Certainly! Let's solve this problem step by step.
### Given:
- The rate of change of iodide ion [tex]\([- \Delta [I^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\]
### 1. Value of \(\Delta[I_3^-] / \Delta t\)[/tex]
To determine the value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex], we need to understand the stoichiometry of the balanced chemical equation. Assuming a 1:1 stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(I_3^{-}\)[/tex], the rate at which [tex]\(I_3^{-}\)[/tex] is produced will be equal to the rate at which [tex]\(I^{-}\)[/tex] is consumed, because the change in concentration of [tex]\(I_3^{-}\)[/tex] directly depends on the loss of [tex]\(I^{-}\)[/tex].
Therefore:
[tex]\[-\Delta [I^{-}] / \Delta t = \Delta [I_3^{-}] / \Delta t\][/tex]
Hence:
[tex]\[\Delta [I_3^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
### 2. Average Rate of Consumption of [tex]\(H^{+}\)[/tex]
To find the average rate of consumption of [tex]\(H^{+}\)[/tex], we need to know the stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(H^{+}\)[/tex] in the balanced chemical reaction. Assuming a hypothetical balanced equation shows that for every mole of [tex]\(I^{-}\)[/tex] consumed, 3 moles of [tex]\(H^{+}\)[/tex] are also consumed (this is a typical stoichiometric ratio in such reactions):
Thus, the rate of consumption of [tex]\(H^{+}\)[/tex] will be three times the rate of consumption of [tex]\(I^{-}\)[/tex]:
[tex]\[\text{Rate of consumption of } H^{+} = 3 \left(\text{Rate of consumption of } I^{-}\right) \][/tex]
[tex]\[
\text{Rate of consumption of } H^{+} = 3 \times 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
[tex]\[
\text{Rate of consumption of } H^{+} = 1.44 \times 10^{-3} \, \text{M/s}
\][/tex]
Thus, the average rate of consumption of [tex]\(H^{+}\)[/tex] during that time interval is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
### Summary:
1. The value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex] is [tex]\(4.8 \times 10^{-4} \, \text{M/s}\)[/tex].
2. The average rate of consumption of [tex]\(H^{+}\)[/tex] is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
