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\begin{tabular}{cccc}
\hline
400 & 0.0120 & 0.0160 & 0.0099 \\
500 & 0.0101 & 0.0197 & 0.0049 \\
607 & 0.0086 & 0.0729 & 0.0054 \\
201 & 0.0072 & 0.0256 & 0.0064 \\
The oxidation & 0.0061 & 0.0278 & 0.0070 \\
\hline
\end{tabular}

The oxidation of iodide ion by arsenic acid, [tex]$H_3AsO_4$[/tex], is described by the balanced equation.

If [tex]$-\Delta\left[I^{-}\right] / \Delta t=4.8 \times 10^{-4} M / s$[/tex], what is the value of [tex][tex]$\Delta\left[I_3^{-}\right] / \Delta t$[/tex][/tex] during the same time interval?

What is the average rate of consumption of [tex]$H^{+}$[/tex] during that time interval?


Sagot :

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].