The rate constant for a certain reaction is = 4.30×10−3 s^-1. If the initial reactant concentration was 0.400 M , what will the concentration be after 19.0 minutes?

Rate would be first order therefore I used.

A=A initial*e^-kt

I keep getting .0030 M but supposely I am wrong can anyone help me or tell me what I am doing wrong?

19 minutes is 1140 s, so kt = 4.90

The molarity should decrease to 0.400*e^-4.9 = .0030 (M)

I agree with your answer

To solve this problem, you are on the right track by using the first-order reaction equation:

A = A_initial * e^(-kt),

where A is the concentration at a given time, A_initial is the initial concentration, k is the rate constant, t is time, and e is the base of natural logarithm.

First, let's convert the time from minutes to seconds since the rate constant is given in s^-1. There are 60 seconds in one minute, so 19.0 minutes is equal to:

t = 19.0 minutes * 60 seconds/minute = 1140 seconds.

Now let's substitute the given values into the equation:

A = 0.400 M * e^(-4.30×10^(-3) s^(-1) * 1140 s).

Calculating this expression using a calculator or software, you should get:

A ≈ 0.285 M.

Therefore, the concentration of A after 19.0 minutes is approximately 0.285 M, not 0.0030 M.

Make sure to double-check your calculations and use the correct units for time.