Question

(A) 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 concentration decreases linearly with time. The time for a certain concentration change can be calculated using the integrated rate law.

Answer 1

For a zero-order reaction, the integrated rate law is given by: \[ [A] = [A_0] - kt \] where: - \( [A] \) is the concentration of A at time \( t \), - \( [A_0] \) is the initial concentration, - \( k \) is the rate constant, and - \( t \) is the time taken. We are given: - \( [A_0] = 0.10 \, \text{M} \), - \( [A] = 0.075 \, \text{M} \), - \( k = 0.0030 \, \text{mol L}^{-1} \, \text{s}^{-1} \). Substituting these values into the rate law equation: \[ 0.075 = 0.10 - (0.0030)(t) \] Rearranging to solve for \( t \): \[ 0.0030t = 0.10 - 0.075 = 0.025 \] \[ t = \frac{0.025}{0.0030} = 8.33 \, \text{seconds} \] Thus, it will take approximately \( 8.33 \, \text{seconds} \) for the concentration of A to fall from 0.10 M to 0.075 M. \bigskip