A zero-order reaction has a constant rate of 3.40×10−4 M/s. If after 80.0 seconds the concentration has dropped to 6.00×10−2 M, what was the initial concentration?

Isn't the integrated rate equation for a zero order reaction the following:

[A] = [Ao]kt?
Plug and chug.

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To solve this problem, we can use the integrated rate equation for a zero-order reaction, which is:

[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:
Rate constant, k = 3.40×10^(-4) M/s
Time, t = 80.0 seconds
Concentration at time t, [A]t = 6.00×10^(-2) M

We can rearrange the equation to solve for the initial concentration [A]0:

[A]0 = [A]t + kt

Substitute the given values:

[A]0 = 6.00×10^(-2) M + (3.40×10^(-4) M/s × 80.0 s)
[A]0 = 6.00×10^(-2) M + 2.72×10^(-2) M
[A]0 = 8.72×10^(-2) M

Therefore, the initial concentration was 8.72×10^(-2) M.

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