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

Answer 1

To solve the problem, we need to determine the time required for the concentration of A in a zero-order reaction to decrease from 0.10 M to 0.075 M, given the rate constant \( k = 0.0030 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \).

1. Understand the Zero-Order Kinetics:
For a zero-order reaction \( A \to P \), the rate law is \( \text{Rate} = k \), and the integrated rate law is:

\( [A]_t = [A]_0 - kt \),
where \( [A]_0 \) is the initial concentration, \( [A]_t \) is the concentration at time \( t \), and \( k \) is the rate constant.

2. Rearrange for Time \( t \):
Rearrange the equation to solve for \( t \):

\( t = \frac{[A]_0 - [A]_t}{k} \).

3. Substitute the Given Values:
\( [A]_0 = 0.10 \, \text{M} \), \( [A]_t = 0.075 \, \text{M} \), and \( k = 0.0030 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \). Plugging in:

\( t = \frac{0.10 - 0.075}{0.0030} = \frac{0.025}{0.0030} \).

4. Calculate \( t \):
\( t = \frac{0.025}{0.0030} \approx 8.33 \, \text{s} \).

Final Answer:
It will take approximately \( 8.33 \, \text{s} \) for the concentration of A to fall from 0.10 M to 0.075 M.