Question

The rate constant for a zero order reaction A $\to$ P is 0.0030 mol L$^{-1}$s$^{-1}$. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?

Hint:

For zero-order reactions, the time to reach a particular concentration is directly proportional to the change in concentration.

Answer 1

For a zero-order reaction, the rate law is: \[ \text{[A]} = \text{[A]}_0 - kt \] Where \(\text{[A]}_0\) is the initial concentration, \(\text{[A]}\) is the final concentration, \(k\) is the rate constant, and \(t\) is the time taken.
Substitute the given values into the equation: \[ 0.075 = 0.10 - (0.0030 \, \text{mol L}^{-1} \, \text{s}^{-1}) \times t \] Solving for \(t\): \[ t = \frac{0.10 - 0.075}{0.0030} = \frac{0.025}{0.0030} = 8.33 \, \text{seconds} \] Thus, it will take 8.33 seconds for the concentration of A to fall from 0.10 M to 0.075 M.