Answer :
Let's go through this step-by-step and fill in the provided table for Trial 1 only.
### Step-by-Step Solution:
Given Data:
1. Concentration of iodide solution ([tex]\(I^{-}\)[/tex]): [tex]\(100.0 \, \text{mM}\)[/tex]
2. Concentration of thiosulfate solution ([tex]\(S_{2}O_{3}^{2-}\)[/tex]): [tex]\(31.5 \, \text{mM}\)[/tex]
3. Concentration of hydrogen peroxide solution ([tex]\(H_{2}O_{2}\)[/tex]): [tex]\(176.5 \, \text{mM}\)[/tex]
4. Temperature of iodide solution: [tex]\(25.0^{\circ} \, \text{C}\)[/tex]
5. Volume of iodide solution used: [tex]\(10.0 \, \text{mL}\)[/tex]
6. Volume of thiosulfate solution used: [tex]\(1.0 \, \text{mL}\)[/tex]
7. Volume of DI water used: [tex]\(10.0 \, \text{mL}\)[/tex]
8. Time for the reaction: [tex]\(17.7 \, \text{s}\)[/tex]
Step 1: Calculate the Total Volume of the Reaction Mixture
[tex]\[ \text{Total volume} = \text{Volume of iodide solution} + \text{Volume of thiosulfate solution} + \text{Volume of DI water} \][/tex]
[tex]\[ \text{Total volume} = 10.0 \, \text{mL} + 1.0 \, \text{mL} + 10.0 \, \text{mL} = 21.0 \, \text{mL} \][/tex]
Step 2: Calculate the Initial Concentration of Iodide in the Reaction Mixture
[tex]\[ \text{Initial concentration of iodide} = \left(\frac{\text{Concentration of iodide} \times \text{Volume of iodide solution}}{\text{Total volume}}\right) \][/tex]
[tex]\[ \text{Initial concentration of iodide} = \left(\frac{100.0 \, \text{mM} \times 10.0 \, \text{mL}}{21.0 \, \text{mL}}\right) \approx 47.62 \, \text{mM} \][/tex]
Step 3: Calculate the Initial Concentration of Thiosulfate in the Reaction Mixture
[tex]\[ \text{Initial concentration of thiosulfate} = \left(\frac{\text{Concentration of thiosulfate} \times \text{Volume of thiosulfate}}{\text{Total volume}}\right) \][/tex]
[tex]\[ \text{Initial concentration of thiosulfate} = \left(\frac{31.5 \, \text{mM} \times 1.0 \, \text{mL}}{21.0 \, \text{mL}}\right) \approx 1.5 \, \text{mM} \][/tex]
Step 4: Calculate the Initial Rate
[tex]\[ \text{Initial Rate} = \left(\frac{\text{Initial concentration of iodide}}{\text{Time}}\right) \][/tex]
[tex]\[ \text{Initial Rate} = \left(\frac{47.62 \, \text{mM}}{17.7 \, \text{s}}\right) \approx 2.69 \, \text{mM/s} \][/tex]
Now, we place these values in the table provided:
\begin{tabular}{|c|c|c|}
\hline & Trial 1 & Trial 2 \\
\hline Concentration of iodide solution [tex]$( m M$[/tex] ) & 100.0 & 100.0 \\
\hline Concentration of thiosulfate solution ( [tex]$mM$[/tex] ) & 31.5 & 31.5 \\
\hline Concentration of hydrogen peroxide solution ( [tex]$mM$[/tex] ) & 176.5 & 176.5 \\
\hline Temperature of iodide solution [tex]$\left({ }^{\circ} C \right)$[/tex] & 25.0 & \\
\hline Volume of iodide solution [tex]$\left( I ^{-}\right)$[/tex] used [tex]$( mL )$[/tex] & 10.0 & \\
\hline Volume of thiosulfate solution [tex]$\left( S _2 O _3{ }^{2-}\right)$[/tex] used [tex]$( m L )$[/tex] & 1.0 & \\
\hline Volume of DI water used [tex]$( mL )$[/tex] & 10.0 & \\
\hline \multicolumn{3}{|l|}{ Volume of hydrogen peroxide solution [tex]$\left( H _2 O _2 \right)$[/tex] used [tex]$( m L )$[/tex]} \\
\hline Time [tex]$( s )$[/tex] & 17.7 & \\
\hline Observations & \begin{tabular}{l}
The liquid turns \\
blue.
\end{tabular} & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of iodide in reaction (m M) } & 47.62 & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of thiosulfate in reaction ( [tex]$mM$[/tex] ) } & 1.5 & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of hydrogen peroxide in reaction ( m M ) } & & \\
\hline Initial Rate [tex]$(m M / s )$[/tex] & 2.69 & \\
\hline
\end{tabular}
For Trial 2, the values are not provided, so they cannot be filled out. However, the steps would be the same for a different set of volumes and times.
