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

The objective is to determine the duration required for the concentration of reactant A in a zero-order reaction to decrease from 0.10 M to 0.075 M, given a rate constant \( k = 0.0030 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \).

1. Zero-Order Reaction Principles:
For a zero-order reaction \( A \to P \), the rate is constant and independent of reactant concentration, defined by the rate law \( \text{Rate} = k \). The corresponding integrated rate law is:

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

2. Isolate Time \( t \):
The integrated rate law can be rearranged to solve for time \( t \):

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

3. Data Substitution:
Using the provided values: \( [A]_0 = 0.10 \, \text{M} \), \( [A]_t = 0.075 \, \text{M} \), and \( k = 0.0030 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \). Substituting these into the equation:

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

4. Time Calculation:
Performing the calculation:

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

Conclusion:
The time required for the concentration of A to decrease from 0.10 M to 0.075 M is approximately \( 8.33 \, \text{s} \).