Answer:- Answer is 69.7 second.
Solution:- For solving problems based on first order equations is:
[tex]ln[A]=-kt+ln[A_0][/tex]
where, [tex][A_0][/tex] 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, [tex][A_0][/tex] = 100
[A] = 70
[tex]k=5.12*10^-^3s^-^1[/tex]
t = ?
Let's plug in the values in the equation and solve it for t:
[tex]ln[70]=-5.12*10^-^3(t)+ln[100][/tex]
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)
[tex]t=\frac{0.357}{0.00512}[/tex]
t = 69.7 seconds
So, it would take 69.7 seconds for 30% substrate to be consumed.