### Step-by-Step Solution:
Given Data:
1. Concentration of iodide solution ([tex]\(I^{-}\)[/tex]): [tex]\(100.0 \, \text{mM}\)[/tex]
2. Concentration of thiosulfate solution ([tex]\(S_{2}O_{3}^{2-}\)[/tex]): [tex]\(31.5 \, \text{mM}\)[/tex]
3. Concentration of hydrogen peroxide solution ([tex]\(H_{2}O_{2}\)[/tex]): [tex]\(176.5 \, \text{mM}\)[/tex]
4. Temperature of iodide solution: [tex]\(25.0^{\circ} \, \text{C}\)[/tex]
5. Volume of iodide solution used: [tex]\(10.0 \, \text{mL}\)[/tex]
6. Volume of thiosulfate solution used: [tex]\(1.0 \, \text{mL}\)[/tex]
7. Volume of DI water used: [tex]\(10.0 \, \text{mL}\)[/tex]
8. Time for the reaction: [tex]\(17.7 \, \text{s}\)[/tex]
Step 1: Calculate the Total Volume of the Reaction Mixture
[tex]\[ \text{Total volume} = \text{Volume of iodide solution} + \text{Volume of thiosulfate solution} + \text{Volume of DI water} \][/tex]
[tex]\[ \text{Total volume} = 10.0 \, \text{mL} + 1.0 \, \text{mL} + 10.0 \, \text{mL} = 21.0 \, \text{mL} \][/tex]
Step 2: Calculate the Initial Concentration of Iodide in the Reaction Mixture
[tex]\[ \text{Initial concentration of iodide} = \left(\frac{\text{Concentration of iodide} \times \text{Volume of iodide solution}}{\text{Total volume}}\right) \][/tex]
[tex]\[ \text{Initial concentration of iodide} = \left(\frac{100.0 \, \text{mM} \times 10.0 \, \text{mL}}{21.0 \, \text{mL}}\right) \approx 47.62 \, \text{mM} \][/tex]
Step 3: Calculate the Initial Concentration of Thiosulfate in the Reaction Mixture
[tex]\[ \text{Initial concentration of thiosulfate} = \left(\frac{\text{Concentration of thiosulfate} \times \text{Volume of thiosulfate}}{\text{Total volume}}\right) \][/tex]
[tex]\[ \text{Initial concentration of thiosulfate} = \left(\frac{31.5 \, \text{mM} \times 1.0 \, \text{mL}}{21.0 \, \text{mL}}\right) \approx 1.5 \, \text{mM} \][/tex]
Step 4: Calculate the Initial Rate
[tex]\[ \text{Initial Rate} = \left(\frac{\text{Initial concentration of iodide}}{\text{Time}}\right) \][/tex]
[tex]\[ \text{Initial Rate} = \left(\frac{47.62 \, \text{mM}}{17.7 \, \text{s}}\right) \approx 2.69 \, \text{mM/s} \][/tex]
Now, we place these values in the table provided:
\begin{tabular}{|c|c|c|}
\hline & Trial 1 & Trial 2 \\
\hline Concentration of iodide solution [tex]$( m M$[/tex] ) & 100.0 & 100.0 \\
\hline Concentration of thiosulfate solution ( [tex]$mM$[/tex] ) & 31.5 & 31.5 \\
\hline Concentration of hydrogen peroxide solution ( [tex]$mM$[/tex] ) & 176.5 & 176.5 \\
\hline Temperature of iodide solution [tex]$\left({ }^{\circ} C \right)$[/tex] & 25.0 & \\
\hline Volume of iodide solution [tex]$\left( I ^{-}\right)$[/tex] used [tex]$( mL )$[/tex] & 10.0 & \\
\hline Volume of thiosulfate solution [tex]$\left( S _2 O _3{ }^{2-}\right)$[/tex] used [tex]$( m L )$[/tex] & 1.0 & \\
\hline Volume of DI water used [tex]$( mL )$[/tex] & 10.0 & \\
\hline \multicolumn{3}{|l|}{ Volume of hydrogen peroxide solution [tex]$\left( H _2 O _2 \right)$[/tex] used [tex]$( m L )$[/tex]} \\
\hline Time [tex]$( s )$[/tex] & 17.7 & \\
\hline Observations & \begin{tabular}{l}
The liquid turns \\
blue.
\end{tabular} & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of iodide in reaction (m M) } & 47.62 & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of thiosulfate in reaction ( [tex]$mM$[/tex] ) } & 1.5 & \\
\hline \multicolumn{3}{|l|}{ Initial concentration of hydrogen peroxide in reaction ( m M ) } & & \\
\hline Initial Rate [tex]$(m M / s )$[/tex] & 2.69 & \\
\hline
\end{tabular}
For Trial 2, the values are not provided, so they cannot be filled out. However, the steps would be the same for a different set of volumes and times.
