A zero-order reaction has a constant rate of 3.20×10^−4 M/s. If after 35.0 seconds the concentration has dropped to 9.00×10^2M , what was the initial concentration?
[A] = [A]o - kt
solve for [A]o. You have k, t, and [A].
9.00 x 10^-2= (-3.20 x 10^-4)(35)+A
9.00 x 10^-2= (-.0112)+A
9.00 x 10^-2+.0112=A
A= .1012
To determine the initial concentration of the zero-order reaction, we can use the integrated rate law for zero-order reactions. The integrated rate law for a zero-order reaction is given by the equation:
[A]t = [A]0 - kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
Given that the rate constant (k) is equal to 3.20×10^−4 M/s and the concentration at 35.0 seconds is 9.00×10^2 M, we can plug these values into the equation to solve for [A]0.
9.00×10^2 M = [A]0 - (3.20×10^−4 M/s)(35.0 s)
Let's calculate the right side of the equation:
(A) [(3.20×10^−4 M/s)(35.0 s)] ≈ 1.12×10^−2 M
Now, let's rearrange the equation to solve for [A]0:
[A]0 = 9.00×10^2 M + 1.12×10^−2 M
(A) [A]0 ≈ 9.01×10^2 M
Therefore, the initial concentration of the zero-order reaction is approximately 9.01×10^2 M.