Question

Consider the below chemical reaction that occurs via first-order kinetics with a rate constant of 5.12 x 10–3 s–1 at a particular temperature. how long will it take for 30% of substrate a to be consumed? a → b 83.7 s

Answer 1

Answer:- Answer is 69.7 second.

Solution:- For solving problems based on first order equations is:

`ln[A]=-kt+ln[A_0]`

where,
`[A_0]` is the initial concentration and [A] is final concentration.

k is rate constant and t is the time.

It asks to calculate the time in which 30% of the substrate would be consumed. Let's say the initial amount of the substrate is 100 then 30 is used and remaining would be, 100 - 30 = 70

So,
`[A_0]` = 100

[A] = 70

`k=5.12*10^-^3s^-^1`

t = ?

Let's plug in the values in the equation and solve it for t:

`ln[70]=-5.12*10^-^3(t)+ln[100]`

4.248 = -0.00512(t) + 4.605

4.248 - 4.605 = -0.00512(t)

-0.357 = -0.00512(t)

we have negative sign on both sides, so it is canceled out.

0.357 = 0.00512(t)

`t=(0.357)/(0.00512)`

t = 69.7 seconds

So, it would take 69.7 seconds for 30% substrate to be consumed.