To determine the rate constant [tex]\( k \)[/tex] for a zero-order reaction, we use the integrated rate law for zero-order reactions:
[tex]\[ [A]_t = -kt + [A]_0 \][/tex]
Where:
- [tex]\([A]_t\)[/tex] is the concentration of the reactant at time [tex]\( t \)[/tex]
- [tex]\( k \)[/tex] is the rate constant
- [tex]\( t \)[/tex] is the time
- [tex]\([A]_0\)[/tex] is the initial concentration of the reactant
Given:
- Initial concentration [tex]\([A]_0\)[/tex] = 0.100 M at [tex]\( t = 110 \)[/tex] s
- Concentration [tex]\([A]_t\)[/tex] = 3.50 \times 10^{-2} \) M at [tex]\( t = 395 \)[/tex] s
We need to find the change in concentration and change in time first.
1. Calculate the change in concentration ([tex]\(\Delta [A]\)[/tex]):
[tex]\[ \Delta [A] = [A]_0 - [A]_t \][/tex]
[tex]\[ \Delta [A] = 0.100 \, \text{M} - 3.50 \times 10^{-2} \, \text{M} \][/tex]
[tex]\[ \Delta [A] = 0.100 \, \text{M} - 0.035 \, \text{M} \][/tex]
[tex]\[ \Delta [A] = 0.065 \, \text{M} \][/tex]
2. Calculate the change in time ([tex]\(\Delta t\)[/tex]):
[tex]\[ \Delta t = t_2 - t_1 \][/tex]
[tex]\[ \Delta t = 395 \, \text{s} - 110 \, \text{s} \][/tex]
[tex]\[ \Delta t = 285 \, \text{s} \][/tex]
3. Now, we use the formula rearranged for the rate constant ([tex]\( k \)[/tex]):
[tex]\[ k = \frac{\Delta [A]}{\Delta t} \][/tex]
Substitute the values:
[tex]\[ k = \frac{0.065 \, \text{M}}{285 \, \text{s}} \][/tex]
Perform the division to find [tex]\( k \)[/tex]:
[tex]\[ k \approx 0.000228 \, \text{M/s} \][/tex]
Therefore, the rate constant for the reaction is:
[tex]\[ k = 0.000228 \, \text{M} \cdot \text{s}^{-1} \][/tex]
Make sure to include the appropriate units to indicate the multiplication explicitly:
[tex]\[ k = 0.000228 \, \text{M} \cdot \text{s}^{-1} \][/tex]
So the final answer is:
[tex]\[ k = 0.000228 \, \text{M} \cdot \text{s}^{-1} \][/tex]
