To determine the expected time for experiment 4, we need to find the rate law expression and the rate constant [tex]\( k \)[/tex] for the given reaction:
[tex]\[ S_2O_8^{2-}(aq) + 2I^-(aq) \rightarrow I_2(aq) + 2SO_4^{2-}(aq) \][/tex]
We will use the data from the provided experiments and hypothesize that the rate law is of the form:
[tex]\[ \text{rate} = k[S_2O_8^{2-}]^m[I^-]^n \][/tex]
We have information from three experiments, which allows us to estimate the rate constant [tex]\( k \)[/tex]. Let’s analyze the given data:
#### Experiment 1:
- Initial [tex]\([S_2O_8^{2-}]\)[/tex]: 0.0400 M
- Initial [tex]\([I^-]\)[/tex]: 0.0800 M
- Time: 39 sec
#### Experiment 2:
- Initial [tex]\([S_2O_8^{2-}]\)[/tex]: 0.0400 M
- Initial [tex]\([I^-]\)[/tex]: 0.0400 M
- Time: 78 sec
#### Experiment 3:
- Initial [tex]\([S_2O_8^{2-}]\)[/tex]: 0.0100 M
- Initial [tex]\([I^-]\)[/tex]: 0.0800 M
- Time: 156 sec
First, let’s calculate the rate constant [tex]\( k \)[/tex] for each experiment using the rate law expression. To compute [tex]\( k \)[/tex], we rearrange the rate law to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\text{Time}}{[S_2O_8^{2-}] [I^-]} \][/tex]
##### For Experiment 1:
[tex]\[ k_1 = \frac{39}{0.0400 \times 0.0800} = 12187.5 \][/tex]
##### For Experiment 2:
[tex]\[ k_2 = \frac{78}{0.0400 \times 0.0400} = 48750.0 \][/tex]
##### For Experiment 3:
[tex]\[ k_3 = \frac{156}{0.0100 \times 0.0800} = 195000.0 \][/tex]
Next, we average these rate constants to obtain a more accurate value for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{12187.5 + 48750.0 + 195000.0}{3} = 85312.5 \][/tex]
Now, using this averaged rate constant [tex]\( k \)[/tex], we can determine the expected time for experiment 4:
#### For Experiment 4:
- Initial [tex]\([S_2O_8^{2-}]\)[/tex]: 0.0200 M
- Initial [tex]\([I^-]\)[/tex]: 0.0200 M
Using the rate law expression and the averaged [tex]\( k \)[/tex]:
[tex]\[
\text{Rate} = k \cdot [S_2O_8^{2-}] \cdot [I^-]
\][/tex]
[tex]\[
\text{Rate} = 85312.5 \times 0.0200 \times 0.0200
\][/tex]
[tex]\[
\text{Rate} = 34.125
\][/tex]
We use this rate to find the expected time for experiment 4:
[tex]\[
\text{Time} = \frac{1}{\text{Rate}}
\][/tex]
[tex]\[
\text{Time} \approx \frac{1}{34.125}
\][/tex]
[tex]\[
\text{Time} \approx 4.875 \text{ seconds}
\][/tex]
The expected time is:
[tex]\[ \boxed{4.875 \text{ sec}} \][/tex]
