$

where

[A0]=100

[At]=6.25

[6.25% of 100 = 6.25]

k = 9.60X10⁻³s⁻¹

Putting values

$ln\frac{100}{6.25}=9.6X10^{-3}t$

taking log of 100/6.25

100/6.25 = 16

ln(16) = 2.7726

Time = 2.7726 / 0.0096 = 288.81  seconds

Answer 2

Answer:

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

Explanation:

Integrated rate law of first order kinetic is given as:

$\log [A]=\log [A_o]-\frac{kt}{2.303}$

Where:

$[A_o]$= Initial concentration of reactant

[A] =concentration left after time t .

k = Rate constant

We are given :

$k=9.60\times 10^{-3} s^{-1}$

$[A_o]=x$

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

Time taken during the reaction = t=?

$\log [0.0625 x]=\log[x]-\frac{9.60\times 10^{-3} s^{-1}\tmes t}{2.303}$

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.