Answer:
the original concentration ([A₀]) is approximately 12.59.
Step-by-step explanation:
To solve this problem, we can use the first-order reaction equation and the concept of half-life.
The first-order reaction equation is given by the equation:
ln([A]/[A₀]) = -kt
Where [A] is the concentration at a given time, [A₀] is the initial concentration, k is the rate constant, and t is the time.
The half-life (t₁/₂) is the time it takes for the reactant concentration to decrease by half. In this case, we are given that the half-life is 32 seconds.
Using the given information, we can set up the equation:
ln([A]/[A₀]) = -k(32)
Since the concentration after 2 minutes is given as 0.062, we can write it as [A]/[A₀] = 0.062.
Substituting these values into the equation, we get:
ln(0.062) = -k(32)
Now, let's solve for k:
k = -(ln(0.062))/32
Next, we can use the value of k to find the initial concentration [A₀].
Substituting the known values into the first-order reaction equation:
ln([A]/[A₀]) = -k(t)
ln([A]/[A₀]) = -(ln(0.062))/32 * (2 minutes * 60 seconds/minute) (converting minutes to seconds)
Simplifying:
ln([A]/[A₀]) = -(ln(0.062))/32 * 120
Now, let's solve for [A₀]:
[A₀] ≈ [A] / e^(-k*t)
[A₀] ≈ 0.062 / e^(-(ln(0.062))/32 * 120)
Using a calculator, we find:
[A₀] ≈ 12.59