If it takes 30 seconds for a reactant concentration to decrease from 1.0 M to 0.5 M in a first-order chemical reaction, then what is the rate constant for the reaction?

well, it went down 1/2 M in 1/2 minute.

To find the rate constant for a first-order chemical reaction, we can make use of the integrated rate equation for a first-order reaction:

ln([A]t/[A]0) = -kt

Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and ln denotes the natural logarithm.

In this case, we are given that the concentration of the reactant decreases from 1.0 M to 0.5 M in 30 seconds. Let's substitute the given values into the equation:

ln(0.5/1.0) = -k * 30

Simplifying further:

ln(0.5) = -k * 30

Now we can isolate the rate constant (k) by rearranging the equation:

k = -ln(0.5) / 30

Using a calculator to evaluate the natural logarithm:

k ≈ -0.693 / 30

Finally, calculating the value:

k ≈ -0.0231 s^-1

Therefore, the rate constant for the reaction is approximately -0.0231 s^-1. Note that the negative sign indicates that the reactant concentration is decreasing over time.

so 0.046s^-1 ??